null hypothesis for sample means

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Chapter 13: Inferential Statistics

Understanding Null Hypothesis Testing

Learning Objectives

Explain the purpose of null hypothesis testing, including the role of sampling error.
Describe the basic logic of null hypothesis testing.
Describe the role of relationship strength and sample size in determining statistical significance and make reasonable judgments about statistical significance based on these two factors.

The Purpose of Null Hypothesis Testing

As we have seen, psychological research typically involves measuring one or more variables for a sample and computing descriptive statistics for that sample. In general, however, the researcher’s goal is not to draw conclusions about that sample but to draw conclusions about the population that the sample was selected from. Thus researchers must use sample statistics to draw conclusions about the corresponding values in the population. These corresponding values in the population are called parameters . Imagine, for example, that a researcher measures the number of depressive symptoms exhibited by each of 50 clinically depressed adults and computes the mean number of symptoms. The researcher probably wants to use this sample statistic (the mean number of symptoms for the sample) to draw conclusions about the corresponding population parameter (the mean number of symptoms for clinically depressed adults).

Unfortunately, sample statistics are not perfect estimates of their corresponding population parameters. This is because there is a certain amount of random variability in any statistic from sample to sample. The mean number of depressive symptoms might be 8.73 in one sample of clinically depressed adults, 6.45 in a second sample, and 9.44 in a third—even though these samples are selected randomly from the same population. Similarly, the correlation (Pearson’s r ) between two variables might be +.24 in one sample, −.04 in a second sample, and +.15 in a third—again, even though these samples are selected randomly from the same population. This random variability in a statistic from sample to sample is called sampling error . (Note that the term error here refers to random variability and does not imply that anyone has made a mistake. No one “commits a sampling error.”)

One implication of this is that when there is a statistical relationship in a sample, it is not always clear that there is a statistical relationship in the population. A small difference between two group means in a sample might indicate that there is a small difference between the two group means in the population. But it could also be that there is no difference between the means in the population and that the difference in the sample is just a matter of sampling error. Similarly, a Pearson’s r value of −.29 in a sample might mean that there is a negative relationship in the population. But it could also be that there is no relationship in the population and that the relationship in the sample is just a matter of sampling error.

In fact, any statistical relationship in a sample can be interpreted in two ways:

There is a relationship in the population, and the relationship in the sample reflects this.
There is no relationship in the population, and the relationship in the sample reflects only sampling error.

The purpose of null hypothesis testing is simply to help researchers decide between these two interpretations.

The Logic of Null Hypothesis Testing

Null hypothesis testing is a formal approach to deciding between two interpretations of a statistical relationship in a sample. One interpretation is called the null hypothesis (often symbolized H 0 and read as “H-naught”). This is the idea that there is no relationship in the population and that the relationship in the sample reflects only sampling error. Informally, the null hypothesis is that the sample relationship “occurred by chance.” The other interpretation is called the alternative hypothesis (often symbolized as H 1 ). This is the idea that there is a relationship in the population and that the relationship in the sample reflects this relationship in the population.

Again, every statistical relationship in a sample can be interpreted in either of these two ways: It might have occurred by chance, or it might reflect a relationship in the population. So researchers need a way to decide between them. Although there are many specific null hypothesis testing techniques, they are all based on the same general logic. The steps are as follows:

Assume for the moment that the null hypothesis is true. There is no relationship between the variables in the population.
Determine how likely the sample relationship would be if the null hypothesis were true.
If the sample relationship would be extremely unlikely, then reject the null hypothesis in favour of the alternative hypothesis. If it would not be extremely unlikely, then retain the null hypothesis .

Following this logic, we can begin to understand why Mehl and his colleagues concluded that there is no difference in talkativeness between women and men in the population. In essence, they asked the following question: “If there were no difference in the population, how likely is it that we would find a small difference of d = 0.06 in our sample?” Their answer to this question was that this sample relationship would be fairly likely if the null hypothesis were true. Therefore, they retained the null hypothesis—concluding that there is no evidence of a sex difference in the population. We can also see why Kanner and his colleagues concluded that there is a correlation between hassles and symptoms in the population. They asked, “If the null hypothesis were true, how likely is it that we would find a strong correlation of +.60 in our sample?” Their answer to this question was that this sample relationship would be fairly unlikely if the null hypothesis were true. Therefore, they rejected the null hypothesis in favour of the alternative hypothesis—concluding that there is a positive correlation between these variables in the population.

A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the p value . A low p value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A high p value means that the sample result would be likely if the null hypothesis were true and leads to the retention of the null hypothesis. But how low must the p value be before the sample result is considered unlikely enough to reject the null hypothesis? In null hypothesis testing, this criterion is called α (alpha) and is almost always set to .05. If there is less than a 5% chance of a result as extreme as the sample result if the null hypothesis were true, then the null hypothesis is rejected. When this happens, the result is said to be statistically significant . If there is greater than a 5% chance of a result as extreme as the sample result when the null hypothesis is true, then the null hypothesis is retained. This does not necessarily mean that the researcher accepts the null hypothesis as true—only that there is not currently enough evidence to conclude that it is true. Researchers often use the expression “fail to reject the null hypothesis” rather than “retain the null hypothesis,” but they never use the expression “accept the null hypothesis.”

The Misunderstood p Value

The p value is one of the most misunderstood quantities in psychological research (Cohen, 1994) [1] . Even professional researchers misinterpret it, and it is not unusual for such misinterpretations to appear in statistics textbooks!

The most common misinterpretation is that the p value is the probability that the null hypothesis is true—that the sample result occurred by chance. For example, a misguided researcher might say that because the p value is .02, there is only a 2% chance that the result is due to chance and a 98% chance that it reflects a real relationship in the population. But this is incorrect . The p value is really the probability of a result at least as extreme as the sample result if the null hypothesis were true. So a p value of .02 means that if the null hypothesis were true, a sample result this extreme would occur only 2% of the time.

You can avoid this misunderstanding by remembering that the p value is not the probability that any particular hypothesis is true or false. Instead, it is the probability of obtaining the sample result if the null hypothesis were true.

Role of Sample Size and Relationship Strength

Recall that null hypothesis testing involves answering the question, “If the null hypothesis were true, what is the probability of a sample result as extreme as this one?” In other words, “What is the p value?” It can be helpful to see that the answer to this question depends on just two considerations: the strength of the relationship and the size of the sample. Specifically, the stronger the sample relationship and the larger the sample, the less likely the result would be if the null hypothesis were true. That is, the lower the p value. This should make sense. Imagine a study in which a sample of 500 women is compared with a sample of 500 men in terms of some psychological characteristic, and Cohen’s d is a strong 0.50. If there were really no sex difference in the population, then a result this strong based on such a large sample should seem highly unlikely. Now imagine a similar study in which a sample of three women is compared with a sample of three men, and Cohen’s d is a weak 0.10. If there were no sex difference in the population, then a relationship this weak based on such a small sample should seem likely. And this is precisely why the null hypothesis would be rejected in the first example and retained in the second.

