sample hypothesis testing examples

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S.3.3 hypothesis testing examples.

Example: Right-Tailed Test
Example: Left-Tailed Test
Example: Two-Tailed Test

Brinell Hardness Scores

An engineer measured the Brinell hardness of 25 pieces of ductile iron that were subcritically annealed. The resulting data were:

The engineer hypothesized that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Therefore, he was interested in testing the hypotheses:

H 0 : μ = 170 H A : μ > 170

The engineer entered his data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. He obtained the following output:

Descriptive Statistics

$\mu$: mean of Brinelli

Null hypothesis H₀: $\mu$ = 170 Alternative hypothesis H₁: $\mu$ > 170

The output tells us that the average Brinell hardness of the n = 25 pieces of ductile iron was 172.52 with a standard deviation of 10.31. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of n = 25, is 2.06). The test statistic t * is 1.22, and the P -value is 0.117.

If the engineer set his significance level α at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t * were greater than 1.7109 (determined using statistical software or a t -table):

t distribution graph for df = 24 and a right tailed test of .05 significance level

Since the engineer's test statistic, t * = 1.22, is not greater than 1.7109, the engineer fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the $\alpha$ = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

If the engineer used the P -value approach to conduct his hypothesis test, he would determine the area under a t n - 1 = t 24 curve and to the right of the test statistic t * = 1.22:

t distribution graph of right tailed test showing the p-value of 0117 for a t-value of 1.22

In the output above, Minitab reports that the P -value is 0.117. Since the P -value, 0.117, is greater than $\alpha$ = 0.05, the engineer fails to reject the null hypothesis. There is insufficient evidence, at the $\alpha$ = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

Note that the engineer obtains the same scientific conclusion regardless of the approach used. This will always be the case.

Height of Sunflowers

A biologist was interested in determining whether sunflower seedlings treated with an extract from Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of 15.7 cm. The biologist treated a random sample of n = 33 seedlings with the extract and subsequently obtained the following heights:

The biologist's hypotheses are:

H 0 : μ = 15.7 H A : μ < 15.7

The biologist entered her data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. She obtained the following output:

$\mu$: mean of Height

Null hypothesis H₀: $\mu$ = 15.7 Alternative hypothesis H₁: $\mu$ < 15.7

The output tells us that the average height of the n = 33 sunflower seedlings was 13.664 with a standard deviation of 2.544. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 13.664 by the square root of n = 33, is 0.443). The test statistic t * is -4.60, and the P -value, 0.000, is to three decimal places.

Minitab Note. Minitab will always report P -values to only 3 decimal places. If Minitab reports the P -value as 0.000, it really means that the P -value is 0.000....something. Throughout this course (and your future research!), when you see that Minitab reports the P -value as 0.000, you should report the P -value as being "< 0.001."

If the biologist set her significance level $\alpha$ at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic t * were less than -1.6939 (determined using statistical software or a t -table):s-3-3

Since the biologist's test statistic, t * = -4.60, is less than -1.6939, the biologist rejects the null hypothesis. That is, the test statistic falls in the "critical region." There is sufficient evidence, at the α = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.

If the biologist used the P -value approach to conduct her hypothesis test, she would determine the area under a t n - 1 = t 32 curve and to the left of the test statistic t * = -4.60:

t-distribution for left tailed test with significance level of 0.05 shown in left tail

In the output above, Minitab reports that the P -value is 0.000, which we take to mean < 0.001. Since the P -value is less than 0.001, it is clearly less than $\alpha$ = 0.05, and the biologist rejects the null hypothesis. There is sufficient evidence, at the $\alpha$ = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.

t-distribution graph for left tailed test with a t-value of -4.60 and left tail area of 0.000

Note again that the biologist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

Gum Thickness

A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of n = 10 pieces of gum and measured their thickness. He obtained:

The quality control specialist's hypotheses are:

H 0 : μ = 7.5 H A : μ ≠ 7.5

The quality control specialist entered his data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. He obtained the following output:

$\mu$: mean of Thickness

Null hypothesis H₀: $\mu$ = 7.5 Alternative hypothesis H₁: $\mu \ne$ 7.5

The output tells us that the average thickness of the n = 10 pieces of gums was 7.55 one-hundredths of an inch with a standard deviation of 0.1027. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 0.1027 by the square root of n = 10, is 0.0325). The test statistic t * is 1.54, and the P -value is 0.158.

If the quality control specialist sets his significance level $\alpha$ at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t * were less than -2.2616 or greater than 2.2616 (determined using statistical software or a t -table):

t-distribution graph of two tails with a significance level of .05 and t values of -2.2616 and 2.2616

Since the quality control specialist's test statistic, t * = 1.54, is not less than -2.2616 nor greater than 2.2616, the quality control specialist fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the $\alpha$ = 0.05 level, to conclude that the mean thickness of all of the manufacturer's spearmint gum differs from 7.5 one-hundredths of an inch.

If the quality control specialist used the P -value approach to conduct his hypothesis test, he would determine the area under a t n - 1 = t 9 curve, to the right of 1.54 and to the left of -1.54:

t-distribution graph for a two tailed test with t values of -1.54 and 1.54, the corresponding p-values are 0.0789732 on both tails

In the output above, Minitab reports that the P -value is 0.158. Since the P -value, 0.158, is greater than $\alpha$ = 0.05, the quality control specialist fails to reject the null hypothesis. There is insufficient evidence, at the $\alpha$ = 0.05 level, to conclude that the mean thickness of all pieces of spearmint gum differs from 7.5 one-hundredths of an inch.

Note that the quality control specialist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

In our review of hypothesis tests, we have focused on just one particular hypothesis test, namely that concerning the population mean $\mu$. The important thing to recognize is that the topics discussed here — the general idea of hypothesis tests, errors in hypothesis testing, the critical value approach, and the P -value approach — generally extend to all of the hypothesis tests you will encounter.

How it works

Hypothesis Testing – A Complete Guide with Examples

Published by Alvin Nicolas at August 14th, 2021 , Revised On October 26, 2023

In statistics, hypothesis testing is a critical tool. It allows us to make informed decisions about populations based on sample data. Whether you are a researcher trying to prove a scientific point, a marketer analysing A/B test results, or a manufacturer ensuring quality control, hypothesis testing plays a pivotal role. This guide aims to introduce you to the concept and walk you through real-world examples.

What is a Hypothesis and a Hypothesis Testing?

A hypothesis is considered a belief or assumption that has to be accepted, rejected, proved or disproved. In contrast, a research hypothesis is a research question for a researcher that has to be proven correct or incorrect through investigation.

What is Hypothesis Testing?

Hypothesis testing is a scientific method used for making a decision and drawing conclusions by using a statistical approach. It is used to suggest new ideas by testing theories to know whether or not the sample data supports research. A research hypothesis is a predictive statement that has to be tested using scientific methods that join an independent variable to a dependent variable.

Example: The academic performance of student A is better than student B

Characteristics of the Hypothesis to be Tested

A hypothesis should be:

Clear and precise
Capable of being tested
Able to relate to a variable
Stated in simple terms
Consistent with known facts
Limited in scope and specific
Tested in a limited timeframe
Explain the facts in detail

What is a Null Hypothesis and Alternative Hypothesis?

A null hypothesis is a hypothesis when there is no significant relationship between the dependent and the participants’ independent variables .

In simple words, it’s a hypothesis that has been put forth but hasn’t been proved as yet. A researcher aims to disprove the theory. The abbreviation “Ho” is used to denote a null hypothesis.

If you want to compare two methods and assume that both methods are equally good, this assumption is considered the null hypothesis.

Example: In an automobile trial, you feel that the new vehicle’s mileage is similar to the previous model of the car, on average. You can write it as: Ho: there is no difference between the mileage of both vehicles. If your findings don’t support your hypothesis and you get opposite results, this outcome will be considered an alternative hypothesis.

If you assume that one method is better than another method, then it’s considered an alternative hypothesis. The alternative hypothesis is the theory that a researcher seeks to prove and is typically denoted by H1 or HA.

If you support a null hypothesis, it means you’re not supporting the alternative hypothesis. Similarly, if you reject a null hypothesis, it means you are recommending the alternative hypothesis.

Example: In an automobile trial, you feel that the new vehicle’s mileage is better than the previous model of the vehicle. You can write it as; Ha: the two vehicles have different mileage. On average/ the fuel consumption of the new vehicle model is better than the previous model.

If a null hypothesis is rejected during the hypothesis test, even if it’s true, then it is considered as a type-I error. On the other hand, if you don’t dismiss a hypothesis, even if it’s false because you could not identify its falseness, it’s considered a type-II error.

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How to Conduct Hypothesis Testing?

Here is a step-by-step guide on how to conduct hypothesis testing.

Step 1: State the Null and Alternative Hypothesis

Once you develop a research hypothesis, it’s important to state it is as a Null hypothesis (Ho) and an Alternative hypothesis (Ha) to test it statistically.

A null hypothesis is a preferred choice as it provides the opportunity to test the theory. In contrast, you can accept the alternative hypothesis when the null hypothesis has been rejected.

Example: You want to identify a relationship between obesity of men and women and the modern living style. You develop a hypothesis that women, on average, gain weight quickly compared to men. Then you write it as: Ho: Women, on average, don’t gain weight quickly compared to men. Ha: Women, on average, gain weight quickly compared to men.

Step 2: Data Collection

Hypothesis testing follows the statistical method, and statistics are all about data. It’s challenging to gather complete information about a specific population you want to study. You need to gather the data obtained through a large number of samples from a specific population.

Example: Suppose you want to test the difference in the rate of obesity between men and women. You should include an equal number of men and women in your sample. Then investigate various aspects such as their lifestyle, eating patterns and profession, and any other variables that may influence average weight. You should also determine your study’s scope, whether it applies to a specific group of population or worldwide population. You can use available information from various places, countries, and regions.

Step 3: Select Appropriate Statistical Test

There are many types of statistical tests , but we discuss the most two common types below, such as One-sided and two-sided tests.

Note: Your choice of the type of test depends on the purpose of your study

One-sided Test

In the one-sided test, the values of rejecting a null hypothesis are located in one tail of the probability distribution. The set of values is less or higher than the critical value of the test. It is also called a one-tailed test of significance.

Example: If you want to test that all mangoes in a basket are ripe. You can write it as: Ho: All mangoes in the basket, on average, are ripe. If you find all ripe mangoes in the basket, the null hypothesis you developed will be true.

Two-sided Test

In the two-sided test, the values of rejecting a null hypothesis are located on both tails of the probability distribution. The set of values is less or higher than the first critical value of the test and higher than the second critical value test. It is also called a two-tailed test of significance.

Example: Nothing can be explicitly said whether all mangoes are ripe in the basket. If you reject the null hypothesis (Ho: All mangoes in the basket, on average, are ripe), then it means all mangoes in the basket are not likely to be ripe. A few mangoes could be raw as well.

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Step 4: Select the Level of Significance

When you reject a null hypothesis, even if it’s true during a statistical hypothesis, it is considered the significance level . It is the probability of a type one error. The significance should be as minimum as possible to avoid the type-I error, which is considered severe and should be avoided.

If the significance level is minimum, then it prevents the researchers from false claims.

The significance level is denoted by P, and it has given the value of 0.05 (P=0.05)

If the P-Value is less than 0.05, then the difference will be significant. If the P-value is higher than 0.05, then the difference is non-significant.

Example: Suppose you apply a one-sided test to test whether women gain weight quickly compared to men. You get to know about the average weight between men and women and the factors promoting weight gain.

Step 5: Find out Whether the Null Hypothesis is Rejected or Supported

After conducting a statistical test, you should identify whether your null hypothesis is rejected or accepted based on the test results. It would help if you observed the P-value for this.

Example: If you find the P-value of your test is less than 0.5/5%, then you need to reject your null hypothesis (Ho: Women, on average, don’t gain weight quickly compared to men). On the other hand, if a null hypothesis is rejected, then it means the alternative hypothesis might be true (Ha: Women, on average, gain weight quickly compared to men. If you find your test’s P-value is above 0.5/5%, then it means your null hypothesis is true.

Step 6: Present the Outcomes of your Study

The final step is to present the outcomes of your study . You need to ensure whether you have met the objectives of your research or not.

In the discussion section and conclusion , you can present your findings by using supporting evidence and conclude whether your null hypothesis was rejected or supported.

In the result section, you can summarise your study’s outcomes, including the average difference and P-value of the two groups.

If we talk about the findings, our study your results will be as follows:

Example: In the study of identifying whether women gain weight quickly compared to men, we found the P-value is less than 0.5. Hence, we can reject the null hypothesis (Ho: Women, on average, don’t gain weight quickly than men) and conclude that women may likely gain weight quickly than men.

Did you know in your academic paper you should not mention whether you have accepted or rejected the null hypothesis?

Always remember that you either conclude to reject Ho in favor of Haor do not reject Ho . It would help if you never rejected Ha or even accept Ha .

Suppose your null hypothesis is rejected in the hypothesis testing. If you conclude reject Ho in favor of Haor do not reject Ho, then it doesn’t mean that the null hypothesis is true. It only means that there is a lack of evidence against Ho in favour of Ha. If your null hypothesis is not true, then the alternative hypothesis is likely to be true.

Example: We found that the P-value is less than 0.5. Hence, we can conclude reject Ho in favour of Ha (Ho: Women, on average, don’t gain weight quickly than men) reject Ho in favour of Ha. However, rejected in favour of Ha means (Ha: women may likely to gain weight quickly than men)

Frequently Asked Questions

What are the 3 types of hypothesis test.

The 3 types of hypothesis tests are:

One-Sample Test : Compare sample data to a known population value.
Two-Sample Test : Compare means between two sample groups.
ANOVA : Analyze variance among multiple groups to determine significant differences.

What is a hypothesis?

A hypothesis is a proposed explanation or prediction about a phenomenon, often based on observations. It serves as a starting point for research or experimentation, providing a testable statement that can either be supported or refuted through data and analysis. In essence, it’s an educated guess that drives scientific inquiry.

What are null hypothesis?

A null hypothesis (often denoted as H0) suggests that there is no effect or difference in a study or experiment. It represents a default position or status quo. Statistical tests evaluate data to determine if there’s enough evidence to reject this null hypothesis.

What is the probability value?

The probability value, or p-value, is a measure used in statistics to determine the significance of an observed effect. It indicates the probability of obtaining the observed results, or more extreme, if the null hypothesis were true. A small p-value (typically <0.05) suggests evidence against the null hypothesis, warranting its rejection.

What is p value?

The p-value is a fundamental concept in statistical hypothesis testing. It represents the probability of observing a test statistic as extreme, or more so, than the one calculated from sample data, assuming the null hypothesis is true. A low p-value suggests evidence against the null, possibly justifying its rejection.