Of course, sometimes the result can be weak and the sample large, or the result can be strong and the sample small. In these cases, the two considerations trade off against each other so that a weak result can be statistically significant if the sample is large enough and a strong relationship can be statistically significant even if the sample is small. Table 13.1 shows roughly how relationship strength and sample size combine to determine whether a sample result is statistically significant. The columns of the table represent the three levels of relationship strength: weak, medium, and strong. The rows represent four sample sizes that can be considered small, medium, large, and extra large in the context of psychological research. Thus each cell in the table represents a combination of relationship strength and sample size. If a cell contains the word Yes , then this combination would be statistically significant for both Cohen’s d and Pearson’s r . If it contains the word No , then it would not be statistically significant for either. There is one cell where the decision for d and r would be different and another where it might be different depending on some additional considerations, which are discussed in Section 13.2 “Some Basic Null Hypothesis Tests”

Although Table 13.1 provides only a rough guideline, it shows very clearly that weak relationships based on medium or small samples are never statistically significant and that strong relationships based on medium or larger samples are always statistically significant. If you keep this lesson in mind, you will often know whether a result is statistically significant based on the descriptive statistics alone. It is extremely useful to be able to develop this kind of intuitive judgment. One reason is that it allows you to develop expectations about how your formal null hypothesis tests are going to come out, which in turn allows you to detect problems in your analyses. For example, if your sample relationship is strong and your sample is medium, then you would expect to reject the null hypothesis. If for some reason your formal null hypothesis test indicates otherwise, then you need to double-check your computations and interpretations. A second reason is that the ability to make this kind of intuitive judgment is an indication that you understand the basic logic of this approach in addition to being able to do the computations.

Statistical Significance Versus Practical Significance

Table 13.1 illustrates another extremely important point. A statistically significant result is not necessarily a strong one. Even a very weak result can be statistically significant if it is based on a large enough sample. This is closely related to Janet Shibley Hyde’s argument about sex differences (Hyde, 2007) [2] . The differences between women and men in mathematical problem solving and leadership ability are statistically significant. But the word significant can cause people to interpret these differences as strong and important—perhaps even important enough to influence the college courses they take or even who they vote for. As we have seen, however, these statistically significant differences are actually quite weak—perhaps even “trivial.”

This is why it is important to distinguish between the statistical significance of a result and the practical significance of that result. Practical significance refers to the importance or usefulness of the result in some real-world context. Many sex differences are statistically significant—and may even be interesting for purely scientific reasons—but they are not practically significant. In clinical practice, this same concept is often referred to as “clinical significance.” For example, a study on a new treatment for social phobia might show that it produces a statistically significant positive effect. Yet this effect still might not be strong enough to justify the time, effort, and other costs of putting it into practice—especially if easier and cheaper treatments that work almost as well already exist. Although statistically significant, this result would be said to lack practical or clinical significance.

Key Takeaways

Null hypothesis testing is a formal approach to deciding whether a statistical relationship in a sample reflects a real relationship in the population or is just due to chance.
The logic of null hypothesis testing involves assuming that the null hypothesis is true, finding how likely the sample result would be if this assumption were correct, and then making a decision. If the sample result would be unlikely if the null hypothesis were true, then it is rejected in favour of the alternative hypothesis. If it would not be unlikely, then the null hypothesis is retained.
The probability of obtaining the sample result if the null hypothesis were true (the p value) is based on two considerations: relationship strength and sample size. Reasonable judgments about whether a sample relationship is statistically significant can often be made by quickly considering these two factors.
Statistical significance is not the same as relationship strength or importance. Even weak relationships can be statistically significant if the sample size is large enough. It is important to consider relationship strength and the practical significance of a result in addition to its statistical significance.
Discussion: Imagine a study showing that people who eat more broccoli tend to be happier. Explain for someone who knows nothing about statistics why the researchers would conduct a null hypothesis test.
The correlation between two variables is r = −.78 based on a sample size of 137.
The mean score on a psychological characteristic for women is 25 ( SD = 5) and the mean score for men is 24 ( SD = 5). There were 12 women and 10 men in this study.
In a memory experiment, the mean number of items recalled by the 40 participants in Condition A was 0.50 standard deviations greater than the mean number recalled by the 40 participants in Condition B.
In another memory experiment, the mean scores for participants in Condition A and Condition B came out exactly the same!
A student finds a correlation of r = .04 between the number of units the students in his research methods class are taking and the students’ level of stress.

Long Descriptions

“Null Hypothesis” long description: A comic depicting a man and a woman talking in the foreground. In the background is a child working at a desk. The man says to the woman, “I can’t believe schools are still teaching kids about the null hypothesis. I remember reading a big study that conclusively disproved it years ago.” [Return to “Null Hypothesis”]

“Conditional Risk” long description: A comic depicting two hikers beside a tree during a thunderstorm. A bolt of lightning goes “crack” in the dark sky as thunder booms. One of the hikers says, “Whoa! We should get inside!” The other hiker says, “It’s okay! Lightning only kills about 45 Americans a year, so the chances of dying are only one in 7,000,000. Let’s go on!” The comic’s caption says, “The annual death rate among people who know that statistic is one in six.” [Return to “Conditional Risk”]

Media Attributions

Null Hypothesis by XKCD CC BY-NC (Attribution NonCommercial)
Conditional Risk by XKCD CC BY-NC (Attribution NonCommercial)
Cohen, J. (1994). The world is round: p < .05. American Psychologist, 49 , 997–1003. ↵
Hyde, J. S. (2007). New directions in the study of gender similarities and differences. Current Directions in Psychological Science, 16 , 259–263. ↵

Values in a population that correspond to variables measured in a study.

The random variability in a statistic from sample to sample.

A formal approach to deciding between two interpretations of a statistical relationship in a sample.

The idea that there is no relationship in the population and that the relationship in the sample reflects only sampling error.

The idea that there is a relationship in the population and that the relationship in the sample reflects this relationship in the population.

When the relationship found in the sample would be extremely unlikely, the idea that the relationship occurred “by chance” is rejected.

When the relationship found in the sample is likely to have occurred by chance, the null hypothesis is not rejected.

The probability that, if the null hypothesis were true, the result found in the sample would occur.

How low the p value must be before the sample result is considered unlikely in null hypothesis testing.

When there is less than a 5% chance of a result as extreme as the sample result occurring and the null hypothesis is rejected.

Research Methods in Psychology - 2nd Canadian Edition Copyright © 2015 by Paul C. Price, Rajiv Jhangiani, & I-Chant A. Chiang is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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What is The Null Hypothesis & When Do You Reject The Null Hypothesis

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Editor at Simply Psychology

BA (Hons) Psychology, Princeton University

Julia Simkus is a graduate of Princeton University with a Bachelor of Arts in Psychology. She is currently studying for a Master's Degree in Counseling for Mental Health and Wellness in September 2023. Julia's research has been published in peer reviewed journals.

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On This Page:

A null hypothesis is a statistical concept suggesting no significant difference or relationship between measured variables. It’s the default assumption unless empirical evidence proves otherwise.

The null hypothesis states no relationship exists between the two variables being studied (i.e., one variable does not affect the other).

The null hypothesis is the statement that a researcher or an investigator wants to disprove.

Testing the null hypothesis can tell you whether your results are due to the effects of manipulating the dependent variable or due to random chance.

How to Write a Null Hypothesis

Null hypotheses (H0) start as research questions that the investigator rephrases as statements indicating no effect or relationship between the independent and dependent variables.

It is a default position that your research aims to challenge or confirm.

For example, if studying the impact of exercise on weight loss, your null hypothesis might be:

There is no significant difference in weight loss between individuals who exercise daily and those who do not.

Examples of Null Hypotheses

When do we reject the null hypothesis .

We reject the null hypothesis when the data provide strong enough evidence to conclude that it is likely incorrect. This often occurs when the p-value (probability of observing the data given the null hypothesis is true) is below a predetermined significance level.

If the collected data does not meet the expectation of the null hypothesis, a researcher can conclude that the data lacks sufficient evidence to back up the null hypothesis, and thus the null hypothesis is rejected.

Rejecting the null hypothesis means that a relationship does exist between a set of variables and the effect is statistically significant ( p > 0.05).

If the data collected from the random sample is not statistically significance , then the null hypothesis will be accepted, and the researchers can conclude that there is no relationship between the variables.