What is a t test?

A t-test is a statistical test used to compare the means of two groups. It determines if observed differences between the groups are statistically significant or if they likely occurred by chance. Commonly applied in research, there are different t-tests, including independent, paired, and one-sample, tailored to various data scenarios.

When to reject null hypothesis?

Reject the null hypothesis when the test statistic falls into a predefined rejection region or when the p-value is less than the chosen significance level (commonly 0.05). This suggests that the observed data is unlikely under the null hypothesis, indicating evidence for the alternative hypothesis. Always consider the study’s context.

Unit 12: Significance tests (hypothesis testing)

About this unit.

Significance tests give us a formal process for using sample data to evaluate the likelihood of some claim about a population value. Learn how to conduct significance tests and calculate p-values to see how likely a sample result is to occur by random chance. You'll also see how we use p-values to make conclusions about hypotheses.

The idea of significance tests

Simple hypothesis testing (Opens a modal)
Idea behind hypothesis testing (Opens a modal)
Examples of null and alternative hypotheses (Opens a modal)
P-values and significance tests (Opens a modal)
Comparing P-values to different significance levels (Opens a modal)
Estimating a P-value from a simulation (Opens a modal)
Using P-values to make conclusions (Opens a modal)
Simple hypothesis testing Get 3 of 4 questions to level up!
Writing null and alternative hypotheses Get 3 of 4 questions to level up!
Estimating P-values from simulations Get 3 of 4 questions to level up!

Error probabilities and power

Introduction to Type I and Type II errors (Opens a modal)
Type 1 errors (Opens a modal)
Examples identifying Type I and Type II errors (Opens a modal)
Introduction to power in significance tests (Opens a modal)
Examples thinking about power in significance tests (Opens a modal)
Consequences of errors and significance (Opens a modal)
Type I vs Type II error Get 3 of 4 questions to level up!
Error probabilities and power Get 3 of 4 questions to level up!

Tests about a population proportion

Constructing hypotheses for a significance test about a proportion (Opens a modal)
Conditions for a z test about a proportion (Opens a modal)
Reference: Conditions for inference on a proportion (Opens a modal)
Calculating a z statistic in a test about a proportion (Opens a modal)
Calculating a P-value given a z statistic (Opens a modal)
Making conclusions in a test about a proportion (Opens a modal)
Writing hypotheses for a test about a proportion Get 3 of 4 questions to level up!
Conditions for a z test about a proportion Get 3 of 4 questions to level up!
Calculating the test statistic in a z test for a proportion Get 3 of 4 questions to level up!
Calculating the P-value in a z test for a proportion Get 3 of 4 questions to level up!
Making conclusions in a z test for a proportion Get 3 of 4 questions to level up!

Tests about a population mean

Writing hypotheses for a significance test about a mean (Opens a modal)
Conditions for a t test about a mean (Opens a modal)
Reference: Conditions for inference on a mean (Opens a modal)
When to use z or t statistics in significance tests (Opens a modal)
Example calculating t statistic for a test about a mean (Opens a modal)
Using TI calculator for P-value from t statistic (Opens a modal)
Using a table to estimate P-value from t statistic (Opens a modal)
Comparing P-value from t statistic to significance level (Opens a modal)
Free response example: Significance test for a mean (Opens a modal)
Writing hypotheses for a test about a mean Get 3 of 4 questions to level up!
Conditions for a t test about a mean Get 3 of 4 questions to level up!
Calculating the test statistic in a t test for a mean Get 3 of 4 questions to level up!
Calculating the P-value in a t test for a mean Get 3 of 4 questions to level up!
Making conclusions in a t test for a mean Get 3 of 4 questions to level up!

9.4 Full Hypothesis Test Examples

Tests on means, example 9.8.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

H 0 : μ = 16.43 H a : μ < 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: X ¯ X ¯ = the mean time to swim the 25-yard freestyle.

Distribution for the test: X ¯ X ¯ is normal (population standard deviation is known: σ = 0.8)

X ¯ ~ N ( μ , σ X n ) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) X ¯ ~ N ( 16.43 , 0.8 15 )

μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15.

Calculate the p -value using the normal distribution for a mean:

p -value = P ( x ¯ x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16.

μ = 16.43 comes from H 0 . Our assumption is μ = 16.43.

Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p -value:

α = 0.05 p -value = 0.0187 α > p -value

Make a decision: Since α > α > p -value, reject H 0 .

This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds.

Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles.

The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Example 9.9

Jasmine has just begun her new job on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of 108 dollars with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least 100 dollars against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jasmine has met this requirement at the significance level of 95%?

H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not "too high" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis.
Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67
Critical value: t a = 1.753 t a = 1.753 with n-1 degrees of freedom= 15

The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t t ( t a ) ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jasmine's performance meets company standards.

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, state your conclusion, and identify the Type I errors.

Example 9.10

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.

Again we will follow the steps in our analysis of this problem.

STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus:

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

This problem has already set the level of significance at 99%. The decision seems an appropriate one and shows the thought process when setting the significance level. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table or the t-table at 0.005 column and infinite degrees of freedom. We draw the graph and mark these points.

STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph.

STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis.

STEP 5 : Reach a Conclusion

Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of significance we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of significance we conclude that the machine is under filling the bottles and is in need of repair”.

Try It 9.10

A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2).

Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 .

Hypothesis Test for Proportions

Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p .

The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size.

When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np μ = np and σ = npq σ = npq . Remember that q = 1 – p q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p .

Again, we begin with the standardizing formula modified because this is the distribution of a binomial.

Substituting p 0 p 0 , the hypothesized value of p , we have:

This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms:

The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test.

Example 9.11

The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance.

STEP 1 : Set the null and alternative hypothesis.

H 0 : p = 0.50 H a : p ≠ 0.50

The words "is the same or different from" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.)

STEP 2 : Decide the level of significance and draw the graph showing the critical value

The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations.

STEP 3 : Calculate the sample parameters and critical value of the test statistic.

The test statistic is a normal distribution, Z, for testing proportions and is:

For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations.

STEP 4 : Compare the test statistic and the critical value.

The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations.

STEP 5 : Reach a conclusion

The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data.

Try It 9.11

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

Example 9.12

Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones.

Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions.

Try It 9.12

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

Example 9.13

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05.

Let’s follow a four-step process to answer this statistical question.

H 0 : μ ≤ 1
H a : μ > 1
Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
Do the calculations and draw the graph .
State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

Try It 9.13

The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows:

205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205

Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal.

Example 9.14

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

H 0 : p ≤ 0.00034
H a : p > 0.00034

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np' = 420,019(0.00034) = 142.8, nq' = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p' = 0.00034. Thus we will be able to generalize our results to the population.

Try It 9.14

In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

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11 Hypothesis Testing with One Sample

Student learning outcomes.

By the end of this chapter, the student should be able to:

Be able to identify and develop the null and alternative hypothesis
Identify the consequences of Type I and Type II error.
Be able to perform an one-tailed and two-tailed hypothesis test using the critical value method
Be able to perform a hypothesis test using the p-value method
Be able to write conclusions based on hypothesis tests.

Introduction

Now we are down to the bread and butter work of the statistician: developing and testing hypotheses. It is important to put this material in a broader context so that the method by which a hypothesis is formed is understood completely. Using textbook examples often clouds the real source of statistical hypotheses.

Statistical testing is part of a much larger process known as the scientific method. This method was developed more than two centuries ago as the accepted way that new knowledge could be created. Until then, and unfortunately even today, among some, “knowledge” could be created simply by some authority saying something was so, ipso dicta . Superstition and conspiracy theories were (are?) accepted uncritically.

The scientific method, briefly, states that only by following a careful and specific process can some assertion be included in the accepted body of knowledge. This process begins with a set of assumptions upon which a theory, sometimes called a model, is built. This theory, if it has any validity, will lead to predictions; what we call hypotheses.

As an example, in Microeconomics the theory of consumer choice begins with certain assumption concerning human behavior. From these assumptions a theory of how consumers make choices using indifference curves and the budget line. This theory gave rise to a very important prediction, namely, that there was an inverse relationship between price and quantity demanded. This relationship was known as the demand curve. The negative slope of the demand curve is really just a prediction, or a hypothesis, that can be tested with statistical tools.

Unless hundreds and hundreds of statistical tests of this hypothesis had not confirmed this relationship, the so-called Law of Demand would have been discarded years ago. This is the role of statistics, to test the hypotheses of various theories to determine if they should be admitted into the accepted body of knowledge; how we understand our world. Once admitted, however, they may be later discarded if new theories come along that make better predictions.

Not long ago two scientists claimed that they could get more energy out of a process than was put in. This caused a tremendous stir for obvious reasons. They were on the cover of Time and were offered extravagant sums to bring their research work to private industry and any number of universities. It was not long until their work was subjected to the rigorous tests of the scientific method and found to be a failure. No other lab could replicate their findings. Consequently they have sunk into obscurity and their theory discarded. It may surface again when someone can pass the tests of the hypotheses required by the scientific method, but until then it is just a curiosity. Many pure frauds have been attempted over time, but most have been found out by applying the process of the scientific method.

This discussion is meant to show just where in this process statistics falls. Statistics and statisticians are not necessarily in the business of developing theories, but in the business of testing others’ theories. Hypotheses come from these theories based upon an explicit set of assumptions and sound logic. The hypothesis comes first, before any data are gathered. Data do not create hypotheses; they are used to test them. If we bear this in mind as we study this section the process of forming and testing hypotheses will make more sense.

One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about the value of a specific parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year.

A statistician will make a decision about these claims. This process is called ” hypothesis testing .” A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis.

In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests.

Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

Table 1 presents the various hypotheses in the relevant pairs. For example, if the null hypothesis is equal to some value, the alternative has to be not equal to that value.

NOTE

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:

$\mu$

We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:

Outcomes and the Type I and Type II Errors

The four possible outcomes in the table are:

Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities.

$\alpha$

By way of example, the American judicial system begins with the concept that a defendant is “presumed innocent”. This is the status quo and is the null hypothesis. The judge will tell the jury that they can not find the defendant guilty unless the evidence indicates guilt beyond a “reasonable doubt” which is usually defined in criminal cases as 95% certainty of guilt. If the jury cannot accept the null, innocent, then action will be taken, jail time. The burden of proof always lies with the alternative hypothesis. (In civil cases, the jury needs only to be more than 50% certain of wrongdoing to find culpability, called “a preponderance of the evidence”).

The example above was for a test of a mean, but the same logic applies to tests of hypotheses for all statistical parameters one may wish to test.

The following are examples of Type I and Type II errors.

Type I error : Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe.

Type II error : Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe.

Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)

This is a situation described as “accepting a false null”.

Type I error : The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error : The emergency crew does not know if the victim is alive when, in fact, the victim is dead.

The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.)

Distribution Needed for Hypothesis Testing

Particular distributions are associated with hypothesis testing.We will perform hypotheses tests of a population mean using a normal distribution or a Student’s t -distribution. (Remember, use a Student’s t -distribution when the population standard deviation is unknown and the sample size is small, where small is considered to be less than 30 observations.) We perform tests of a population proportion using a normal distribution when we can assume that the distribution is normally distributed. We consider this to be true if the sample proportion, p ‘ , times the sample size is greater than 5 and 1- p ‘ times the sample size is also greater then 5. This is the same rule of thumb we used when developing the formula for the confidence interval for a population proportion.

Hypothesis Test for the Mean

Going back to the standardizing formula we can derive the test statistic for testing hypotheses concerning means.

$Z_c=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

This gives us the decision rule for testing a hypothesis for a two-tailed test:

P-Value Approach

Both decision rules will result in the same decision and it is a matter of preference which one is used.

One and Two-tailed Tests

$\mu\neq100$

The claim would be in the alternative hypothesis. The burden of proof in hypothesis testing is carried in the alternative. This is because failing to reject the null, the status quo, must be accomplished with 90 or 95 percent significance that it cannot be maintained. Said another way, we want to have only a 5 or 10 percent probability of making a Type I error, rejecting a good null; overthrowing the status quo.

Figure 5 shows the two possible cases and the form of the null and alternative hypothesis that give rise to them.

Effects of Sample Size on Test Statistic

$\sigma$

Table 3 summarizes test statistics for varying sample sizes and population standard deviation known and unknown.

A Systematic Approach for Testing A Hypothesis

A systematic approach to hypothesis testing follows the following steps and in this order. This template will work for all hypotheses that you will ever test.

Set up the null and alternative hypothesis. This is typically the hardest part of the process. Here the question being asked is reviewed. What parameter is being tested, a mean, a proportion, differences in means, etc. Is this a one-tailed test or two-tailed test? Remember, if someone is making a claim it will always be a one-tailed test.
Decide the level of significance required for this particular case and determine the critical value. These can be found in the appropriate statistical table. The levels of confidence typical for the social sciences are 90, 95 and 99. However, the level of significance is a policy decision and should be based upon the risk of making a Type I error, rejecting a good null. Consider the consequences of making a Type I error.
Take a sample(s) and calculate the relevant parameters: sample mean, standard deviation, or proportion. Using the formula for the test statistic from above in step 2, now calculate the test statistic for this particular case using the parameters you have just calculated.
Compare the calculated test statistic and the critical value. Marking these on the graph will give a good visual picture of the situation. There are now only two situations:

a. The test statistic is in the tail: Cannot Accept the null, the probability that this sample mean (proportion) came from the hypothesized distribution is too small to believe that it is the real home of these sample data.

b. The test statistic is not in the tail: Cannot Reject the null, the sample data are compatible with the hypothesized population parameter.

Reach a conclusion. It is best to articulate the conclusion two different ways. First a formal statistical conclusion such as “With a 95 % level of significance we cannot accept the null hypotheses that the population mean is equal to XX (units of measurement)”. The second statement of the conclusion is less formal and states the action, or lack of action, required. If the formal conclusion was that above, then the informal one might be, “The machine is broken and we need to shut it down and call for repairs”.

All hypotheses tested will go through this same process. The only changes are the relevant formulas and those are determined by the hypothesis required to answer the original question.

Full Hypothesis Test Examples

Tests on means.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey’s mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05.

Solution – Example 6

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean . Set the null and alternative hypothesis:

In this case there is an implied challenge or claim. This is that the goggles will reduce the swimming time. The effect of this is to set the hypothesis as a one-tailed test. The claim will always be in the alternative hypothesis because the burden of proof always lies with the alternative. Remember that the status quo must be defeated with a high degree of confidence, in this case 95 % confidence. The null and alternative hypotheses are thus:

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The “<” tells you this is left-tailed. Determine the distribution needed:

Distribution for the test statistic:

The sample size is less than 30 and we do not know the population standard deviation so this is a t-test and the proper formula is:

$t_c=\frac{\bar{x}-{\mu_0}}{\frac{s}{\sqrt{n}}}$

Our step 2, setting the level of significance, has already been determined by the problem, .05 for a 95 % significance level. It is worth thinking about the meaning of this choice. The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.) For this case the only concern with a Type I error would seem to be that Jeffery’s dad may fail to bet on his son’s victory because he does not have appropriate confidence in the effect of the goggles.

To find the critical value we need to select the appropriate test statistic. We have concluded that this is a t-test on the basis of the sample size and that we are interested in a population mean. We can now draw the graph of the t-distribution and mark the critical value (Figure 6). For this problem the degrees of freedom are n-1, or 14. Looking up 14 degrees of freedom at the 0.05 column of the t-table we find 1.761. This is the critical value and we can put this on our graph.

Step 3 is the calculation of the test statistic using the formula we have selected.

$t_c=\frac{16-16.43}{\frac{0.8}{\sqrt{15}}}$

We find that the calculated test statistic is 2.08, meaning that the sample mean is 2.08 standard deviations away from the hypothesized mean of 16.43.

Step 4 has us compare the test statistic and the critical value and mark these on the graph. We see that the test statistic is in the tail and thus we move to step 4 and reach a conclusion. The probability that an average time of 16 minutes could come from a distribution with a population mean of 16.43 minutes is too unlikely for us to accept the null hypothesis. We cannot accept the null.

Step 5 has us state our conclusions first formally and then less formally. A formal conclusion would be stated as: “With a 95% level of significance we cannot accept the null hypothesis that the swimming time with goggles comes from a distribution with a population mean time of 16.43 minutes.” Less formally, “With 95% significance we believe that the goggles improves swimming speed”

If we wished to use the p-value system of reaching a conclusion we would calculate the statistic and take the additional step to find the probability of being 2.08 standard deviations from the mean on a t-distribution. This value is .0187. Comparing this to the α-level of .05 we see that we cannot accept the null. The p-value has been put on the graph as the shaded area beyond -2.08 and it shows that it is smaller than the hatched area which is the alpha level of 0.05. Both methods reach the same conclusion that we cannot accept the null hypothesis.

Jane has just begun her new job as on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of $108 with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least $100 against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jane has met this requirement at the significance level of 95%?

Solution – Example 7

STEP 1 : Set the Null and Alternative Hypothesis.

STEP 2 : Decide the level of significance and draw the graph (Figure 7) showing the critical value.

STEP 3 : Calculate sample parameters and the test statistic.

$t_c=\frac{108-100}{\frac{12}{\sqrt{16}}} = 2.67$

STEP 4 : Compare test statistic and the critical values

STEP 5 : Reach a Conclusion

The test statistic is a Student’s t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jane’s performance meets company standards.

Again we will follow the steps in our analysis of this problem.

Solution – Example 8

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

$Z_c=\frac{\bar{x}-{\mu_0}}{\frac{s}{\sqrt{n}}} = Z_c=\frac{7.91-8}{\frac{.173}{\sqrt{35}}}=-3.07$

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Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as $H_{0}$. Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as $H_{1}$ or $H_{a}$. For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, $\alpha$ or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$. $\overline{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation and n is the size of the sample.
t = $\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$. s is the sample standard deviation.
$\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}$. $O_{i}$ is the observed value and $E_{i}$ is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

One sample: z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.
Two samples: z = $\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$.

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

One sample: t = $\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$.
Two samples: t = $\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$.

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

$H_{0}$: The population parameter is ≤ some value

$H_{1}$: The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

$H_{0}$: The population parameter is ≥ some value

$H_{1}$: The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

$H_{0}$: the population parameter = some value

$H_{1}$: the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
Step 2: Set up the alternative hypothesis.
Step 3: Choose the correct significance level, $\alpha$, and find the critical value.
Step 4: Calculate the correct test statistic (z, t or $\chi$) and p-value.
Step 5: Compare the test statistic with the critical value or compare the p-value with $\alpha$ to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as $H_{0}$: $\mu$ = 100.

Step 2: The alternative hypothesis is given by $H_{1}$: $\mu$ > 100.

Step 3: As this is a one-tailed test, $\alpha$ = 100% - 95% = 5%. This can be used to determine the critical value.

1 - $\alpha$ = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.

$\mu$ = 100, $\overline{x}$ = 112.5, n = 30, $\sigma$ = 15

z = $\frac{112.5-100}{\frac{15}{\sqrt{30}}}$ = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Probability and Statistics
Data Handling

Important Notes on Hypothesis Testing

Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
It involves the setting up of a null hypothesis and an alternate hypothesis.
There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. $H_{0}$: $\mu$ = 90, $H_{1}$: $\mu$ > 90 $\overline{x}$ = 110, $\mu$ = 90, n = 5, s = 18. $\alpha$ = 0.05 Using the t-distribution table, the critical value is 2.132 t = $\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$ t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. $H_{0}$: $\mu$ = 80, $H_{1}$: $\mu$ ≠ 80 $\overline{x}$ = 88, $\mu$ = 80, n = 36, $\sigma$ = 10. $\alpha$ = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$ z = $\frac{88-80}{\frac{10}{\sqrt{36}}}$ = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. $H_{0}$: $\mu$ = 90, $H_{1}$: $\mu$ < 90 $\overline{x}$ = 110, $\mu$ = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = $\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$ t = $\frac{82-90}{\frac{18}{\sqrt{6}}}$ t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

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FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = $\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$ and for two samples is z = $\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$.

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide $\alpha$ by 2. This is done as there are two rejection regions in the curve.

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9.E: Hypothesis Testing with One Sample (Exercises)

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These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

9.1: Introduction

9.2: null and alternative hypotheses.

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, $H_{0}$, and the alternative hypothesis. $H_{a}$, in terms of the appropriate parameter $(\mu \text{or} p)$.

The mean number of years Americans work before retiring is 34.
At most 60% of Americans vote in presidential elections.
The mean starting salary for San Jose State University graduates is at least $100,000 per year.
Twenty-nine percent of high school seniors get drunk each month.
Fewer than 5% of adults ride the bus to work in Los Angeles.
The mean number of cars a person owns in her lifetime is not more than ten.
About half of Americans prefer to live away from cities, given the choice.
Europeans have a mean paid vacation each year of six weeks.
The chance of developing breast cancer is under 11% for women.
Private universities' mean tuition cost is more than $20,000 per year.
$H_{0}: \mu = 34; H_{a}: \mu \neq 34$
$H_{0}: p \leq 0.60; H_{a}: p > 0.60$
$H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000$
$H_{0}: p = 0.29; H_{a}: p \neq 0.29$
$H_{0}: p = 0.05; H_{a}: p < 0.05$
$H_{0}: \mu \leq 10; H_{a}: \mu > 10$
$H_{0}: p = 0.50; H_{a}: p \neq 0.50$
$H_{0}: \mu = 6; H_{a}: \mu \neq 6$
$H_{0}: p ≥ 0.11; H_{a}: p < 0.11$
$H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000$

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:

$p < 0.30$
$p \leq 0.30$
$p \geq 0.30$
$p > 0.30$

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:

$p = 0.20$
$p > 0.20$
$p < 0.20$
$p \leq 0.20$

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:

$H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5$
$H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5$
$H_{0}: \mu = 4.75, H_{a}: \mu > 4.75$
$H_{0}: \mu = 4.5, H_{a}: \mu > 4.5$

9.3: Outcomes and the Type I and Type II Errors

State the Type I and Type II errors in complete sentences given the following statements.

The mean number of cars a person owns in his or her lifetime is not more than ten.
Private universities mean tuition cost is more than $20,000 per year.
Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

For statements a-j in Exercise 9.109 , answer the following in complete sentences.

State a consequence of committing a Type I error.
State a consequence of committing a Type II error.

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?

To conclude the drug is safe when in, fact, it is unsafe.
Not to conclude the drug is safe when, in fact, it is safe.
To conclude the drug is safe when, in fact, it is safe.
Not to conclude the drug is unsafe when, in fact, it is unsafe.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

at least 20%, when in fact, it is less than 20%.
20%, when in fact, it is 20%.
less than 20%, when in fact, it is at least 20%.
less than 20%, when in fact, it is less than 20%.

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

is more than seven hours.
is at most seven hours.
is at least seven hours.
is less than seven hours.

to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

9.4: Distribution Needed for Hypothesis Testing

$N\left(7.24, \frac{1.93}{\sqrt{22}}\right)$
$N\left(7.24, 1.93\right)$

9.5: Rare Events, the Sample, Decision and Conclusion

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

Is this a test of one mean or proportion?
State the null and alternative hypotheses. $H_{0}$ : ____________________ $H_{a}$ : ____________________
Is this a right-tailed, left-tailed, or two-tailed test?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
$x =$ ________________
$n =$ ________________
$p′ =$ _____________
Calculate $\sigma_{x} =$ __________. Show the formula set-up.
State the distribution to use for the hypothesis test.
Find the $p\text{-value}$.
Reason for the decision:
Conclusion (write out in a complete sentence):

9.6: Additional Information and Full Hypothesis Test Examples

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student's $t$ - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using $\alpha = 0.05$, is the data highly inconsistent with the claim?

$H_{0}: \mu \geq 50,000$
$H_{a}: \mu < 50,000$
Let $\bar{X} =$ the average lifespan of a brand of tires.
normal distribution
$z = -2.315$
$p\text{-value} = 0.0103$
Check student’s solution.
alpha: 0.05
Decision: Reject the null hypothesis.
Reason for decision: The $p\text{-value}$ is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
$(43,537, 49,463)$

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

$H_{0}: \mu = $1.00$
$H_{a}: \mu \neq $1.00$
Let $\bar{X} =$ the average cost of a daily newspaper.
$z = –0.866$
$p\text{-value} = 0.3865$
$\alpha: 0.01$
Decision: Do not reject the null hypothesis.
Reason for decision: The $p\text{-value}$ is greater than 0.01.
Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
$($0.84, $1.06)$

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let $x =$ the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

$H_{0}: \mu = 10$
$H_{a}: \mu \neq 10$
Let $\bar{X}$ the mean number of sick days an employee takes per year.
Student’s t -distribution
$t = –1.12$
$p\text{-value} = 0.300$
$\alpha: 0.05$
Reason for decision: The $p\text{-value}$ is greater than 0.05.
Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
$(4.9443, 11.806)$

In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

$H_{0}: p \geq 0.6$
$H_{a}: p < 0.6$
Let $P′ =$ the proportion of students who feel more enriched as a result of taking Elementary Statistics.
normal for a single proportion
$p\text{-value} = 0.1308$
Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

The “plus-4s” confidence interval is $(0.411, 0.648)$

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

$H_{0}: \mu = 4$
$H_{a}: \mu \neq 4$
Let $\bar{X}$ the average I.Q. of a set of brown trout.
two-tailed Student's t-test
$t = 1.95$
$p\text{-value} = 0.076$
Reason for decision: The $p\text{-value}$ is greater than 0.05
Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
$(3.8865,5.9468)$

According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

$H_{a}: p < 0.13$
Let $P′ =$ the proportion of Americans who have seen or sensed angels
–2.688
$p\text{-value} = 0.0036$
Reason for decision: The $p\text{-value}$e is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.

The“plus-4s” confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.

$H_{0}: \mu \geq 129$
$H_{a}: \mu < 129$
Let $\bar{X} =$ the average time in seconds that Terri finishes Lap 4.
Student's t -distribution
$t = 1.209$
Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
$(128.63, 130.37)$

Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

$H_{0}: p = 0.60$
$H_{a}: p < 0.60$
Let $P′ =$ the proportion of family members who shed tears at a reunion.
–1.71
Reason for decision: $p\text{-value} < \alpha$
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the $p\text{-value}$ and alpha are quite close, so other tests should be done.
We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. $(0.3829, 0.6171)$. The“plus-4s” confidence interval (see chapter 8) is $(0.3861, 0.6139)$

Note that here the “large-sample” $1 - \text{PropZTest}$ provides the approximate $p\text{-value}$ of 0.0438. Whenever a $p\text{-value}$ based on a normal approximation is close to the level of significance, the exact $p\text{-value}$ based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

Ninety-five high school and college students I did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

$H_{0}: \mu \geq 22$
$H_{a}: \mu < 22$
Let $\bar{X} =$ the mean number of bubbles per blow.
–2.667
$p\text{-value} = 0.00486$
Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
$(18.501, 21.499)$

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

5 stores @ $2.02
15 stores @ $0.25
3 stores @ $1.29
6 stores @ $0.35
4 stores @ $2.27
7 stores @ $1.50
5 stores @ $1.89
8 stores @ 0.75.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

$H_{0}: \mu \leq 1$
$H_{a}: \mu > 1$
Let $\bar{X} =$ the mean cost in dollars of macaroni and cheese in a certain town.
Student's $t$-distribution
$t = 0.340$
$p\text{-value} = 0.36756$
Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
$(0.8291, 1.241)$

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

HAMLET, Prince of Denmark and student of Statistics
POLONIUS, Hamlet’s tutor
HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)

(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

$H_{0}: p = 0.01$
$H_{a}: p > 0.01$
Let $P′ =$ the proportion of errors generated
Normal for a single proportion
Decision: Reject the null hypothesis
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

The“plus-4s” confidence interval is $(0.004, 0.144)$.

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

$H_{0}: p = 0.50$
$H_{a}: p < 0.50$
Let $P′ =$ the proportion of friends that has a pierced ear.
–1.70
$p\text{-value} = 0.0448$
Reason for decision: The $p\text{-value}$ is less than 0.05. (However, they are very close.)
Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
Confidence Interval: $(0.245, 0.515)$: The “plus-4s” confidence interval is $(0.259, 0.519)$.

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.

$H_{0}: p = 0.40$
$H_{a}: p < 0.40$
Let $P′ =$ the proportion of schoolmates who fear public speaking.
–1.01
$p\text{-value} = 0.1563$
Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
Confidence Interval: $(0.3241, 0.4240)$: The “plus-4s” confidence interval is $(0.3257, 0.4250)$.

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

$H_{0}: p = 0.14$
$H_{a}: p < 0.14$
Let $P′ =$ the proportion of NYC residents that smoke.
–0.2756
$p\text{-value} = 0.3914$
At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
Confidence Interval: $(0.0502, 0.2070)$: The “plus-4s” confidence interval (see chapter 8) is $(0.0676, 0.2297)$.