You need to perform a statistical test on your data in order to evaluate how consistent it is with the null hypothesis. A p-value is one statistical measurement used to validate a hypothesis against observed data.

Calculating the p-value is a critical part of null-hypothesis significance testing because it quantifies how strongly the sample data contradicts the null hypothesis.

The level of statistical significance is often expressed as a p -value between 0 and 1. The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.

Probability and statistical significance in ab testing. Statistical significance in a b experiments

Usually, a researcher uses a confidence level of 95% or 99% (p-value of 0.05 or 0.01) as general guidelines to decide if you should reject or keep the null.

When your p-value is less than or equal to your significance level, you reject the null hypothesis.

In other words, smaller p-values are taken as stronger evidence against the null hypothesis. Conversely, when the p-value is greater than your significance level, you fail to reject the null hypothesis.

In this case, the sample data provides insufficient data to conclude that the effect exists in the population.

Because you can never know with complete certainty whether there is an effect in the population, your inferences about a population will sometimes be incorrect.

When you incorrectly reject the null hypothesis, it’s called a type I error. When you incorrectly fail to reject it, it’s called a type II error.

Why Do We Never Accept The Null Hypothesis?

The reason we do not say “accept the null” is because we are always assuming the null hypothesis is true and then conducting a study to see if there is evidence against it. And, even if we don’t find evidence against it, a null hypothesis is not accepted.

A lack of evidence only means that you haven’t proven that something exists. It does not prove that something doesn’t exist.

It is risky to conclude that the null hypothesis is true merely because we did not find evidence to reject it. It is always possible that researchers elsewhere have disproved the null hypothesis, so we cannot accept it as true, but instead, we state that we failed to reject the null.

One can either reject the null hypothesis, or fail to reject it, but can never accept it.

Why Do We Use The Null Hypothesis?

We can never prove with 100% certainty that a hypothesis is true; We can only collect evidence that supports a theory. However, testing a hypothesis can set the stage for rejecting or accepting this hypothesis within a certain confidence level.

The null hypothesis is useful because it can tell us whether the results of our study are due to random chance or the manipulation of a variable (with a certain level of confidence).

A null hypothesis is rejected if the measured data is significantly unlikely to have occurred and a null hypothesis is accepted if the observed outcome is consistent with the position held by the null hypothesis.

Rejecting the null hypothesis sets the stage for further experimentation to see if a relationship between two variables exists.

Hypothesis testing is a critical part of the scientific method as it helps decide whether the results of a research study support a particular theory about a given population. Hypothesis testing is a systematic way of backing up researchers’ predictions with statistical analysis.

It helps provide sufficient statistical evidence that either favors or rejects a certain hypothesis about the population parameter.

Purpose of a Null Hypothesis

The primary purpose of the null hypothesis is to disprove an assumption.
Whether rejected or accepted, the null hypothesis can help further progress a theory in many scientific cases.
A null hypothesis can be used to ascertain how consistent the outcomes of multiple studies are.

Do you always need both a Null Hypothesis and an Alternative Hypothesis?

The null (H0) and alternative (Ha or H1) hypotheses are two competing claims that describe the effect of the independent variable on the dependent variable. They are mutually exclusive, which means that only one of the two hypotheses can be true.

While the null hypothesis states that there is no effect in the population, an alternative hypothesis states that there is statistical significance between two variables.

The goal of hypothesis testing is to make inferences about a population based on a sample. In order to undertake hypothesis testing, you must express your research hypothesis as a null and alternative hypothesis. Both hypotheses are required to cover every possible outcome of the study.

What is the difference between a null hypothesis and an alternative hypothesis?

The alternative hypothesis is the complement to the null hypothesis. The null hypothesis states that there is no effect or no relationship between variables, while the alternative hypothesis claims that there is an effect or relationship in the population.

It is the claim that you expect or hope will be true. The null hypothesis and the alternative hypothesis are always mutually exclusive, meaning that only one can be true at a time.

What are some problems with the null hypothesis?

One major problem with the null hypothesis is that researchers typically will assume that accepting the null is a failure of the experiment. However, accepting or rejecting any hypothesis is a positive result. Even if the null is not refuted, the researchers will still learn something new.

Why can a null hypothesis not be accepted?

We can either reject or fail to reject a null hypothesis, but never accept it. If your test fails to detect an effect, this is not proof that the effect doesn’t exist. It just means that your sample did not have enough evidence to conclude that it exists.

We can’t accept a null hypothesis because a lack of evidence does not prove something that does not exist. Instead, we fail to reject it.

Failing to reject the null indicates that the sample did not provide sufficient enough evidence to conclude that an effect exists.

If the p-value is greater than the significance level, then you fail to reject the null hypothesis.

Is a null hypothesis directional or non-directional?

A hypothesis test can either contain an alternative directional hypothesis or a non-directional alternative hypothesis. A directional hypothesis is one that contains the less than (“<“) or greater than (“>”) sign.

A nondirectional hypothesis contains the not equal sign (“≠”). However, a null hypothesis is neither directional nor non-directional.

A null hypothesis is a prediction that there will be no change, relationship, or difference between two variables.

The directional hypothesis or nondirectional hypothesis would then be considered alternative hypotheses to the null hypothesis.

Gill, J. (1999). The insignificance of null hypothesis significance testing. Political research quarterly , 52 (3), 647-674.

Krueger, J. (2001). Null hypothesis significance testing: On the survival of a flawed method. American Psychologist , 56 (1), 16.

Masson, M. E. (2011). A tutorial on a practical Bayesian alternative to null-hypothesis significance testing. Behavior research methods , 43 , 679-690.

Nickerson, R. S. (2000). Null hypothesis significance testing: a review of an old and continuing controversy. Psychological methods , 5 (2), 241.

Rozeboom, W. W. (1960). The fallacy of the null-hypothesis significance test. Psychological bulletin , 57 (5), 416.

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Cross-Cultural Research Methodology In Psychology

What Is Internal Validity In Research?

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Idea behind hypothesis testing

Examples of null and alternative hypotheses

Writing null and alternative hypotheses
P-values and significance tests
Comparing P-values to different significance levels
Estimating a P-value from a simulation
Estimating P-values from simulations
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Video transcript

Module 9: Hypothesis Testing With One Sample

Null and alternative hypotheses, learning outcomes.

Describe hypothesis testing in general and in practice

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

H 0 : The null hypothesis: It is a statement about the population that either is believed to be true or is used to put forth an argument unless it can be shown to be incorrect beyond a reasonable doubt.

H a : The alternative hypothesis : It is a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

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.

After you have determined which hypothesis the sample supports, you make adecision. There are two options for a decision . They are “reject H 0 ” if the sample information favors the alternative hypothesis or “do not reject H 0 ” or “decline to reject H 0 ” if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ 30

H a : More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.

H 0 : The drug reduces cholesterol by 25%. p = 0.25

H a : The drug does not reduce cholesterol by 25%. p ≠ 0.25

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:

H 0 : μ = 2.0

H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 66 H a : μ __ 66

H 0 : μ = 66
H a : μ ≠ 66

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:

H 0 : μ ≥ 5

H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 45 H a : μ __ 45

H 0 : μ ≥ 45
H a : μ < 45

In an issue of U.S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.

H 0 : p ≤ 0.066

H a : p > 0.066

On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : p __ 0.40 H a : p __ 0.40

H 0 : p = 0.40
H a : p > 0.40

Concept Review

In a hypothesis test , sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis , typically denoted with H 0 . The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤ or ≥) Always write the alternative hypothesis , typically denoted with H a or H 1 , using less than, greater than, or not equals symbols, i.e., (≠, >, or <). If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties.