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

$H_{0}: \mu = 69,110$
$H_{0}: \mu > 69,110$
Let $\bar{X} =$ the mean salary in dollars for California registered nurses.
$t = 1.719$
$p\text{-value}: 0.0466$
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
$($68,757, $73,485)$

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

After conducting the test, your decision and conclusion are

Reject $H_{0}$: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
Do not reject $H_{0}$: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
Do not reject $H_{0}$: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
Reject $H_{0}$: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.

At a 1% level of significance, an appropriate conclusion is:

There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.

At a significance level of $a = 0.05$, what is the correct conclusion?

There is enough evidence to conclude that the mean number of hours is more than 4.75
There is enough evidence to conclude that the mean number of hours is more than 4.5
There is not enough evidence to conclude that the mean number of hours is more than 4.5
There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

State the null and alternate hypothesis.

State the $p\text{-value}$.

State $\alpha$.

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

$H_{0}: p = 0.488$ $H_{a}: p \neq 0.488$
$p\text{-value} = 0.0114$
$\alpha = 0.05$
Reject the null hypothesis.
At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using $\alpha = 0.05$, is the AAA proportion accurate?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the $\alpha = 0.05$ level in Kentucky? Are the results applicable across the country? Why?

$H_{0}: p = 0.517$ $H_{0}: p \neq 0.517$
$p\text{-value} = 0.9203$.
$\alpha = 0.05$.
Do not reject the null hypothesis.
At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use $\alpha = 0.01$ level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the $\alpha = 0.05 level$, can it be concluded that the mean rainfall was below the reported average? What if $\alpha = 0.01$? Assume the amount of summer rainfall follows a normal distribution.

$H_{0}: \mu \geq 11.52$ $H_{a}: \mu < 11.52$
$p\text{-value} = 0.000002$ which is almost 0.
At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
We would make the same conclusion if alpha was 1% because the $p\text{-value}$ is almost 0.

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the $\alpha = 0.10$ level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals

3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1

At the $\alpha = 0.05$ level can it be concluded that the sample mean is higher than 5.8 visits per year?

$H_{0}: \mu \leq 5.8$ $H_{a}: \mu > 5.8$
$p\text{-value} = 0.9987$
At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2

At $\alpha = 0.05$ level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

$H_{0}: \mu \geq 150$ $H_{0}: \mu < 150$
$p\text{-value} = 0.0622$
$\alpha = 0.01$
At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
The student academic group’s claim appears to be correct.

9.7: Hypothesis Testing of a Single Mean and Single Proportion

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Everything you need to know about the normal distribution, an in-depth explanation of cumulative distribution function, a complete guide to chi-square test, a complete guide on hypothesis testing in statistics, understanding the fundamentals of arithmetic and geometric progression, the definitive guide to understand spearman’s rank correlation, a comprehensive guide to understand mean squared error, all you need to know about the empirical rule in statistics, the complete guide to skewness and kurtosis, a holistic look at bernoulli distribution.

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Lesson 10 of 24 By Avijeet Biswal

A Complete Guide on Hypothesis Testing in Statistics

In today’s data-driven world , decisions are based on data all the time. Hypothesis plays a crucial role in that process, whether it may be making business decisions, in the health sector, academia, or in quality improvement. Without hypothesis & hypothesis tests, you risk drawing the wrong conclusions and making bad decisions. In this tutorial, you will look at Hypothesis Testing in Statistics.

What Is Hypothesis Testing in Statistics?

Hypothesis Testing is a type of statistical analysis in which you put your assumptions about a population parameter to the test. It is used to estimate the relationship between 2 statistical variables.

Let's discuss few examples of statistical hypothesis from real-life -

A teacher assumes that 60% of his college's students come from lower-middle-class families.
A doctor believes that 3D (Diet, Dose, and Discipline) is 90% effective for diabetic patients.

Now that you know about hypothesis testing, look at the two types of hypothesis testing in statistics.

Hypothesis Testing Formula

Z = ( x̅ – μ0 ) / (σ /√n)

Here, x̅ is the sample mean,
μ0 is the population mean,
σ is the standard deviation,
n is the sample size.

How Hypothesis Testing Works?

An analyst performs hypothesis testing on a statistical sample to present evidence of the plausibility of the null hypothesis. Measurements and analyses are conducted on a random sample of the population to test a theory. Analysts use a random population sample to test two hypotheses: the null and alternative hypotheses.

The null hypothesis is typically an equality hypothesis between population parameters; for example, a null hypothesis may claim that the population means return equals zero. The alternate hypothesis is essentially the inverse of the null hypothesis (e.g., the population means the return is not equal to zero). As a result, they are mutually exclusive, and only one can be correct. One of the two possibilities, however, will always be correct.

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Null Hypothesis and Alternate Hypothesis

The Null Hypothesis is the assumption that the event will not occur. A null hypothesis has no bearing on the study's outcome unless it is rejected.

H0 is the symbol for it, and it is pronounced H-naught.

The Alternate Hypothesis is the logical opposite of the null hypothesis. The acceptance of the alternative hypothesis follows the rejection of the null hypothesis. H1 is the symbol for it.

Let's understand this with an example.

A sanitizer manufacturer claims that its product kills 95 percent of germs on average.

To put this company's claim to the test, create a null and alternate hypothesis.

H0 (Null Hypothesis): Average = 95%.

Alternative Hypothesis (H1): The average is less than 95%.

Another straightforward example to understand this concept is determining whether or not a coin is fair and balanced. The null hypothesis states that the probability of a show of heads is equal to the likelihood of a show of tails. In contrast, the alternate theory states that the probability of a show of heads and tails would be very different.

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Hypothesis Testing Calculation With Examples

Let's consider a hypothesis test for the average height of women in the United States. Suppose our null hypothesis is that the average height is 5'4". We gather a sample of 100 women and determine that their average height is 5'5". The standard deviation of population is 2.

To calculate the z-score, we would use the following formula:

z = ( x̅ – μ0 ) / (σ /√n)

z = (5'5" - 5'4") / (2" / √100)

z = 0.5 / (0.045)

We will reject the null hypothesis as the z-score of 11.11 is very large and conclude that there is evidence to suggest that the average height of women in the US is greater than 5'4".

Steps of Hypothesis Testing

Step 1: specify your null and alternate hypotheses.

It is critical to rephrase your original research hypothesis (the prediction that you wish to study) as a null (Ho) and alternative (Ha) hypothesis so that you can test it quantitatively. Your first hypothesis, which predicts a link between variables, is generally your alternate hypothesis. The null hypothesis predicts no link between the variables of interest.

Step 2: Gather Data

For a statistical test to be legitimate, sampling and data collection must be done in a way that is meant to test your hypothesis. You cannot draw statistical conclusions about the population you are interested in if your data is not representative.

Step 3: Conduct a Statistical Test

Other statistical tests are available, but they all compare within-group variance (how to spread out the data inside a category) against between-group variance (how different the categories are from one another). If the between-group variation is big enough that there is little or no overlap between groups, your statistical test will display a low p-value to represent this. This suggests that the disparities between these groups are unlikely to have occurred by accident. Alternatively, if there is a large within-group variance and a low between-group variance, your statistical test will show a high p-value. Any difference you find across groups is most likely attributable to chance. The variety of variables and the level of measurement of your obtained data will influence your statistical test selection.

Step 4: Determine Rejection Of Your Null Hypothesis

Your statistical test results must determine whether your null hypothesis should be rejected or not. In most circumstances, you will base your judgment on the p-value provided by the statistical test. In most circumstances, your preset level of significance for rejecting the null hypothesis will be 0.05 - that is, when there is less than a 5% likelihood that these data would be seen if the null hypothesis were true. In other circumstances, researchers use a lower level of significance, such as 0.01 (1%). This reduces the possibility of wrongly rejecting the null hypothesis.

Step 5: Present Your Results

The findings of hypothesis testing will be discussed in the results and discussion portions of your research paper, dissertation, or thesis. You should include a concise overview of the data and a summary of the findings of your statistical test in the results section. You can talk about whether your results confirmed your initial hypothesis or not in the conversation. Rejecting or failing to reject the null hypothesis is a formal term used in hypothesis testing. This is likely a must for your statistics assignments.

Types of Hypothesis Testing

To determine whether a discovery or relationship is statistically significant, hypothesis testing uses a z-test. It usually checks to see if two means are the same (the null hypothesis). Only when the population standard deviation is known and the sample size is 30 data points or more, can a z-test be applied.

A statistical test called a t-test is employed to compare the means of two groups. To determine whether two groups differ or if a procedure or treatment affects the population of interest, it is frequently used in hypothesis testing.

Chi-Square

You utilize a Chi-square test for hypothesis testing concerning whether your data is as predicted. To determine if the expected and observed results are well-fitted, the Chi-square test analyzes the differences between categorical variables from a random sample. The test's fundamental premise is that the observed values in your data should be compared to the predicted values that would be present if the null hypothesis were true.

Hypothesis Testing and Confidence Intervals

Both confidence intervals and hypothesis tests are inferential techniques that depend on approximating the sample distribution. Data from a sample is used to estimate a population parameter using confidence intervals. Data from a sample is used in hypothesis testing to examine a given hypothesis. We must have a postulated parameter to conduct hypothesis testing.

Bootstrap distributions and randomization distributions are created using comparable simulation techniques. The observed sample statistic is the focal point of a bootstrap distribution, whereas the null hypothesis value is the focal point of a randomization distribution.

A variety of feasible population parameter estimates are included in confidence ranges. In this lesson, we created just two-tailed confidence intervals. There is a direct connection between these two-tail confidence intervals and these two-tail hypothesis tests. The results of a two-tailed hypothesis test and two-tailed confidence intervals typically provide the same results. In other words, a hypothesis test at the 0.05 level will virtually always fail to reject the null hypothesis if the 95% confidence interval contains the predicted value. A hypothesis test at the 0.05 level will nearly certainly reject the null hypothesis if the 95% confidence interval does not include the hypothesized parameter.

Simple and Composite Hypothesis Testing

Depending on the population distribution, you can classify the statistical hypothesis into two types.

Simple Hypothesis: A simple hypothesis specifies an exact value for the parameter.

Composite Hypothesis: A composite hypothesis specifies a range of values.

A company is claiming that their average sales for this quarter are 1000 units. This is an example of a simple hypothesis.

Suppose the company claims that the sales are in the range of 900 to 1000 units. Then this is a case of a composite hypothesis.

One-Tailed and Two-Tailed Hypothesis Testing

The One-Tailed test, also called a directional test, considers a critical region of data that would result in the null hypothesis being rejected if the test sample falls into it, inevitably meaning the acceptance of the alternate hypothesis.

In a one-tailed test, the critical distribution area is one-sided, meaning the test sample is either greater or lesser than a specific value.

In two tails, the test sample is checked to be greater or less than a range of values in a Two-Tailed test, implying that the critical distribution area is two-sided.

If the sample falls within this range, the alternate hypothesis will be accepted, and the null hypothesis will be rejected.

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Right Tailed Hypothesis Testing

If the larger than (>) sign appears in your hypothesis statement, you are using a right-tailed test, also known as an upper test. Or, to put it another way, the disparity is to the right. For instance, you can contrast the battery life before and after a change in production. Your hypothesis statements can be the following if you want to know if the battery life is longer than the original (let's say 90 hours):

The null hypothesis is (H0 <= 90) or less change.
A possibility is that battery life has risen (H1) > 90.

The crucial point in this situation is that the alternate hypothesis (H1), not the null hypothesis, decides whether you get a right-tailed test.

Left Tailed Hypothesis Testing

Alternative hypotheses that assert the true value of a parameter is lower than the null hypothesis are tested with a left-tailed test; they are indicated by the asterisk "<".

Suppose H0: mean = 50 and H1: mean not equal to 50

According to the H1, the mean can be greater than or less than 50. This is an example of a Two-tailed test.

In a similar manner, if H0: mean >=50, then H1: mean <50

Here the mean is less than 50. It is called a One-tailed test.

Type 1 and Type 2 Error

A hypothesis test can result in two types of errors.

Type 1 Error: A Type-I error occurs when sample results reject the null hypothesis despite being true.

Type 2 Error: A Type-II error occurs when the null hypothesis is not rejected when it is false, unlike a Type-I error.

Suppose a teacher evaluates the examination paper to decide whether a student passes or fails.

H0: Student has passed

H1: Student has failed

Type I error will be the teacher failing the student [rejects H0] although the student scored the passing marks [H0 was true].

Type II error will be the case where the teacher passes the student [do not reject H0] although the student did not score the passing marks [H1 is true].

Level of Significance

The alpha value is a criterion for determining whether a test statistic is statistically significant. In a statistical test, Alpha represents an acceptable probability of a Type I error. Because alpha is a probability, it can be anywhere between 0 and 1. In practice, the most commonly used alpha values are 0.01, 0.05, and 0.1, which represent a 1%, 5%, and 10% chance of a Type I error, respectively (i.e. rejecting the null hypothesis when it is in fact correct).

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A p-value is a metric that expresses the likelihood that an observed difference could have occurred by chance. As the p-value decreases the statistical significance of the observed difference increases. If the p-value is too low, you reject the null hypothesis.

Here you have taken an example in which you are trying to test whether the new advertising campaign has increased the product's sales. The p-value is the likelihood that the null hypothesis, which states that there is no change in the sales due to the new advertising campaign, is true. If the p-value is .30, then there is a 30% chance that there is no increase or decrease in the product's sales. If the p-value is 0.03, then there is a 3% probability that there is no increase or decrease in the sales value due to the new advertising campaign. As you can see, the lower the p-value, the chances of the alternate hypothesis being true increases, which means that the new advertising campaign causes an increase or decrease in sales.

Why is Hypothesis Testing Important in Research Methodology?

Hypothesis testing is crucial in research methodology for several reasons:

Provides evidence-based conclusions: It allows researchers to make objective conclusions based on empirical data, providing evidence to support or refute their research hypotheses.
Supports decision-making: It helps make informed decisions, such as accepting or rejecting a new treatment, implementing policy changes, or adopting new practices.
Adds rigor and validity: It adds scientific rigor to research using statistical methods to analyze data, ensuring that conclusions are based on sound statistical evidence.
Contributes to the advancement of knowledge: By testing hypotheses, researchers contribute to the growth of knowledge in their respective fields by confirming existing theories or discovering new patterns and relationships.

Limitations of Hypothesis Testing

Hypothesis testing has some limitations that researchers should be aware of:

It cannot prove or establish the truth: Hypothesis testing provides evidence to support or reject a hypothesis, but it cannot confirm the absolute truth of the research question.
Results are sample-specific: Hypothesis testing is based on analyzing a sample from a population, and the conclusions drawn are specific to that particular sample.
Possible errors: During hypothesis testing, there is a chance of committing type I error (rejecting a true null hypothesis) or type II error (failing to reject a false null hypothesis).
Assumptions and requirements: Different tests have specific assumptions and requirements that must be met to accurately interpret results.