Formula Review

H 0 and H a are contradictory.

OpenStax, Statistics, Null and Alternative Hypotheses. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:58/Introductory_Statistics . License : CC BY: Attribution
Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
Simple hypothesis testing | Probability and Statistics | Khan Academy. Authored by : Khan Academy. Located at : https://youtu.be/5D1gV37bKXY . License : All Rights Reserved . License Terms : Standard YouTube License

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Hypothesis Testing with One Sample

Null and Alternative Hypotheses

OpenStaxCollege

[latexpage]

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

H a : The alternative hypothesis: It is a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are “reject H 0 ” if the sample information favors the alternative hypothesis or “do not reject H 0 ” or “decline to reject H 0 ” if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.

H 0 : The drug reduces cholesterol by 25%. p = 0.25

H a : The drug does not reduce cholesterol by 25%. p ≠ 0.25

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:

H 0 : μ = 2.0

H 0 : μ = 66
H a : μ ≠ 66

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:

H 0 : μ ≥ 5

H 0 : μ ≥ 45
H a : μ < 45

In an issue of U. S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.

H 0 : p ≤ 0.066

H 0 : p = 0.40
H a : p > 0.40

<!– ??? –>

Bring to class a newspaper, some news magazines, and some Internet articles . In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

Chapter Review

Formula Review

H 0 and H a are contradictory.

If α ≤ p -value, then do not reject H 0 .

If α > p -value, then reject H 0 .

α is preconceived. Its value is set before the hypothesis test starts. The p -value is calculated from the data.

You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. What is the random variable? Describe in words.

The random variable is the mean Internet speed in Megabits per second.

You are testing that the mean speed of your cable Internet connection is more than three Megabits per second. State the null and alternative hypotheses.

The American family has an average of two children. What is the random variable? Describe in words.

The random variable is the mean number of children an American family has.

The mean entry level salary of an employee at a company is 💲58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses.

A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words.

The random variable is the proportion of people picked at random in Times Square visiting the city.

A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses.

In a population of fish, approximately 42% are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses.

Suppose that a recent article stated that the mean time spent in jail by a first–time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was 3 years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of jail time has increased, what would the null and alternative hypotheses be? The distribution of the population is normal.

A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time on death row could likely be 15 years, what would the null and alternative hypotheses be?

H 0 : __________
H a : __________
H 0 : μ = 15
H a : μ ≠ 15

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. If you were conducting 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, what would the null and alternative hypotheses be?

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 ( μ 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 : μ = 34; H a : μ ≠ 34
H 0 : p ≤ 0.60; H a : p > 0.60
H 0 : μ ≥ 100,000; H a : μ < 100,000
H 0 : p = 0.29; H a : p ≠ 0.29
H 0 : p = 0.05; H a : p < 0.05
H 0 : μ ≤ 10; H a : μ > 10
H 0 : p = 0.50; H a : p ≠ 0.50
H 0 : μ = 6; H a : μ ≠ 6
H 0 : p ≥ 0.11; H a : p < 0.11
H 0 : μ ≤ 20,000; H a : μ > 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 > 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

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 o : $\overline{x}$ = 4.5, H a : $\overline{x}$ > 4.5
H o : μ ≥ 4.5, H a : μ < 4.5
H o : μ = 4.75, H a : μ > 4.75
H o : μ = 4.5, H a : μ > 4.5

Data from the National Institute of Mental Health. Available online at http://www.nimh.nih.gov/publicat/depression.cfm.

Null Hypothesis Examples

ThoughtCo / Hilary Allison

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Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
B.A., Physics and Mathematics, Hastings College

In statistical analysis, the null hypothesis assumes there is no meaningful relationship between two variables. Testing the null hypothesis can tell you whether your results are due to the effect of manipulating a dependent variable or due to chance. It's often used in conjunction with an alternative hypothesis, which assumes there is, in fact, a relationship between two variables.

The null hypothesis is among the easiest hypothesis to test using statistical analysis, making it perhaps the most valuable hypothesis for the scientific method. By evaluating a null hypothesis in addition to another hypothesis, researchers can support their conclusions with a higher level of confidence. Below are examples of how you might formulate a null hypothesis to fit certain questions.

What Is the Null Hypothesis?

The null hypothesis states there is no relationship between the measured phenomenon (the dependent variable ) and the independent variable , which is the variable an experimenter typically controls or changes. You do not need to believe that the null hypothesis is true to test it. On the contrary, you will likely suspect there is a relationship between a set of variables. One way to prove that this is the case is to reject the null hypothesis. Rejecting a hypothesis does not mean an experiment was "bad" or that it didn't produce results. In fact, it is often one of the first steps toward further inquiry.

To distinguish it from other hypotheses , the null hypothesis is written as H 0 (which is read as “H-nought,” "H-null," or "H-zero"). A significance test is used to determine the likelihood that the results supporting the null hypothesis are not due to chance. A confidence level of 95% or 99% is common. Keep in mind, even if the confidence level is high, there is still a small chance the null hypothesis is not true, perhaps because the experimenter did not account for a critical factor or because of chance. This is one reason why it's important to repeat experiments.

Examples of the Null Hypothesis

To write a null hypothesis, first start by asking a question. Rephrase that question in a form that assumes no relationship between the variables. In other words, assume a treatment has no effect. Write your hypothesis in a way that reflects this.

Other Types of Hypotheses

In addition to the null hypothesis, the alternative hypothesis is also a staple in traditional significance tests . It's essentially the opposite of the null hypothesis because it assumes the claim in question is true. For the first item in the table above, for example, an alternative hypothesis might be "Age does have an effect on mathematical ability."

Key Takeaways

In hypothesis testing, the null hypothesis assumes no relationship between two variables, providing a baseline for statistical analysis.
Rejecting the null hypothesis suggests there is evidence of a relationship between variables.
By formulating a null hypothesis, researchers can systematically test assumptions and draw more reliable conclusions from their experiments.
Difference Between Independent and Dependent Variables
Examples of Independent and Dependent Variables
What Is a Hypothesis? (Science)
What 'Fail to Reject' Means in a Hypothesis Test
Definition of a Hypothesis
Null Hypothesis Definition and Examples
Scientific Method Vocabulary Terms
Null Hypothesis and Alternative Hypothesis
Hypothesis Test for the Difference of Two Population Proportions
How to Conduct a Hypothesis Test
What Is a P-Value?
What Are the Elements of a Good Hypothesis?
What Is the Difference Between Alpha and P-Values?
Understanding Path Analysis
Hypothesis Test Example
An Example of a Hypothesis Test

13.1 Understanding Null Hypothesis Testing

Learning objectives.

Explain the purpose of null hypothesis testing, including the role of sampling error.
Describe the basic logic of null hypothesis testing.
Describe the role of relationship strength and sample size in determining statistical significance and make reasonable judgments about statistical significance based on these two factors.

The Purpose of Null Hypothesis Testing

As we have seen, psychological research typically involves measuring one or more variables in a sample and computing descriptive statistics for that sample. In general, however, the researcher’s goal is not to draw conclusions about that sample but to draw conclusions about the population that the sample was selected from. Thus researchers must use sample statistics to draw conclusions about the corresponding values in the population. These corresponding values in the population are called parameters . Imagine, for example, that a researcher measures the number of depressive symptoms exhibited by each of 50 adults with clinical depression and computes the mean number of symptoms. The researcher probably wants to use this sample statistic (the mean number of symptoms for the sample) to draw conclusions about the corresponding population parameter (the mean number of symptoms for adults with clinical depression).