After reading this tutorial, you would have a much better understanding of hypothesis testing, one of the most important concepts in the field of Data Science . The majority of hypotheses are based on speculation about observed behavior, natural phenomena, or established theories.

If you are interested in statistics of data science and skills needed for such a career, you ought to explore Simplilearn’s Post Graduate Program in Data Science.

If you have any questions regarding this ‘Hypothesis Testing In Statistics’ tutorial, do share them in the comment section. Our subject matter expert will respond to your queries. Happy learning!

1. What is hypothesis testing in statistics with example?

Hypothesis testing is a statistical method used to determine if there is enough evidence in a sample data to draw conclusions about a population. It involves formulating two competing hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and then collecting data to assess the evidence. An example: testing if a new drug improves patient recovery (Ha) compared to the standard treatment (H0) based on collected patient data.

2. What is hypothesis testing and its types?

Hypothesis testing is a statistical method used to make inferences about a population based on sample data. It involves formulating two hypotheses: the null hypothesis (H0), which represents the default assumption, and the alternative hypothesis (Ha), which contradicts H0. The goal is to assess the evidence and determine whether there is enough statistical significance to reject the null hypothesis in favor of the alternative hypothesis.

Types of hypothesis testing:

One-sample test: Used to compare a sample to a known value or a hypothesized value.
Two-sample test: Compares two independent samples to assess if there is a significant difference between their means or distributions.
Paired-sample test: Compares two related samples, such as pre-test and post-test data, to evaluate changes within the same subjects over time or under different conditions.
Chi-square test: Used to analyze categorical data and determine if there is a significant association between variables.
ANOVA (Analysis of Variance): Compares means across multiple groups to check if there is a significant difference between them.

3. What are the steps of hypothesis testing?

The steps of hypothesis testing are as follows:

Formulate the hypotheses: State the null hypothesis (H0) and the alternative hypothesis (Ha) based on the research question.
Set the significance level: Determine the acceptable level of error (alpha) for making a decision.
Collect and analyze data: Gather and process the sample data.
Compute test statistic: Calculate the appropriate statistical test to assess the evidence.
Make a decision: Compare the test statistic with critical values or p-values and determine whether to reject H0 in favor of Ha or not.
Draw conclusions: Interpret the results and communicate the findings in the context of the research question.

4. What are the 2 types of hypothesis testing?

One-tailed (or one-sided) test: Tests for the significance of an effect in only one direction, either positive or negative.
Two-tailed (or two-sided) test: Tests for the significance of an effect in both directions, allowing for the possibility of a positive or negative effect.

The choice between one-tailed and two-tailed tests depends on the specific research question and the directionality of the expected effect.

5. What are the 3 major types of hypothesis?

The three major types of hypotheses are:

Null Hypothesis (H0): Represents the default assumption, stating that there is no significant effect or relationship in the data.
Alternative Hypothesis (Ha): Contradicts the null hypothesis and proposes a specific effect or relationship that researchers want to investigate.
Nondirectional Hypothesis: An alternative hypothesis that doesn't specify the direction of the effect, leaving it open for both positive and negative possibilities.

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Avijeet is a Senior Research Analyst at Simplilearn. Passionate about Data Analytics, Machine Learning, and Deep Learning, Avijeet is also interested in politics, cricket, and football.

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10: Hypothesis Testing with One Sample

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One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year.

10.1: Prelude to Hypothesis Testing A statistician will make a decision about claims via a process called "hypothesis testing." A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analysis of the data, to reject the null hypothesis.
10.2.1: Null and Alternative Hypotheses (Exercises)
10.3.1: Outcomes and the Type I and Type II Errors (Exercises)
10.4.1: Distribution Needed for Hypothesis Testing (Exercises)
10.5.1: Rare Events, the Sample, Decision and Conclusion (Exercises)
10.6: Additional Information and Full Hypothesis Test Examples The hypothesis test itself has an established process. This can be summarized as follows: Determine H0 and Ha. Remember, they are contradictory. Determine the random variable. Determine the distribution for the test. Draw a graph, calculate the test statistic, and use the test statistic to calculate the p-value. (A z-score and a t-score are examples of test statistics.) Compare the preconceived α with the p-value, make a decision (reject or do not reject H0), and write a clear conclusion.
10.7: Hypothesis Testing of a Single Mean and Single Proportion (Worksheet) A statistics Worksheet: The student will select the appropriate distributions to use in each case. The student will conduct hypothesis tests and interpret the results.
10.8: Hypothesis Testing with One Sample (Exercises) These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

Contributors and Attributions

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8 Hypothesis Testing Examples in Real Life

Hypothesis testing refers to the systematic and scientific method of examining whether the hypothesis set by the researcher is valid or not. Hypothesis testing verifies that the findings of an experiment are valid and the particular results did not happen by chance. If the particular results have occurred by chance then that experiment can not be repeated and its findings won’t be reliable. For example, if you conduct a study that finds that a particular drug is responsible for the blood pressure problem in diabetic patients. But, when you repeat this experiment and it does not give the same results, no one would trust this experiment’s findings. Hence, hypothesis testing is a very crucial step to verify the experimental findings. The main criterion of hypothesis testing is to check whether the null hypothesis is rejected or retained. The null hypothesis assumes that there does not exist any relationship between the variables under investigation, while the alternate hypothesis confirms the association between the variables under investigation. If the null hypothesis is rejected, it means that alternative hypotheses (research hypothesis) are accepted, and if the null hypothesis is accepted, the alternate hypothesis is rejected automatically. In this article, we’ll learn about hypothesis testing and various real-life examples of hypothesis testing.

Understanding Hypothesis Testing

The hypothesis testing broadly involves the following steps,

Step 1 : Formulate the research hypothesis and the null hypothesis of the experiment.
Step 2: Set the characteristics of the comparison distribution.
Step3: Set the criterion for decision making, i.e., cut off sample score for the comparison to reject or retain the null hypothesis.
Step:4 Set the outcome of the sample on the comparison distribution.
Step 5: Make a decision whether to reject or retain the null hypothesis.

Let us understand these steps through the following example,

Suppose the researcher wants to examine whether the memorizing power of students improves after consuming caffeine or not. To examine this he conducts experiments, the experiment involves two groups say group A (experimental group) and group B (control group). Group A consumed the coffee before the memory test, while group B consumed the water before the memory test. The average normally distributed score of the people of the experimental group has a standard deviation of 4 and a mean of 19. On the basis of the score, the researcher can state that there is an association between the two variables, i.e., the memory power and the caffeine, but the researcher can not predict any particular direction, i.e., which out of the experimental group and the control group had performed better in the memory tests. Hence, the level of significance value, i.e., 5 per cent will help to draw the conclusion. Following is the stepwise hypothesis testing of this example,

Step 1: Formulating Null hypothesis and alternate hypothesis

There exist two sample populations, i.e., group A and group B.

Group A: People who consumed coffee before the experiment

Group B: People who consumed water before the experiment.

On the basis of this, the null hypothesis and the alternative hypothesis would be as follows.

Alternate Hypothesis: Group A will perform differently from Group B, i.e., there exists an association between the two variables.

Null Hypothesis: There will not be any difference between the performance of both groups, i.e., Group A and Group B both will perform similarly.

Step 2: Characteristics of the comparison distribution

The characteristics of the comparison distribution in this example are given below,

Population Mean = 19

Standard Deviation= 4, normally distributed.

Step 3: Cut off score

In this test the direction of effect is not stated, i.e., it is a two-tailed test. In the case of a two-tailed test, the cut off sample scores is equal to +1.96 and -1.99 at the 5 per cent level.

Step 4: Outcome of Sample Score

The sample score is then converted into the Z value. Using the appropriate method of conversion this value is turned out to be equal to 2.

Step 5: Decision Making

The Z score value of 2 is far more than the cut off Z value, ie., +1.96, hence the result is significant, ie., rejection of the null hypothesis, i.e., there exists an association between the memory power and the consumption of the coffee before the test.

Click here , to understand hypothesis testing in detail.

Hypothesis Testing Real Life Examples

Following are some real-life examples of hypothesis testing.

1. To Check the Manufacturing Processes

Hypothesis testing finds its application in the manufacturing processes such as in determining whether the implication of the new technique or process in the manufacturing plant caused the anomalies in the quality of the product or not. Let us suppose, that manufacturing plant X decides to verify that the particular method results in an increase in the defective products per quarter, say this number to be 200. Now, to verify this the researcher needs to calculate the mean of the number of defective products produced before the start and the end of the quarter.

Following is the representation of the Hypothesis testing of this example,

Null Hypothesis (Ho) : The average of the defective products produced is the same before and after the implementation of the new manufacturing method, i.e., μ after = μ before

Alternative Hypothesis (Ha) : The average number of defective products produced are different before and after the implementation of the new manufacturing method, i.e., μ after ≠ μ before

If the resultant p-value of the hypothesis testing comes lesser than the significant value, i.e., α = .05, then the null hypothesis is rejected and it can be concluded that the changes in the method of production lead to the rise in the number of defective products production per quarter.

2. To Plan the Marketing Strategies

Many businesses often use hypothesis testing to determine the impact of the newly implemented marketing techniques, campaigns or other tactics on the sales of the product. For example, the marketing department of the company assumed that if they spend more the digital advertisements it would lead to a rise in sales. To verify this assumption, the marketing department may raise the digital advertisement budget for a particular period, and then analyse the collected data at the end of that period. They have to perform hypothesis testing to verify their assumption. Here,

Null Hypothesis (Ho) : The average sales are the same before and after the rise in the digital advertisement budget, i.e., μafter = μbefore

Alternative Hypothesis (Ha) : The average sales increase after the rise in the digital advertisement budget, i.e., μafter > μbefore

If the P-value is smaller than the significant value (say .05), then the null hypothesis can be rejected by the marketing department, and they can conclude that the rise in the digital advertisement budget can result in a rise in the sales of the product.

3. In Clinical Trials

Many pharmacists and doctors use hypothesis testing for clinical trials. The impact of the new clinical methods, medicines or procedures on the condition of the patients is analysed through hypothesis testing. For example, a pharmacist believes that the new medicine is resulting in the rise of blood pressure in diabetic patients. To test this assumption, the researcher has to measure the blood pressure of the sample patients (patients under investigation) before and after the intake of the new medicine for nearly a particular period say one month. The following procedure of the hypothesis testing is then followed,

Null Hypothesis (H0) : The average blood pressure is the same after and before the consumption of the medicine, i.e., μafter = μbefore

Alternative Hypothesis (Ha): The average blood pressure after the consumption of the medicine is less than the average blood pressure before the consumption of the medicine, i.e., μafter < μbefore

If the p-value of the hypothesis test is less than the significance value (say .o5), the null hypothesis is rejected, i.e., it can be concluded that the new drug is responsible for the rise in the blood pressure of diabetic patients.

4. In Testing Effectiveness of Essential Oils

Essential oils are gaining popularity nowadays due to their various benefits. Various essential oils such as ylang-ylang, lavender, and chamomile claim to reduce anxiety. You might like to test the true healing powers of all these essential oils. Suppose you assume that the lavender essential oil has the ability to reduce stress and anxiety. To check this assumption you may conduct the hypothesis testing by restating the hypothesis as follows,

Null Hypothesis (Ho) : Lavender essential has no effect on reducing anxiety.

Alternative Hypothesis (Ha): Lavender oil helps in reducing anxiety.

In this experiment, group A, i.e., the experimental group are provided with the lavender oil, while group B, i.e., the control group is provided with the placebo. The data is then collected using the various statistical tools and the stress level of both the groups, i.e., the experimental and the control group is then analysed. After the calculation, the significance level, and the p-value are found to be 0.25, and 0.05 respectively. The p values are less than the significance values, hence the null hypothesis is rejected, and it can be concluded that the lavender oil helps in reducing the stress among the people.

5. In Testing Fertilizer’s Impact on Plants

Nowadays, hypothesis testing is also used to examine the impact of pesticides, fertilizers, and other chemicals on the growth of plants or animals. Let us suppose a researcher wants to check his assumption that the particular fertilizer may result in the faster growth of the plant in a month than its usual growth of 10 inches. To verify this assumption he consistently gave that fertilizer to the plant for nearly a month. Following is the mathematical procedure of the hypothesis testing in this case,

Null Hypothesis (H0): The fertilizer does not have any influence on the growth of the plant. i.e., μ = 20 inches

Alternative Hypothesis (Ha): The fertiliser results in the faster growth of the plant, i.e., μ > 20 inches

Now, if the p-value of the hypothesis testing comes smaller than the level of significance, say .05, then the null hypothesis can be rejected, and you can conclude that the particular fertilizer is responsible for the faster plant growth.

6. In Testing the Effectiveness of Vitamin E

Suppose the researcher assumes that Vitamin E helps in the faster growth of the Hair. He conduct an experiment in which the experimental group is provided with vitamin E for three months while the controlled group is provided with the placebo. The results are then analysed after the duration of three months. To verify his assumption he restates the hypothesis as follows,

Null Hypothesis (H0) : There is no association between the Vitamin E and the hair growth of the sample group, i.e., μafter = μbefore

Alternative Hypothesis (Ha) : The group of people who consumed the vitamin E shows faster hair growth than the average hair growth of them before the consumption of the Vitamin E provided other variables remains constant. Here, μafter > μbefore.

After performing the statistical analysis, the significance level and the p-value in this scenario are o.o5, and 0.20 respectively. Hence, the researcher can conclude that the consumption of vitamin E results in faster hair growth.

7. In Testing the Teaching Strategy

Suppose the two teachers say Mr X and Mr Y argue about the best teaching strategy. Mr X says that children will perform better in the annual exams if they are given the weekly tests, while Mr Y argues that the weekly test would not impact the performance of the children in the annual exams and it is waste of time. Now, to verify who is right between the both, we may conduct hypothesis testing. The researcher may formulate the hypothesis as follows,

Null Hypothesis (Ho): There is no association between the weekly tests on the performance of the children in the annual exams, i.e., the average marks scored by the children when they were given the weekly exams and when not, were the same. (μafter = μbefore)

Alternative Hypothesis (Ha): The children will perform better in the annual exams, when they have to give the weekly tests, rather than just giving the annual exams, i.e., μafter > μbefore.

Now, if the p-value of the hypothesis testing comes smaller than the level of significance, say .05, then the null hypothesis can be rejected, and the researcher can conclude that the children will perform better in the annual exams if the weekly examination system would be implemented.