Unfortunately, sample statistics are not perfect estimates of their corresponding population parameters. This is because there is a certain amount of random variability in any statistic from sample to sample. The mean number of depressive symptoms might be 8.73 in one sample of adults with clinical depression, 6.45 in a second sample, and 9.44 in a third—even though these samples are selected randomly from the same population. Similarly, the correlation (Pearson’s r ) between two variables might be +.24 in one sample, −.04 in a second sample, and +.15 in a third—again, even though these samples are selected randomly from the same population. This random variability in a statistic from sample to sample is called sampling error . (Note that the term error here refers to random variability and does not imply that anyone has made a mistake. No one “commits a sampling error.”)

In fact, any statistical relationship in a sample can be interpreted in two ways:

There is a relationship in the population, and the relationship in the sample reflects this.
There is no relationship in the population, and the relationship in the sample reflects only sampling error.

The purpose of null hypothesis testing is simply to help researchers decide between these two interpretations.

The Logic of Null Hypothesis Testing

Assume for the moment that the null hypothesis is true. There is no relationship between the variables in the population.
Determine how likely the sample relationship would be if the null hypothesis were true.
If the sample relationship would be extremely unlikely, then reject the null hypothesis in favor of the alternative hypothesis. If it would not be extremely unlikely, then retain the null hypothesis .

Following this logic, we can begin to understand why Mehl and his colleagues concluded that there is no difference in talkativeness between women and men in the population. In essence, they asked the following question: “If there were no difference in the population, how likely is it that we would find a small difference of d = 0.06 in our sample?” Their answer to this question was that this sample relationship would be fairly likely if the null hypothesis were true. Therefore, they retained the null hypothesis—concluding that there is no evidence of a sex difference in the population. We can also see why Kanner and his colleagues concluded that there is a correlation between hassles and symptoms in the population. They asked, “If the null hypothesis were true, how likely is it that we would find a strong correlation of +.60 in our sample?” Their answer to this question was that this sample relationship would be fairly unlikely if the null hypothesis were true. Therefore, they rejected the null hypothesis in favor of the alternative hypothesis—concluding that there is a positive correlation between these variables in the population.

A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the p value . A low p value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A p value that is not low means that the sample result would be likely if the null hypothesis were true and leads to the retention of the null hypothesis. But how low must the p value be before the sample result is considered unlikely enough to reject the null hypothesis? In null hypothesis testing, this criterion is called α (alpha) and is almost always set to .05. If there is a 5% chance or less of a result as extreme as the sample result if the null hypothesis were true, then the null hypothesis is rejected. When this happens, the result is said to be statistically significant . If there is greater than a 5% chance of a result as extreme as the sample result when the null hypothesis is true, then the null hypothesis is retained. This does not necessarily mean that the researcher accepts the null hypothesis as true—only that there is not currently enough evidence to reject it. Researchers often use the expression “fail to reject the null hypothesis” rather than “retain the null hypothesis,” but they never use the expression “accept the null hypothesis.”

The Misunderstood p Value

“Null Hypothesis” retrieved from http://imgs.xkcd.com/comics/null_hypothesis.png (CC-BY-NC 2.5)

Role of Sample Size and Relationship Strength

Statistical Significance Versus Practical Significance

“Conditional Risk” retrieved from http://imgs.xkcd.com/comics/conditional_risk.png (CC-BY-NC 2.5)

Key Takeaways

Null hypothesis testing is a formal approach to deciding whether a statistical relationship in a sample reflects a real relationship in the population or is just due to chance.
The logic of null hypothesis testing involves assuming that the null hypothesis is true, finding how likely the sample result would be if this assumption were correct, and then making a decision. If the sample result would be unlikely if the null hypothesis were true, then it is rejected in favor of the alternative hypothesis. If it would not be unlikely, then the null hypothesis is retained.
The probability of obtaining the sample result if the null hypothesis were true (the p value) is based on two considerations: relationship strength and sample size. Reasonable judgments about whether a sample relationship is statistically significant can often be made by quickly considering these two factors.
Statistical significance is not the same as relationship strength or importance. Even weak relationships can be statistically significant if the sample size is large enough. It is important to consider relationship strength and the practical significance of a result in addition to its statistical significance.
Discussion: Imagine a study showing that people who eat more broccoli tend to be happier. Explain for someone who knows nothing about statistics why the researchers would conduct a null hypothesis test.
The correlation between two variables is r = −.78 based on a sample size of 137.
The mean score on a psychological characteristic for women is 25 ( SD = 5) and the mean score for men is 24 ( SD = 5). There were 12 women and 10 men in this study.
In a memory experiment, the mean number of items recalled by the 40 participants in Condition A was 0.50 standard deviations greater than the mean number recalled by the 40 participants in Condition B.
In another memory experiment, the mean scores for participants in Condition A and Condition B came out exactly the same!
A student finds a correlation of r = .04 between the number of units the students in his research methods class are taking and the students’ level of stress.
Cohen, J. (1994). The world is round: p < .05. American Psychologist, 49 , 997–1003. ↵
Hyde, J. S. (2007). New directions in the study of gender similarities and differences. Current Directions in Psychological Science, 16 , 259–263. ↵

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Teach yourself statistics

Hypothesis Test: Difference Between Means

This lesson explains how to conduct a hypothesis test for the difference between two means. The test procedure, called the two-sample t-test , is appropriate when the following conditions are met:

The sampling method for each sample is simple random sampling .
The samples are independent .
Each population is at least 20 times larger than its respective sample .
The population distribution is normal.
The population data are symmetric , unimodal , without outliers , and the sample size is 15 or less.
The population data are slightly skewed , unimodal, without outliers, and the sample size is 16 to 40.
The sample size is greater than 40, without outliers.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

The table below shows three sets of null and alternative hypotheses. Each makes a statement about the difference d between the mean of one population μ 1 and the mean of another population μ 2 . (In the table, the symbol ≠ means " not equal to ".)

The first set of hypotheses (Set 1) is an example of a two-tailed test , since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests , since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.

When the null hypothesis states that there is no difference between the two population means (i.e., d = 0), the null and alternative hypothesis are often stated in the following form.

H o : μ 1 = μ 2

H a : μ 1 ≠ μ 2

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
Test method. Use the two-sample t-test to determine whether the difference between means found in the sample is significantly different from the hypothesized difference between means.

Analyze Sample Data

Using sample data, find the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

SE = sqrt[ (s 1 2 /n 1 ) + (s 2 2 /n 2 ) ]

DF = (s 1 2 /n 1 + s 2 2 /n 2 ) 2 / { [ (s 1 2 / n 1 ) 2 / (n 1 - 1) ] + [ (s 2 2 / n 2 ) 2 / (n 2 - 1) ] }

t = [ ( x 1 - x 2 ) - d ] / SE

P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, having the degrees of freedom computed above. (See sample problems at the end of this lesson for examples of how this is done.)

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

In this section, two sample problems illustrate how to conduct a hypothesis test of a difference between mean scores. The first problem involves a two-tailed test; the second problem, a one-tailed test.

Problem 1: Two-Tailed Test

Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students.

At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15.

Test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance. (Assume that student performance is approximately normal.)