8. In Verifying the Assumption related to Intelligence

Suppose a principle states that the students studying in her school have an IQ level of above average. To support her statement, the researcher may take a sample of around 50 random students from that school. Let’s say the average IQ score of those children is around 110, and the average IQ score of the mean population is 100 with a standard deviation of 15. The hypothesis testing is given as follows,

Null Hypothesis (Ho) : The population mean IQ score of 100 is a general fact, i.e., μ = 100.

Alternative Hypothesis (Ha): The average IQ score of the students is above average, i.e., μ > 100

It’s a one-tailed test as we are aiming for the ‘greater than’ assumption. Let us suppose the alpha level or we can say the significance level, in this case, is 5 per cent, i.e., 0.05, and this corresponds to the Z score equal to 1.645. The Z score is found by the statistical formula given by (112.5 – 100) / (15/√30) = 4.56. Now, the final step is to compare the values of the expected z score and the calculated z score. Here, the calculated Z score is lesser than the expected Z score, hence, the Null Hypothesis is rejected, i.e., the average IQ score of the children belonging to that school is above average.

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15 Hypothesis Examples

hypothesis definition and example, explained below

A hypothesis is defined as a testable prediction , and is used primarily in scientific experiments as a potential or predicted outcome that scientists attempt to prove or disprove (Atkinson et al., 2021; Tan, 2022).

In my types of hypothesis article, I outlined 13 different hypotheses, including the directional hypothesis (which makes a prediction about an effect of a treatment will be positive or negative) and the associative hypothesis (which makes a prediction about the association between two variables).

This article will dive into some interesting examples of hypotheses and examine potential ways you might test each one.

Hypothesis Examples

1. “inadequate sleep decreases memory retention”.

Field: Psychology

Type: Causal Hypothesis A causal hypothesis explores the effect of one variable on another. This example posits that a lack of adequate sleep causes decreased memory retention. In other words, if you are not getting enough sleep, your ability to remember and recall information may suffer.

How to Test:

To test this hypothesis, you might devise an experiment whereby your participants are divided into two groups: one receives an average of 8 hours of sleep per night for a week, while the other gets less than the recommended sleep amount.

During this time, all participants would daily study and recall new, specific information. You’d then measure memory retention of this information for both groups using standard memory tests and compare the results.

Should the group with less sleep have statistically significant poorer memory scores, the hypothesis would be supported.

Ensuring the integrity of the experiment requires taking into account factors such as individual health differences, stress levels, and daily nutrition.

Relevant Study: Sleep loss, learning capacity and academic performance (Curcio, Ferrara & De Gennaro, 2006)

2. “Increase in Temperature Leads to Increase in Kinetic Energy”

Field: Physics

Type: Deductive Hypothesis The deductive hypothesis applies the logic of deductive reasoning – it moves from a general premise to a more specific conclusion. This specific hypothesis assumes that as temperature increases, the kinetic energy of particles also increases – that is, when you heat something up, its particles move around more rapidly.

This hypothesis could be examined by heating a gas in a controlled environment and capturing the movement of its particles as a function of temperature.

You’d gradually increase the temperature and measure the kinetic energy of the gas particles with each increment. If the kinetic energy consistently rises with the temperature, your hypothesis gets supporting evidence.

Variables such as pressure and volume of the gas would need to be held constant to ensure validity of results.

3. “Children Raised in Bilingual Homes Develop Better Cognitive Skills”

Field: Psychology/Linguistics

Type: Comparative Hypothesis The comparative hypothesis posits a difference between two or more groups based on certain variables. In this context, you might propose that children raised in bilingual homes have superior cognitive skills compared to those raised in monolingual homes.

Testing this hypothesis could involve identifying two groups of children: those raised in bilingual homes, and those raised in monolingual homes.

Cognitive skills in both groups would be evaluated using a standard cognitive ability test at different stages of development. The examination would be repeated over a significant time period for consistency.

If the group raised in bilingual homes persistently scores higher than the other, the hypothesis would thereby be supported.

The challenge for the researcher would be controlling for other variables that could impact cognitive development, such as socio-economic status, education level of parents, and parenting styles.

Relevant Study: The cognitive benefits of being bilingual (Marian & Shook, 2012)

4. “High-Fiber Diet Leads to Lower Incidences of Cardiovascular Diseases”

Field: Medicine/Nutrition

Type: Alternative Hypothesis The alternative hypothesis suggests an alternative to a null hypothesis. In this context, the implied null hypothesis could be that diet has no effect on cardiovascular health, which the alternative hypothesis contradicts by suggesting that a high-fiber diet leads to fewer instances of cardiovascular diseases.

To test this hypothesis, a longitudinal study could be conducted on two groups of participants; one adheres to a high-fiber diet, while the other follows a diet low in fiber.

After a fixed period, the cardiovascular health of participants in both groups could be analyzed and compared. If the group following a high-fiber diet has a lower number of recorded cases of cardiovascular diseases, it would provide evidence supporting the hypothesis.

Control measures should be implemented to exclude the influence of other lifestyle and genetic factors that contribute to cardiovascular health.

Relevant Study: Dietary fiber, inflammation, and cardiovascular disease (King, 2005)

5. “Gravity Influences the Directional Growth of Plants”

Field: Agronomy / Botany

Type: Explanatory Hypothesis An explanatory hypothesis attempts to explain a phenomenon. In this case, the hypothesis proposes that gravity affects how plants direct their growth – both above-ground (toward sunlight) and below-ground (towards water and other resources).

The testing could be conducted by growing plants in a rotating cylinder to create artificial gravity.

Observations on the direction of growth, over a specified period, can provide insights into the influencing factors. If plants consistently direct their growth in a manner that indicates the influence of gravitational pull, the hypothesis is substantiated.

It is crucial to ensure that other growth-influencing factors, such as light and water, are uniformly distributed so that only gravity influences the directional growth.

6. “The Implementation of Gamified Learning Improves Students’ Motivation”

Field: Education

Type: Relational Hypothesis The relational hypothesis describes the relation between two variables. Here, the hypothesis is that the implementation of gamified learning has a positive effect on the motivation of students.

To validate this proposition, two sets of classes could be compared: one that implements a learning approach with game-based elements, and another that follows a traditional learning approach.

The students’ motivation levels could be gauged by monitoring their engagement, performance, and feedback over a considerable timeframe.

If the students engaged in the gamified learning context present higher levels of motivation and achievement, the hypothesis would be supported.

Control measures ought to be put into place to account for individual differences, including prior knowledge and attitudes towards learning.

Relevant Study: Does educational gamification improve students’ motivation? (Chapman & Rich, 2018)

7. “Mathematics Anxiety Negatively Affects Performance”

Field: Educational Psychology

Type: Research Hypothesis The research hypothesis involves making a prediction that will be tested. In this case, the hypothesis proposes that a student’s anxiety about math can negatively influence their performance in math-related tasks.

To assess this hypothesis, researchers must first measure the mathematics anxiety levels of a sample of students using a validated instrument, such as the Mathematics Anxiety Rating Scale.

Then, the students’ performance in mathematics would be evaluated through standard testing. If there’s a negative correlation between the levels of math anxiety and math performance (meaning as anxiety increases, performance decreases), the hypothesis would be supported.

It would be crucial to control for relevant factors such as overall academic performance and previous mathematical achievement.

8. “Disruption of Natural Sleep Cycle Impairs Worker Productivity”

Field: Organizational Psychology

Type: Operational Hypothesis The operational hypothesis involves defining the variables in measurable terms. In this example, the hypothesis posits that disrupting the natural sleep cycle, for instance through shift work or irregular working hours, can lessen productivity among workers.

To test this hypothesis, you could collect data from workers who maintain regular working hours and those with irregular schedules.

Measuring productivity could involve examining the worker’s ability to complete tasks, the quality of their work, and their efficiency.

If workers with interrupted sleep cycles demonstrate lower productivity compared to those with regular sleep patterns, it would lend support to the hypothesis.

Consideration should be given to potential confounding variables such as job type, worker age, and overall health.

9. “Regular Physical Activity Reduces the Risk of Depression”

Field: Health Psychology

Type: Predictive Hypothesis A predictive hypothesis involves making a prediction about the outcome of a study based on the observed relationship between variables. In this case, it is hypothesized that individuals who engage in regular physical activity are less likely to suffer from depression.

Longitudinal studies would suit to test this hypothesis, tracking participants’ levels of physical activity and their mental health status over time.

The level of physical activity could be self-reported or monitored, while mental health status could be assessed using standard diagnostic tools or surveys.

If data analysis shows that participants maintaining regular physical activity have a lower incidence of depression, this would endorse the hypothesis.

However, care should be taken to control other lifestyle and behavioral factors that could intervene with the results.

Relevant Study: Regular physical exercise and its association with depression (Kim, 2022)

10. “Regular Meditation Enhances Emotional Stability”

Type: Empirical Hypothesis In the empirical hypothesis, predictions are based on amassed empirical evidence . This particular hypothesis theorizes that frequent meditation leads to improved emotional stability, resonating with numerous studies linking meditation to a variety of psychological benefits.

Earlier studies reported some correlations, but to test this hypothesis directly, you’d organize an experiment where one group meditates regularly over a set period while a control group doesn’t.

Both groups’ emotional stability levels would be measured at the start and end of the experiment using a validated emotional stability assessment.

If regular meditators display noticeable improvements in emotional stability compared to the control group, the hypothesis gains credit.

You’d have to ensure a similar emotional baseline for all participants at the start to avoid skewed results.

11. “Children Exposed to Reading at an Early Age Show Superior Academic Progress”

Type: Directional Hypothesis The directional hypothesis predicts the direction of an expected relationship between variables. Here, the hypothesis anticipates that early exposure to reading positively affects a child’s academic advancement.

A longitudinal study tracking children’s reading habits from an early age and their consequent academic performance could validate this hypothesis.

Parents could report their children’s exposure to reading at home, while standardized school exam results would provide a measure of academic achievement.

If the children exposed to early reading consistently perform better acadically, it gives weight to the hypothesis.

However, it would be important to control for variables that might impact academic performance, such as socioeconomic background, parental education level, and school quality.

12. “Adopting Energy-efficient Technologies Reduces Carbon Footprint of Industries”

Field: Environmental Science

Type: Descriptive Hypothesis A descriptive hypothesis predicts the existence of an association or pattern related to variables. In this scenario, the hypothesis suggests that industries adopting energy-efficient technologies will resultantly show a reduced carbon footprint.

Global industries making use of energy-efficient technologies could track their carbon emissions over time. At the same time, others not implementing such technologies continue their regular tracking.

After a defined time, the carbon emission data of both groups could be compared. If industries that adopted energy-efficient technologies demonstrate a notable reduction in their carbon footprints, the hypothesis would hold strong.

In the experiment, you would exclude variations brought by factors such as industry type, size, and location.

13. “Reduced Screen Time Improves Sleep Quality”

Type: Simple Hypothesis The simple hypothesis is a prediction about the relationship between two variables, excluding any other variables from consideration. This example posits that by reducing time spent on devices like smartphones and computers, an individual should experience improved sleep quality.

A sample group would need to reduce their daily screen time for a pre-determined period. Sleep quality before and after the reduction could be measured using self-report sleep diaries and objective measures like actigraphy, monitoring movement and wakefulness during sleep.

If the data shows that sleep quality improved post the screen time reduction, the hypothesis would be validated.

Other aspects affecting sleep quality, like caffeine intake, should be controlled during the experiment.

Relevant Study: Screen time use impacts low‐income preschool children’s sleep quality, tiredness, and ability to fall asleep (Waller et al., 2021)

14. Engaging in Brain-Training Games Improves Cognitive Functioning in Elderly

Field: Gerontology

Type: Inductive Hypothesis Inductive hypotheses are based on observations leading to broader generalizations and theories. In this context, the hypothesis deduces from observed instances that engaging in brain-training games can help improve cognitive functioning in the elderly.

A longitudinal study could be conducted where an experimental group of elderly people partakes in regular brain-training games.

Their cognitive functioning could be assessed at the start of the study and at regular intervals using standard neuropsychological tests.

If the group engaging in brain-training games shows better cognitive functioning scores over time compared to a control group not playing these games, the hypothesis would be supported.

15. Farming Practices Influence Soil Erosion Rates

Type: Null Hypothesis A null hypothesis is a negative statement assuming no relationship or difference between variables. The hypothesis in this context asserts there’s no effect of different farming practices on the rates of soil erosion.

Comparing soil erosion rates in areas with different farming practices over a considerable timeframe could help test this hypothesis.

If, statistically, the farming practices do not lead to differences in soil erosion rates, the null hypothesis is accepted.

However, if marked variation appears, the null hypothesis is rejected, meaning farming practices do influence soil erosion rates. It would be crucial to control for external factors like weather, soil type, and natural vegetation.

The variety of hypotheses mentioned above underscores the diversity of research constructs inherent in different fields, each with its unique purpose and way of testing.

While researchers may develop hypotheses primarily as tools to define and narrow the focus of the study, these hypotheses also serve as valuable guiding forces for the data collection and analysis procedures, making the research process more efficient and direction-focused.

Hypotheses serve as a compass for any form of academic research. The diverse examples provided, from Psychology to Educational Studies, Environmental Science to Gerontology, clearly demonstrate how certain hypotheses suit specific fields more aptly than others.

It is important to underline that although these varied hypotheses differ in their structure and methods of testing, each endorses the fundamental value of empiricism in research. Evidence-based decision making remains at the heart of scholarly inquiry, regardless of the research field, thus aligning all hypotheses to the core purpose of scientific investigation.

Testing hypotheses is an essential part of the scientific method . By doing so, researchers can either confirm their predictions, giving further validity to an existing theory, or they might uncover new insights that could potentially shift the field’s understanding of a particular phenomenon. In either case, hypotheses serve as the stepping stones for scientific exploration and discovery.

Atkinson, P., Delamont, S., Cernat, A., Sakshaug, J. W., & Williams, R. A. (2021). SAGE research methods foundations . SAGE Publications Ltd.

Curcio, G., Ferrara, M., & De Gennaro, L. (2006). Sleep loss, learning capacity and academic performance. Sleep medicine reviews , 10 (5), 323-337.

Kim, J. H. (2022). Regular physical exercise and its association with depression: A population-based study short title: Exercise and depression. Psychiatry Research , 309 , 114406.

King, D. E. (2005). Dietary fiber, inflammation, and cardiovascular disease. Molecular nutrition & food research , 49 (6), 594-600.

Marian, V., & Shook, A. (2012, September). The cognitive benefits of being bilingual. In Cerebrum: the Dana forum on brain science (Vol. 2012). Dana Foundation.

Tan, W. C. K. (2022). Research Methods: A Practical Guide For Students And Researchers (Second Edition) . World Scientific Publishing Company.