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ 1 - μ 2 = 0

Alternative hypothesis: μ 1 - μ 2 ≠ 0

Formulate an analysis plan . For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

SE = sqrt[(s 1 2 /n 1 ) + (s 2 2 /n 2 )]

SE = sqrt[(10 2 /30) + (15 2 /25] = sqrt(3.33 + 9)

SE = sqrt(12.33) = 3.51

DF = (10 2 /30 + 15 2 /25) 2 / { [ (10 2 / 30) 2 / (29) ] + [ (15 2 / 25) 2 / (24) ] }

DF = (3.33 + 9) 2 / { [ (3.33) 2 / (29) ] + [ (9) 2 / (24) ] } = 152.03 / (0.382 + 3.375) = 152.03/3.757 = 40.47

t = [ ( x 1 - x 2 ) - d ] / SE = [ (78 - 85) - 0 ] / 3.51 = -7/3.51 = -1.99

where s 1 is the standard deviation of sample 1, s 2 is the standard deviation of sample 2, n 1 is the size of sample 1, n 2 is the size of sample 2, x 1 is the mean of sample 1, x 2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test , the P-value is the probability that a t statistic having 40 degrees of freedom is more extreme than -1.99; that is, less than -1.99 or greater than 1.99.

We use the t Distribution Calculator to find P(t < -1.99) is about 0.027.

If you enter 1.99 as the sample mean in the t Distribution Calculator, you will find the that the P(t ≤ 1.99) is about 0.973. Therefore, P(t > 1.99) is 1 minus 0.973 or 0.027. Thus, the P-value = 0.027 + 0.027 = 0.054.
Interpret results . Since the P-value (0.054) is less than the significance level (0.10), we cannot accept the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, the sample size was much smaller than the population size, and the samples were drawn from a normal population.

Problem 2: One-Tailed Test

The Acme Company has developed a new battery. The engineer in charge claims that the new battery will operate continuously for at least 7 minutes longer than the old battery.

To test the claim, the company selects a simple random sample of 100 new batteries and 100 old batteries. The old batteries run continuously for 190 minutes with a standard deviation of 20 minutes; the new batteries, 200 minutes with a standard deviation of 40 minutes.

Test the engineer's claim that the new batteries run at least 7 minutes longer than the old. Use a 0.05 level of significance. (Assume that there are no outliers in either sample.)

Null hypothesis: μ 1 - μ 2 <= 7

Alternative hypothesis: μ 1 - μ 2 > 7

where μ 1 is battery life for the new battery, and μ 2 is battery life for the old battery.

Formulate an analysis plan . For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

SE = sqrt[(40 2 /100) + (20 2 /100]

SE = sqrt(16 + 4) = 4.472

DF = (40 2 /100 + 20 2 /100) 2 / { [ (40 2 / 100) 2 / (99) ] + [ (20 2 / 100) 2 / (99) ] }

DF = (20) 2 / { [ (16) 2 / (99) ] + [ (2) 2 / (99) ] } = 400 / (2.586 + 0.162) = 145.56

t = [ ( x 1 - x 2 ) - d ] / SE = [(200 - 190) - 7] / 4.472 = 3/4.472 = 0.67

Here is the logic of the analysis: Given the alternative hypothesis (μ 1 - μ 2 > 7), we want to know whether the observed difference in sample means is big enough (i.e., sufficiently greater than 7) to cause us to reject the null hypothesis.

Interpret results . Suppose we replicated this study many times with different samples. If the true difference in population means were actually 7, we would expect the observed difference in sample means to be 10 or less in 75% of our samples. And we would expect to find an observed difference to be more than 10 in 25% of our samples Therefore, the P-value in this analysis is 0.25.

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8.6: Hypothesis Test of a Single Population Mean with Examples

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Steps for performing Hypothesis Test of a Single Population Mean

Step 1: State your hypotheses about the population mean. Step 2: Summarize the data. State a significance level. State and check conditions required for the procedure

Find or identify the sample size, n, the sample mean, $\bar{x}$ and the sample standard deviation, s .

The sampling distribution for the one-mean test statistic is, approximately, T- distribution if the following conditions are met

Sample is random with independent observations .
Sample is large. The population must be Normal or the sample size must be at least 30.

Step 3: Perform the procedure based on the assumption that $H_{0}$ is true

Find the Estimated Standard Error: $SE=\frac{s}{\sqrt{n}}$.
Compute the observed value of the test statistic: $T_{obs}=\frac{\bar{x}-\mu_{0}}{SE}$.
Check the type of the test (right-, left-, or two-tailed)
Find the p-value in order to measure your level of surprise.

Step 4: Make a decision about $H_{0}$ and $H_{a}$

Do you reject or not reject your null hypothesis?

Step 5: Make a conclusion

What does this mean in the context of the data?

The following examples illustrate a left-, right-, and two-tailed test.

Example $\pageindex{1}$.

$H_{0}: \mu = 5, H_{a}: \mu < 5$

Test of a single population mean. $H_{a}$ tells you the test is left-tailed. The picture of the $p$-value is as follows:

$Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.$

Exercise $\PageIndex{1}$

$H_{0}: \mu = 10, H_{a}: \mu < 10$

Assume the $p$-value is 0.0935. What type of test is this? Draw the picture of the $p$-value.

left-tailed test

$alt$

Example $\PageIndex{2}$

$H_{0}: \mu \leq 0.2, H_{a}: \mu > 0.2$

This is a test of a single population proportion. $H_{a}$ tells you the test is right-tailed . The picture of the p -value is as follows:

$Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.$

Exercise $\PageIndex{2}$

$H_{0}: \mu \leq 1, H_{a}: \mu > 1$

Assume the $p$-value is 0.1243. What type of test is this? Draw the picture of the $p$-value.

right-tailed test

$alt$

Example $\PageIndex{3}$

$H_{0}: \mu = 50, H_{a}: \mu \neq 50$

This is a test of a single population mean. $H_{a}$ tells you the test is two-tailed . The picture of the $p$-value is as follows.

$Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.$

Exercise $\PageIndex{3}$

$H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5$

Assume the p -value is 0.2564. What type of test is this? Draw the picture of the $p$-value.

two-tailed test

$alt$

Full Hypothesis Test Examples

Example $\pageindex{4}$.

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

Set up the hypothesis test:

A 5% level of significance means that $\alpha = 0.05$. This is a test of a single population mean .

$H_{0}: \mu = 65 H_{a}: \mu > 65$

Since the instructor thinks the average score is higher, use a "$>$". The "$>$" means the test is right-tailed.

Determine the distribution needed:

Random variable: $\bar{X} =$ average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given $n = 10$ sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's $t$.

Use $t_{df}$. Therefore, the distribution for the test is $t_{9}$ where $n = 10$ and $df = 10 - 1 = 9$.

The sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Calculate the $p$-value using the Student's $t$-distribution:

\[t_{obs} = \dfrac{\bar{x}-\mu_{\bar{x}}}{\left(\dfrac{s}{\sqrt{n}}\right)}=\dfrac{67-65}{\left(\dfrac{3.1972}{\sqrt{10}}\right)}\]

Use the T-table or Excel's t_dist() function to find p-value:

$p\text{-value} = P(\bar{x} > 67) =P(T >1.9782 )= 1-0.9604=0.0396$

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.

$Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.$

Compare $\alpha$ and the $p-\text{value}$:

Since $α = 0.05$ and $p\text{-value} = 0.0396$. $\alpha > p\text{-value}$.

Make a decision: Since $\alpha > p\text{-value}$, reject $H_{0}$.

This means you reject $\mu = 65$. In other words, you believe the average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The $p\text{-value}$ can easily be calculated.

Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for $\mu_{0}$, the name of the list where you put the data, and 1 for Freq: . Arrow down to $\mu$: and arrow over to $> \mu_{0}$. Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the $p\text{-value}$ (p = 0.0396) but it also calculates the test statistic ( t -score) for the sample mean, the sample mean, and the sample standard deviation. $\mu > 65$ is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with $t = 1.9781$ (test statistic) and $p = 0.0396$ ($p\text{-value}$). Make sure when you use Draw that no other equations are highlighted in $Y =$ and the plots are turned off.

Exercise $\PageIndex{4}$

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, find the p -value, state your conclusion, and identify the Type I and Type II errors.

$H_{0}: \mu = 5$
$H_{a}: \mu < 5$
$p = 0.0082$

Because $p < \alpha$, we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.

Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Example $\PageIndex{5}$

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. Assume the population is normal.

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

$H_{0}: \mu \leq 1$
$H_{a}: \mu > 1$
Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
Do the calculations : $p\text{-value} ( = 0.036)$

4. State the Conclusions : Since the $p\text{-value} (= 0.036)$ is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

The hypothesis test itself has an established process. This can be summarized as follows:

Determine $H_{0}$ and $H_{a}$. 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 $p\text{-value}$. (A t -score is an example of test statistics.)
Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use $\alpha$ and not $\beta$. $\beta$ is needed to help determine the sample size of the data that is used in calculating the $p\text{-value}$. Remember that the quantity $1 – \beta$ is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
Data from Bloomberg Businessweek . Available online at www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
Data from Growing by Degrees by Allen and Seaman.
Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
“Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Math Article

Null Hypothesis

In mathematics, Statistics deals with the study of research and surveys on the numerical data. For taking surveys, we have to define the hypothesis. Generally, there are two types of hypothesis. One is a null hypothesis, and another is an alternative hypothesis .

In probability and statistics, the null hypothesis is a comprehensive statement or default status that there is zero happening or nothing happening. For example, there is no connection among groups or no association between two measured events. It is generally assumed here that the hypothesis is true until any other proof has been brought into the light to deny the hypothesis. Let us learn more here with definition, symbol, principle, types and example, in this article.

Table of contents:

Comparison with Alternative Hypothesis

Null Hypothesis Definition

The null hypothesis is a kind of hypothesis which explains the population parameter whose purpose is to test the validity of the given experimental data. This hypothesis is either rejected or not rejected based on the viability of the given population or sample . In other words, the null hypothesis is a hypothesis in which the sample observations results from the chance. It is said to be a statement in which the surveyors wants to examine the data. It is denoted by H 0 .

Null Hypothesis Symbol

In statistics, the null hypothesis is usually denoted by letter H with subscript ‘0’ (zero), such that H 0 . It is pronounced as H-null or H-zero or H-nought. At the same time, the alternative hypothesis expresses the observations determined by the non-random cause. It is represented by H 1 or H a .

Null Hypothesis Principle

The principle followed for null hypothesis testing is, collecting the data and determining the chances of a given set of data during the study on some random sample, assuming that the null hypothesis is true. In case if the given data does not face the expected null hypothesis, then the outcome will be quite weaker, and they conclude by saying that the given set of data does not provide strong evidence against the null hypothesis because of insufficient evidence. Finally, the researchers tend to reject that.

Null Hypothesis Formula

Here, the hypothesis test formulas are given below for reference.

The formula for the null hypothesis is:

H 0 : p = p 0

The formula for the alternative hypothesis is:

H a = p >p 0 , < p 0 ≠ p 0

The formula for the test static is:

Remember that, p 0 is the null hypothesis and p – hat is the sample proportion.

Also, read:

Types of Null Hypothesis

There are different types of hypothesis. They are:

Simple Hypothesis

It completely specifies the population distribution. In this method, the sampling distribution is the function of the sample size.

Composite Hypothesis

The composite hypothesis is one that does not completely specify the population distribution.

Exact Hypothesis

Exact hypothesis defines the exact value of the parameter. For example μ= 50

Inexact Hypothesis

This type of hypothesis does not define the exact value of the parameter. But it denotes a specific range or interval. For example 45< μ <60

Null Hypothesis Rejection

Sometimes the null hypothesis is rejected too. If this hypothesis is rejected means, that research could be invalid. Many researchers will neglect this hypothesis as it is merely opposite to the alternate hypothesis. It is a better practice to create a hypothesis and test it. The goal of researchers is not to reject the hypothesis. But it is evident that a perfect statistical model is always associated with the failure to reject the null hypothesis.

How do you Find the Null Hypothesis?

The null hypothesis says there is no correlation between the measured event (the dependent variable) and the independent variable. We don’t have to believe that the null hypothesis is true to test it. On the contrast, you will possibly assume that there is a connection between a set of variables ( dependent and independent).

When is Null Hypothesis Rejected?

The null hypothesis is rejected using the P-value approach. If the P-value is less than or equal to the α, there should be a rejection of the null hypothesis in favour of the alternate hypothesis. In case, if P-value is greater than α, the null hypothesis is not rejected.

Null Hypothesis and Alternative Hypothesis

Now, let us discuss the difference between the null hypothesis and the alternative hypothesis.

Null Hypothesis Examples

Here, some of the examples of the null hypothesis are given below. Go through the below ones to understand the concept of the null hypothesis in a better way.

If a medicine reduces the risk of cardiac stroke, then the null hypothesis should be “the medicine does not reduce the chance of cardiac stroke”. This testing can be performed by the administration of a drug to a certain group of people in a controlled way. If the survey shows that there is a significant change in the people, then the hypothesis is rejected.

Few more examples are:

1). Are there is 100% chance of getting affected by dengue?

Ans: There could be chances of getting affected by dengue but not 100%.

2). Do teenagers are using mobile phones more than grown-ups to access the internet?

Ans: Age has no limit on using mobile phones to access the internet.

3). Does having apple daily will not cause fever?

Ans: Having apple daily does not assure of not having fever, but increases the immunity to fight against such diseases.

4). Do the children more good in doing mathematical calculations than grown-ups?

Ans: Age has no effect on Mathematical skills.

In many common applications, the choice of the null hypothesis is not automated, but the testing and calculations may be automated. Also, the choice of the null hypothesis is completely based on previous experiences and inconsistent advice. The choice can be more complicated and based on the variety of applications and the diversity of the objectives.

The main limitation for the choice of the null hypothesis is that the hypothesis suggested by the data is based on the reasoning which proves nothing. It means that if some hypothesis provides a summary of the data set, then there would be no value in the testing of the hypothesis on the particular set of data.

Frequently Asked Questions on Null Hypothesis

What is meant by the null hypothesis.

In Statistics, a null hypothesis is a type of hypothesis which explains the population parameter whose purpose is to test the validity of the given experimental data.

What are the benefits of hypothesis testing?

Hypothesis testing is defined as a form of inferential statistics, which allows making conclusions from the entire population based on the sample representative.

When a null hypothesis is accepted and rejected?

The null hypothesis is either accepted or rejected in terms of the given data. If P-value is less than α, then the null hypothesis is rejected in favor of the alternative hypothesis, and if the P-value is greater than α, then the null hypothesis is accepted in favor of the alternative hypothesis.

Why is the null hypothesis important?

The importance of the null hypothesis is that it provides an approximate description of the phenomena of the given data. It allows the investigators to directly test the relational statement in a research study.

How to accept or reject the null hypothesis in the chi-square test?

If the result of the chi-square test is bigger than the critical value in the table, then the data does not fit the model, which represents the rejection of the null hypothesis.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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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:

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

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

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

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

9.1 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.

H 0 : The null hypothesis: It is a statement of no difference between the variables—they are not related. This can often be considered the status quo and as a result if you cannot accept the null it requires some action.

H a : The alternative hypothesis: It is a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 . This is usually what the researcher is trying to prove.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are "reject H 0 " if the sample information favors the alternative hypothesis or "do not reject H 0 " or "decline to reject H 0 " if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

Example 9.1

H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ .30 H a : More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.