Waller, N. A., Zhang, N., Cocci, A. H., D’Agostino, C., Wesolek‐Greenson, S., Wheelock, K., … & Resnicow, K. (2021). Screen time use impacts low‐income preschool children’s sleep quality, tiredness, and ability to fall asleep. Child: care, health and development, 47 (5), 618-626.

Chris Drew (PhD)

Dr. Chris Drew is the founder of the Helpful Professor. He holds a PhD in education and has published over 20 articles in scholarly journals. He is the former editor of the Journal of Learning Development in Higher Education. [Image Descriptor: Photo of Chris]

Chris Drew (PhD) https://helpfulprofessor.com/author/chris-drew-phd/ Social-Emotional Learning (Definition, Examples, Pros & Cons)
Chris Drew (PhD) https://helpfulprofessor.com/author/chris-drew-phd/ What is Educational Psychology?
Chris Drew (PhD) https://helpfulprofessor.com/author/chris-drew-phd/ What is IQ? (Intelligence Quotient)
Chris Drew (PhD) https://helpfulprofessor.com/author/chris-drew-phd/ 5 Top Tips for Succeeding at University

Two Sample t-test: Definition, Formula, and Example

A two sample t-test is used to determine whether or not two population means are equal.

This tutorial explains the following:

The motivation for performing a two sample t-test.
The formula to perform a two sample t-test.
The assumptions that should be met to perform a two sample t-test.
An example of how to perform a two sample t-test.

Two Sample t-test: Motivation

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. Since there are thousands of turtles in each population, it would be too time-consuming and costly to go around and weigh each individual turtle.

Instead, we might take a simple random sample of 15 turtles from each population and use the mean weight in each sample to determine if the mean weight is equal between the two populations:

However, it’s virtually guaranteed that the mean weight between the two samples will be at least a little different. The question is whether or not this difference is statistically significant . Fortunately, a two sample t-test allows us to answer this question.

Two Sample t-test: Formula

A two-sample t-test always uses the following null hypothesis:

H 0 : μ 1 = μ 2 (the two population means are equal)

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

H 1 (two-tailed): μ 1 ≠ μ 2 (the two population means are not equal)
H 1 (left-tailed): μ 1 < μ 2 (population 1 mean is less than population 2 mean)
H 1 (right-tailed): μ 1 > μ 2 (population 1 mean is greater than population 2 mean)

We use the following formula to calculate the test statistic t:

Test statistic: ( x 1 – x 2 ) / s p (√ 1/n 1 + 1/n 2 )

where x 1 and x 2 are the sample means, n 1 and n 2 are the sample sizes, and where s p is calculated as:

s p = √ (n 1 -1)s 1 2 + (n 2 -1)s 2 2 / (n 1 +n 2 -2)

where s 1 2 and s 2 2 are the sample variances.

If the p-value that corresponds to the test statistic t with (n 1 +n 2 -1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

Two Sample t-test: Assumptions

For the results of a two sample t-test to be valid, the following assumptions should be met:

The observations in one sample should be independent of the observations in the other sample.
The data should be approximately normally distributed.
The two samples should have approximately the same variance. If this assumption is not met, you should instead perform Welch’s t-test .
The data in both samples was obtained using a random sampling method .

Two Sample t-test : Example

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. To test this, will perform a two sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose we collect a random sample of turtles from each population with the following information:

Sample size n 1 = 40
Sample mean weight x 1 = 300
Sample standard deviation s 1 = 18.5
Sample size n 2 = 38
Sample mean weight x 2 = 305
Sample standard deviation s 2 = 16.7

Step 2: Define the hypotheses.

We will perform the two sample t-test with the following hypotheses:

H 0 : μ 1 = μ 2 (the two population means are equal)
H 1 : μ 1 ≠ μ 2 (the two population means are not equal)

Step 3: Calculate the test statistic t .

First, we will calculate the pooled standard deviation s p :

s p = √ (n 1 -1)s 1 2 + (n 2 -1)s 2 2 / (n 1 +n 2 -2) = √ (40-1)18.5 2 + (38-1)16.7 2 / (40+38-2) = 17.647

Next, we will calculate the test statistic t :

t = ( x 1 – x 2 ) / s p (√ 1/n 1 + 1/n 2 ) = (300-305) / 17.647(√ 1/40 + 1/38 ) = -1.2508

Step 4: Calculate the p-value of the test statistic t .

According to the T Score to P Value Calculator , the p-value associated with t = -1.2508 and degrees of freedom = n 1 +n 2 -2 = 40+38-2 = 76 is 0.21484 .

Step 5: Draw a conclusion.

Since this p-value is not less than our significance level α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the mean weight of turtles between these two populations is different.

Note: You can also perform this entire two sample t-test by simply using the Two Sample t-test Calculator .

Additional Resources

The following tutorials explain how to perform a two-sample t-test using different statistical programs:

How to Perform a Two Sample t-test in Excel How to Perform a Two Sample t-test in SPSS How to Perform a Two Sample t-test in Stata How to Perform a Two Sample t-test in R How to Perform a Two Sample t-test in Python How to Perform a Two Sample t-test on a TI-84 Calculator

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Statistics Cheat Sheets to Get Before Your Job Interview

Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “Two Sample t-test: Definition, Formula, and Example”

I like the detailed information and simplified in the way I can understand and relate easily. Thank you

It seems a couple of parenthesis is missed at the pooled standard deviation formula. Under square root you have (n1-1)s12 + (n2-1)s22 / (n1+n2-2) but it should be [(n1-1)s12 + (n2-1)s22] / (n1+n2-2) I used square bracket

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What Is Hypothesis Testing?

Step 1: define the hypothesis, step 2: set the criteria, step 3: calculate the statistic, step 4: reach a conclusion, types of errors, the bottom line.

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Hypothesis Testing in Finance: Concept and Examples

Charlene Rhinehart is a CPA , CFE, chair of an Illinois CPA Society committee, and has a degree in accounting and finance from DePaul University.

Your investment advisor proposes you a monthly income investment plan that promises a variable return each month. You will invest in it only if you are assured of an average $180 monthly income. Your advisor also tells you that for the past 300 months, the scheme had investment returns with an average value of $190 and a standard deviation of $75. Should you invest in this scheme? Hypothesis testing comes to the aid for such decision-making.

Key Takeaways

Hypothesis testing is a mathematical tool for confirming a financial or business claim or idea.
Hypothesis testing is useful for investors trying to decide what to invest in and whether the instrument is likely to provide a satisfactory return.
Despite the existence of different methodologies of hypothesis testing, the same four steps are used: define the hypothesis, set the criteria, calculate the statistic, and reach a conclusion.
This mathematical model, like most statistical tools and models, has limitations and is prone to certain errors, necessitating investors also considering other models in conjunction with this one

Hypothesis or significance testing is a mathematical model for testing a claim, idea or hypothesis about a parameter of interest in a given population set, using data measured in a sample set. Calculations are performed on selected samples to gather more decisive information about the characteristics of the entire population, which enables a systematic way to test claims or ideas about the entire dataset.

Here is a simple example: A school principal reports that students in their school score an average of 7 out of 10 in exams. To test this “hypothesis,” we record marks of say 30 students (sample) from the entire student population of the school (say 300) and calculate the mean of that sample. We can then compare the (calculated) sample mean to the (reported) population mean and attempt to confirm the hypothesis.

To take another example, the annual return of a particular mutual fund is 8%. Assume that mutual fund has been in existence for 20 years. We take a random sample of annual returns of the mutual fund for, say, five years (sample) and calculate its mean. We then compare the (calculated) sample mean to the (claimed) population mean to verify the hypothesis.

This article assumes readers' familiarity with concepts of a normal distribution table, formula, p-value and related basics of statistics.

Different methodologies exist for hypothesis testing, but the same four basic steps are involved:

Usually, the reported value (or the claim statistics) is stated as the hypothesis and presumed to be true. For the above examples, the hypothesis will be:

Example A: Students in the school score an average of 7 out of 10 in exams.
Example B: The annual return of the mutual fund is 8% per annum.

This stated description constitutes the “ Null Hypothesis (H 0 ) ” and is assumed to be true – the way a defendant in a jury trial is presumed innocent until proven guilty by the evidence presented in court. Similarly, hypothesis testing starts by stating and assuming a “ null hypothesis ,” and then the process determines whether the assumption is likely to be true or false.

The important point to note is that we are testing the null hypothesis because there is an element of doubt about its validity. Whatever information that is against the stated null hypothesis is captured in the Alternative Hypothesis (H 1 ). For the above examples, the alternative hypothesis will be:

Students score an average that is not equal to 7.
The annual return of the mutual fund is not equal to 8% per annum.

In other words, the alternative hypothesis is a direct contradiction of the null hypothesis.

As in a trial, the jury assumes the defendant's innocence (null hypothesis). The prosecutor has to prove otherwise (alternative hypothesis). Similarly, the researcher has to prove that the null hypothesis is either true or false. If the prosecutor fails to prove the alternative hypothesis, the jury has to let the defendant go (basing the decision on the null hypothesis). Similarly, if the researcher fails to prove an alternative hypothesis (or simply does nothing), then the null hypothesis is assumed to be true.

The decision-making criteria have to be based on certain parameters of datasets.

The decision-making criteria have to be based on certain parameters of datasets and this is where the connection to normal distribution comes into the picture.

As per the standard statistics postulate about sampling distribution , “For any sample size n, the sampling distribution of X̅ is normal if the population X from which the sample is drawn is normally distributed.” Hence, the probabilities of all other possible sample mean that one could select are normally distributed.

For e.g., determine if the average daily return, of any stock listed on XYZ stock market , around New Year's Day is greater than 2%.

H 0 : Null Hypothesis: mean = 2%

H 1 : Alternative Hypothesis: mean > 2% (this is what we want to prove)

Take the sample (say of 50 stocks out of total 500) and compute the mean of the sample.

For a normal distribution, 95% of the values lie within two standard deviations of the population mean. Hence, this normal distribution and central limit assumption for the sample dataset allows us to establish 5% as a significance level. It makes sense as, under this assumption, there is less than a 5% probability (100-95) of getting outliers that are beyond two standard deviations from the population mean. Depending upon the nature of datasets, other significance levels can be taken at 1%, 5% or 10%. For financial calculations (including behavioral finance), 5% is the generally accepted limit. If we find any calculations that go beyond the usual two standard deviations, then we have a strong case of outliers to reject the null hypothesis.

Graphically, it is represented as follows:

In the above example, if the mean of the sample is much larger than 2% (say 3.5%), then we reject the null hypothesis. The alternative hypothesis (mean >2%) is accepted, which confirms that the average daily return of the stocks is indeed above 2%.

However, if the mean of the sample is not likely to be significantly greater than 2% (and remains at, say, around 2.2%), then we CANNOT reject the null hypothesis. The challenge comes on how to decide on such close range cases. To make a conclusion from selected samples and results, a level of significance is to be determined, which enables a conclusion to be made about the null hypothesis. The alternative hypothesis enables establishing the level of significance or the "critical value” concept for deciding on such close range cases.

According to the textbook standard definition , “A critical value is a cutoff value that defines the boundaries beyond which less than 5% of sample means can be obtained if the null hypothesis is true. Sample means obtained beyond a critical value will result in a decision to reject the null hypothesis." In the above example, if we have defined the critical value as 2.1%, and the calculated mean comes to 2.2%, then we reject the null hypothesis. A critical value establishes a clear demarcation about acceptance or rejection.

This step involves calculating the required figure(s), known as test statistics (like mean, z-score , p-value , etc.), for the selected sample. (We'll get to these in a later section.)

With the computed value(s), decide on the null hypothesis. If the probability of getting a sample mean is less than 5%, then the conclusion is to reject the null hypothesis. Otherwise, accept and retain the null hypothesis.

There can be four possible outcomes in sample-based decision-making, with regard to the correct applicability to the entire population:

The “Correct” cases are the ones where the decisions taken on the samples are truly applicable to the entire population. The cases of errors arise when one decides to retain (or reject) the null hypothesis based on the sample calculations, but that decision does not really apply for the entire population. These cases constitute Type 1 ( alpha ) and Type 2 ( beta ) errors, as indicated in the table above.

Selecting the correct critical value allows eliminating the type-1 alpha errors or limiting them to an acceptable range.

Alpha denotes the error on the level of significance and is determined by the researcher. To maintain the standard 5% significance or confidence level for probability calculations, this is retained at 5%.

According to the applicable decision-making benchmarks and definitions:

“This (alpha) criterion is usually set at 0.05 (a = 0.05), and we compare the alpha level to the p-value. When the probability of a Type I error is less than 5% (p < 0.05), we decide to reject the null hypothesis; otherwise, we retain the null hypothesis.”
The technical term used for this probability is the p-value . It is defined as “the probability of obtaining a sample outcome, given that the value stated in the null hypothesis is true. The p-value for obtaining a sample outcome is compared to the level of significance."
A Type II error, or beta error, is defined as the probability of incorrectly retaining the null hypothesis, when in fact it is not applicable to the entire population.

A few more examples will demonstrate this and other calculations.

A monthly income investment scheme exists that promises variable monthly returns. An investor will invest in it only if they are assured of an average $180 monthly income. The investor has a sample of 300 months’ returns which has a mean of $190 and a standard deviation of $75. Should they invest in this scheme?

Let’s set up the problem. The investor will invest in the scheme if they are assured of the investor's desired $180 average return.

H 0 : Null Hypothesis: mean = 180

H 1 : Alternative Hypothesis: mean > 180

Method 1: Critical Value Approach

Identify a critical value X L for the sample mean, which is large enough to reject the null hypothesis – i.e. reject the null hypothesis if the sample mean >= critical value X L

P (identify a Type I alpha error) = P(reject H 0 given that H 0 is true),

This would be achieved when the sample mean exceeds the critical limits.

= P (given that H 0 is true) = alpha

Graphically, it appears as follows:

Taking alpha = 0.05 (i.e. 5% significance level), Z 0.05 = 1.645 (from the Z-table or normal distribution table)

= > X L = 180 +1.645*(75/sqrt(300)) = 187.12

Since the sample mean (190) is greater than the critical value (187.12), the null hypothesis is rejected, and the conclusion is that the average monthly return is indeed greater than $180, so the investor can consider investing in this scheme.

Method 2: Using Standardized Test Statistics

One can also use standardized value z.

Test Statistic, Z = (sample mean – population mean) / (std-dev / sqrt (no. of samples).

Then, the rejection region becomes the following:

Z= (190 – 180) / (75 / sqrt (300)) = 2.309

Our rejection region at 5% significance level is Z> Z 0.05 = 1.645.

Since Z= 2.309 is greater than 1.645, the null hypothesis can be rejected with a similar conclusion mentioned above.

Method 3: P-value Calculation

We aim to identify P (sample mean >= 190, when mean = 180).

= P (Z >= (190- 180) / (75 / sqrt (300))

= P (Z >= 2.309) = 0.0084 = 0.84%

The following table to infer p-value calculations concludes that there is confirmed evidence of average monthly returns being higher than 180:

A new stockbroker (XYZ) claims that their brokerage fees are lower than that of your current stock broker's (ABC). Data available from an independent research firm indicates that the mean and std-dev of all ABC broker clients are $18 and $6, respectively.

A sample of 100 clients of ABC is taken and brokerage charges are calculated with the new rates of XYZ broker. If the mean of the sample is $18.75 and std-dev is the same ($6), can any inference be made about the difference in the average brokerage bill between ABC and XYZ broker?

H 0 : Null Hypothesis: mean = 18

H 1 : Alternative Hypothesis: mean <> 18 (This is what we want to prove.)

Rejection region: Z <= - Z 2.5 and Z>=Z 2.5 (assuming 5% significance level, split 2.5 each on either side).

Z = (sample mean – mean) / (std-dev / sqrt (no. of samples))

= (18.75 – 18) / (6/(sqrt(100)) = 1.25

This calculated Z value falls between the two limits defined by:

- Z 2.5 = -1.96 and Z 2.5 = 1.96.

This concludes that there is insufficient evidence to infer that there is any difference between the rates of your existing broker and the new broker.

Alternatively, The p-value = P(Z< -1.25)+P(Z >1.25)

= 2 * 0.1056 = 0.2112 = 21.12% which is greater than 0.05 or 5%, leading to the same conclusion.

Graphically, it is represented by the following:

Criticism Points for the Hypothetical Testing Method:

A statistical method based on assumptions
Error-prone as detailed in terms of alpha and beta errors
Interpretation of p-value can be ambiguous, leading to confusing results

Hypothesis testing allows a mathematical model to validate a claim or idea with a certain confidence level. However, like the majority of statistical tools and models, it is bound by a few limitations. The use of this model for making financial decisions should be considered with a critical eye, keeping all dependencies in mind. Alternate methods like Bayesian Inference are also worth exploring for similar analysis.

Sage Publications. " Introduction to Hypothesis Testing ," Page 13.

Sage Publications. " Introduction to Hypothesis Testing ," Page 11.

Sage Publications. " Introduction to Hypothesis Testing ," Page 7.

Sage Publications. " Introduction to Hypothesis Testing ," Pages 10-11.

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Hypothesis Testing Solved Examples(Questions and Solutions)

Here is a list hypothesis testing exercises and solutions. Try to solve a question by yourself first before you look at the solution.

Question 1 In the population, the average IQ is 100 with a standard deviation of 15. A team of scientists want to test a new medication to see if it has either a positive or negative effect on intelligence, or not effect at all. A sample of 30 participants who have taken the medication has a mean of 140. Did the medication affect intelligence? View Solution to Question 1

A professor wants to know if her introductory statistics class has a good grasp of basic math. Six students are chosen at random from the class and given a math proficiency test. The professor wants the class to be able to score above 70 on the test. The six students get the following scores:62, 92, 75, 68, 83, 95. Can the professor have 90% confidence that the mean score for the class on the test would be above 70. Solution to Question 2

Question 3 In a packaging plant, a machine packs cartons with jars. It is supposed that a new machine would pack faster on the average than the machine currently used. To test the hypothesis, the time it takes each machine to pack ten cartons are recorded. The result in seconds is as follows.

Do the data provide sufficient evidence to conclude that, on the average, the new machine packs faster? Perform the required hypothesis test at the 5% level of significance. Solution to Question 3

Question 4 We want to compare the heights in inches of two groups of individuals. Here are the measurements: X: 175, 168, 168, 190, 156, 181, 182, 175, 174, 179 Y: 120, 180, 125, 188, 130, 190, 110, 185, 112, 188 Solution to Question 4

Question 5 A clinic provides a program to help their clients lose weight and asks a consumer agency to investigate the effectiveness of the program. The agency takes a sample of 15 people, weighing each person in the sample before the program begins and 3 months later. The results a tabulated below

Determine is the program is effective. Solution to Question 5

Question 6 A sample of 20 students were selected and given a diagnostic module prior to studying for a test. And then they were given the test again after completing the module. . The result of the students scores in the test before and after the test is tabulated below.

We want to see if there is significant improvement in the student’s performance due to this teaching method Solution to Question 6

Question 7 A study was performed to test wether cars get better mileage on premium gas than on regular gas. Each of 10 cars was first filled with regular or premium gas, decided by a coin toss, and the mileage for the tank was recorded. The mileage was recorded again for the same cars using other kind of gasoline. Determine wether cars get significantly better mileage with premium gas.

Mileage with regular gas: 16,20,21,22,23,22,27,25,27,28 Mileage with premium gas: 19, 22,24,24,25,25,26,26,28,32 Solution to Question 7

Question 8 An automatic cutter machine must cut steel strips of 1200 mm length. From a preliminary data, we checked that the lengths of the pieces produced by the machine can be considered as normal random variables with a 3mm standard deviation. We want to make sure that the machine is set correctly. Therefore 16 pieces of the products are randomly selected and weight. The figures were in mm: 1193,1196,1198,1195,1198,1199,1204,1193,1203,1201,1196,1200,1191,1196,1198,1191 Examine wether there is any significant deviation from the required size Solution to Question 8

Question 9 Blood pressure reading of ten patients before and after medication for reducing the blood pressure are as follows

Patient: 1,2,3,4,5,6,7,8,9,10 Before treatment: 86,84,78,90,92,77,89,90,90,86 After treatment: 80,80,92,79,92,82,88,89,92,83

Test the null hypothesis of no effect agains the alternate hypothesis that medication is effective. Execute it with Wilcoxon test Solution to Question 9

Question on ANOVA Sussan Sound predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound or no sound at all. She randomly divides 24 students into three groups of 8 each. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with nose that changes volume periodically. Those in group 3 study with no sound at all. After studying, all students take a 10 point multiple choice test over the material. Their scores are tabulated below.

Group1: Constant sound: 7,4,6,8,6,6,2,9 Group 2: Random sound: 5,5,3,4,4,7,2,2 Group 3: No sound at all: 2,4,7,1,2,1,5,5 Solution to Question 10

Question 11 Using the following three groups of data, perform a one-way analysis of variance using α = 0.05.

Solution to Question 11

Question 12 In a packaging plant, a machine packs cartons with jars. It is supposed that a new machine would pack faster on the average than the machine currently used. To test the hypothesis, the time it takes each machine to pack ten cartons are recorded. The result in seconds is as follows.

New Machine: 42,41,41.3,41.8,42.4,42.8,43.2,42.3,41.8,42.7 Old Machine: 42.7,43.6,43.8,43.3,42.5,43.5,43.1,41.7,44,44.1

Perform an F-test to determine if the null hypothesis should be accepted. Solution to Question 12

Question 13 A random sample 500 U.S adults are questioned about their political affiliation and opinion on a tax reform bill. We need to test if the political affiliation and their opinon on a tax reform bill are dependent, at 5% level of significance. The observed contingency table is given below.

Solution to Question 13

Question 14 Can a dice be considered regular which is showing the following frequency distribution during 1000 throws?

Solution to Question 14

Solution to Question 15

Question 16 A newly developed muesli contains five types of seeds (A, B, C, D and E). The percentage of which is 35%, 25%, 20%, 10% and 10% according to the product information. In a randomly selected muesli, the following volume distribution was found.

Lets us decide about the null hypothesis whether the composition of the sample corresponds to the distribution indicated on the packaging at alpha = 0.1 significance level. Solution to Question 16

Question 17 A research team investigated whether there was any significant correlation between the severity of a certain disease runoff and the age of the patients. During the study, data for n = 200 patients were collected and grouped according to the severity of the disease and the age of the patient. The table below shows the result

Let us decided about the correlation between the age of the patients and the severity of disease progression. Solution to Question 17

Question 18 A publisher is interested in determine which of three book cover is most attractive. He interviews 400 people in each of the three states (California, Illinois and New York), and asks each person which of the cover he or she prefers. The number of preference for each cover is as follows:

Do these data indicate that there are regional differences in people’s preferences concerning these covers? Use the 0.05 level of significance. Solution to Question 18

Question 19 Trees planted along the road were checked for which ones are healthy(H) or diseased (D) and the following arrangement of the trees were obtained:

H H H H D D D H H H H H H H D D H H D D D

Test at the = 0.05 significance wether this arrangement may be regarded as random

Solution to Question 19

Question 20 Suppose we flip a coin n = 15 times and come up with the following arrangements

H T T T H H T T T T H H T H H

(H = head, T = tail)

Test at the alpha = 0.05 significance level whether this arrangement may be regarded as random.

Solution to Question 20

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Hypothesis Testing in Business: Examples

hypothesis testing for business - examples

Are you a product manager or data scientist looking for ways to identify and use most appropriate hypothesis testing for understanding business problems and creating solutions for data-driven decision making? Hypothesis testing is a powerful statistical technique that can help you understand problems during exploratory data analysis (EDA) and identify most appropriate hypotheses / analytical solution. In this blog, we will discuss hypothesis testing with examples from business. We’ll also give you tips on how to use it effectively in your own problem-solving journey. With this knowledge, you’ll be able to confidently create hypotheses, run experiments, and analyze the results to derive meaningful conclusions. So let’s get started!

Before going any further, you may want to check out my detailed blog on hypothesis testing – Hypothesis testing steps & examples .

The picture below represents the key steps you can take to identify appropriate hypothesis tests related to your business problem you are trying to solve.

Table of Contents

Business Objective / Problem Analysis to Asking Key Questions

Here are the steps which you can use to come up with hypothesis tests related to your business problems. You can then use data to perform hypothesis tests and arrive at different conclusions or inferences.

Setting / Identifying business objective : First & foremost, you need to have a business objective which you want to achieve. For example, achieve an increase of 10% revenue in the year ahead.
Identifying key business divisions / units and products & services : Second step is to identify key departments / divisions and related products & services which can help achieve the business objective. For current example, sales can be increased by increase in sales of products and services. For service based companies, it can be increase in sales of existing services and one or more new services. For products based companies, it could be increase in sales of different products.
Identify key personas / stakeholders : For each business division / department, identify key personas or stakeholders who could be accountable for contributing to achievement of business objective. For current example, it could personas / stakeholders who would own the increase in sales of products and / or services.
Are the sales of product A, B and C different?
Are the sales of product A, B and C similar across all the regions, countries, states, etc.?
Are there differences between products and competitors’ products vis-a-vis sales?
Are there any differences between customer queries / complaints across different products (A, B, C)?
Are there any differences between product usage patterns across different products, and for each product?
Are there differences between marketing initiatives run for different products?
Are there differences between teams working on different products?

Hypothesis formulation

Once the questions have been asked / raised, you can create hypotheses from these questions in order to arrive at the answers based on data analysis and create strategy / action plan. Lets take a look at one of the question and how you can formulate hypothesis and perform hypothesis testing. We will also talk about data and analytics aspects.

In order to create strategy around increasing sales revenue, it is very important to understand how has been the sales of different products in past and whether the sales have been different for us to dig deeper into the reasons and create some action plan?

The status quo becomes null hypothesis ([latex]H_0[/latex]. In our current analysis, the status quo is that there is no difference between the sales revenue of different products and that each product is doing equally good and selling well with the customers.

[latex]H_0[/latex]: There is no difference between sales revenue of different products.

The new knowledge for which the null hypothesis can be thrown away can be called as alternate hypothesis, [latex]H_a[/latex]. In current example, the new knowledge or alternate hypothesis is that there is a significant difference between the sales revenue of different products.

[latex]H_a[/latex]: There is a significant difference between sales revenue of different products.

Identifying Test Statistics for Hypothesis Testing

Once the hypothesis has been formulated, the next step is to identify the test statistics which can be used to perform the hypothesis test.

We can perform one-way Anova test to check whether there is a difference between sales based on the product. One-way ANOVA test requires calculation of F-statistics . The factor is product and levels are product A, B and C. Read my blog post on one-way ANOVA test to learn about different aspect of this test. One-Way ANOVA Test: Concepts, Formula & Examples

Apart from Hypothesis test and statistics, one can also set the level of significance based on which one can reject the null hypothesis or otherwise. Generally, it is chosen as 0.05.

Gather Data

Once the hypothesis test and statistics gets chosen, next step is to gather data. You can identify the system which holds the sales data and then gather the data from that system for last 1 year.

Perform Hypothesis Testing

Once the data is gathered, you can use Excel tool or any other statistical packages in Python / R and perform hypothesis testing by doing the following:

Calculating the value of test statistics
Calculate P-value
Comparing the P-value with level of significance
Reject the null hypothesis or otherwise

In conclusion, hypothesis testing is an essential tool for businesses to make data-driven decisions. It involves identifying a problem or question, formulating a hypothesis, identifying the appropriate test statistics, gathering data, and performing hypothesis testing. By following these steps, businesses can gain valuable insights into their operations, identify areas of improvement, and make informed decisions. It is important to note that hypothesis testing is not a one-time process but rather a continuous effort that businesses must undertake to stay ahead of the competition. Examples of hypothesis testing in business can range from identifying the effectiveness of a new marketing campaign to determining the impact of changes in pricing strategies. By analyzing data and performing hypothesis testing, businesses can determine the significance of these changes and make informed decisions that will improve their bottom line.

Ajitesh Kumar

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Hypothesis Testing – A Complete Guide with Examples

What is a Hypothesis and a Hypothesis Testing?

What is Hypothesis Testing?

Characteristics of the Hypothesis to be Tested

What is a Null Hypothesis and Alternative Hypothesis?

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How to Conduct Hypothesis Testing?

Step 1: State the Null and Alternative Hypothesis

Step 2: Data Collection

Step 3: Select Appropriate Statistical Test

One-sided Test

Two-sided Test

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Step 4: Select the Level of Significance

Step 5: Find out Whether the Null Hypothesis is Rejected or Supported

Step 6: Present the Outcomes of your Study

Frequently Asked Questions

What is a hypothesis?

What are null hypothesis?

What is the probability value?

What is p value?

What is a t test?

When to reject null hypothesis?

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Introduction

Null and Alternative Hypotheses

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A Systematic Approach for Testing A Hypothesis

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Hypothesis Testing

What is Hypothesis Testing in Statistics?

Hypothesis Testing Definition

Null Hypothesis

Alternative Hypothesis

Hypothesis Testing P Value

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