Example 9.2

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: H 0 : μ = 2.0 H a : μ ≠ 2.0

H 0 : μ __ 66
H a : μ __ 66

Example 9.3

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: H 0 : μ ≥ 5 H a : μ < 5

H 0 : μ __ 45
H a : μ __ 45

Example 9.4

In an issue of U. S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

H 0 : p __ 0.40
H a : p __ 0.40

Collaborative Exercise

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Book title: Introductory Statistics 2e
Publication date: Dec 13, 2023
Location: Houston, Texas
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Statistics Made Easy

5 Tips for Interpreting P-Values Correctly in Hypothesis Testing

Hypothesis testing is a critical part of statistical analysis and is often the endpoint where conclusions are drawn about larger populations based on a sample or experimental dataset. Central to this process is the p-value. Broadly, the p-value quantifies the strength of evidence against the null hypothesis. Given the importance of the p-value, it is essential to ensure its interpretation is correct. Here are five essential tips for ensuring the p-value from a hypothesis test is understood correctly.

1. Know What the P-value Represents

First, it is essential to understand what a p-value is. In hypothesis testing, the p-value is defined as the probability of observing your data, or data more extreme, if the null hypothesis is true. As a reminder, the null hypothesis states no difference between your data and the expected population.

For example, in a hypothesis test to see if changing a company’s logo drives more traffic to the website, a null hypothesis would state that the new traffic numbers are equal to the old traffic numbers. In this context, the p-value would be the probability that the data you observed, or data more extreme, would occur if this null hypothesis were true.

Therefore, a smaller p-value indicates that what you observed is unlikely to have occurred if the null were true, offering evidence to reject the null hypothesis. Typically, a cut-off value of 0.05 is used where any p-value below this is considered significant evidence against the null.

2. Understand the Directionality of Your Hypothesis

Based on the research question under exploration, there are two types of hypotheses: one-sided and two-sided. A one-sided test specifies a particular direction of effect, such as traffic to a website increasing after a design change. On the other hand, a two-sided test allows the change to be in either direction and is effective when the researcher wants to see any effect of the change.

Either way, determining the statistical significance of a p-value is the same: if the p-value is below a threshold value, it is statistically significant. However, when calculating the p-value, it is important to ensure the correct sided calculations have been completed.

Additionally, the interpretation of the meaning of a p-value will differ based on the directionality of the hypothesis. If a one-sided test is significant, the researchers can use the p-value to support a statistically significant increase or decrease based on the direction of the test. If a two-sided test is significant, the p-value can only be used to say that the two groups are different, but not that one is necessarily greater.

3. Avoid Threshold Thinking

A common pitfall in interpreting p-values is falling into the threshold thinking trap. The most commonly used cut-off value for whether a calculated p-value is statistically significant is 0.05. Typically, a p-value of less than 0.05 is considered statistically significant evidence against the null hypothesis.

However, this is just an arbitrary value. Rigid adherence to this or any other predefined cut-off value can obscure business-relevant effect sizes. For example, a hypothesis test looking at changes in traffic after a website design may find that an increase of 10,000 views is not statistically significant with a p-value of 0.055 since that value is above 0.05. However, the actual increase of 10,000 may be important to the growth of the business.

Therefore, a p-value can be practically significant while not being statistically significant. Both types of significance and the broader context of the hypothesis test should be considered when making a final interpretation.

4. Consider the Power of Your Study

Similarly, some study conditions can result in a non-significant p-value even if practical significance exists. Statistical power is the ability of a study to detect an effect when it truly exists. In other words, it is the probability that the null hypothesis will be rejected when it is false.

Power is impacted by a lot of factors. These include sample size, the effect size you are looking for, and variability within the data. In the example of website traffic after a design change, if the number of visits overall is too small, there may not be enough views to have enough power to detect a difference.

Simple ways to increase the power of a hypothesis test and increase the chances of detecting an effect are increasing the sample size, looking for a smaller effect size, changing the experiment design to control for variables that can increase variability, or adjusting the type of statistical test being run.

5. Be Aware of Multiple Comparisons

Whenever multiple p-values are calculated in a single study due to multiple comparisons, there is an increased risk of false positives. This is because each individual comparison introduces random fluctuations, and each additional comparison compounds these fluctuations.

For example, in a hypothesis test looking at traffic before and after a website redesign, the team may be interested in making more than one comparison. This can include total visits, page views, and average time spent on the website. Since multiple comparisons are being made, there must be a correction made when interpreting the p-value.

The Bonferroni correction is one of the most commonly used methods to account for this increased probability of false positives. In this method, the significance cut-off value, typically 0.05, is divided by the number of comparisons made. The result is used as the new significance cut-off value. Applying this correction mitigates the risk of false positives and improves the reliability of findings from a hypothesis test.

In conclusion, interpreting p-values requires a nuanced understanding of many statistical concepts and careful consideration of the hypothesis test’s context. By following these five tips, the interpretation of the p-value from a hypothesis test can be more accurate and reliable, leading to better data-driven decision-making.

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IMAGES

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Mastering Hypothesis Writing: Expert Tips for 2023
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Null Hypothesis
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Hypothesis Testing: the null and alternative hypotheses
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Testing a null hypothesis

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Null Hypothesis: Definition, Rejecting & Examples
The null hypothesis in statistics states that there is no difference between groups or no relationship between variables. It is one of two mutually exclusive hypotheses about a population in a hypothesis test. When your sample contains sufficient evidence, you can reject the null and conclude that the effect is statistically significant.
Null & Alternative Hypotheses
The null and alternative hypotheses offer competing answers to your research question. When the research question asks "Does the independent variable affect the dependent variable?": The null hypothesis ( H0) answers "No, there's no effect in the population.". The alternative hypothesis ( Ha) answers "Yes, there is an effect in the ...
How to Write a Null Hypothesis (5 Examples)
H 0 (Null Hypothesis): Population parameter =, ≤, ≥ some value. H A (Alternative Hypothesis): Population parameter <, >, ≠ some value. Note that the null hypothesis always contains the equal sign. We interpret the hypotheses as follows: Null hypothesis: The sample data provides no evidence to support some claim being made by an individual.
9.1 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. H 0, the —null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.
9.1: Null and Alternative Hypotheses
Review. In a hypothesis test, sample data is evaluated in order to arrive at a decision about some type of claim.If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis, typically denoted with $H_{0}$.The null is not rejected unless the hypothesis test shows otherwise.
Understanding Null Hypothesis Testing
A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the p value. A low p value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A high p value means that the sample ...
What Is The Null Hypothesis & When To Reject It
Rejecting the null hypothesis means that a relationship does exist between a set of variables and the effect is statistically significant (p > 0.05). If the data collected from the random sample is not statistically significance , then the null hypothesis will be accepted, and the researchers can conclude that there is no relationship between ...
Null hypothesis
A possible null hypothesis is that the mean male score is the same as the mean female score: H 0: μ 1 = μ 2. where H 0 = the null hypothesis, μ 1 = the mean of population 1, and μ 2 = the mean of population 2. A stronger null hypothesis is that the two samples have equal variances and shapes of their respective distributions. Terminology
Examples of null and alternative hypotheses
The null hypothesis is often stated as the assumption that there is no change, no difference between two groups, or no relationship between two variables. The alternative hypothesis, on the other hand, is the statement that there is a change, difference, or relationship. ... The mean from this sample, the mean from the sample, is 7.5 hours ...
Null and Alternative Hypotheses
Concept Review. In a hypothesis test, sample data is evaluated in order to arrive at a decision about some type of claim.If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis, typically denoted with H 0.The null is not rejected unless the hypothesis test shows otherwise.
Lesson 6b: Hypothesis Testing for One-Sample Mean
Our test statistic, 2.11, is greater than our critical value of 1.69 and therefore is in the rejection region. We would reject the null hypothesis. Step 6. State the conclusion in words. There is enough evidence, at a significance level of 5%, to reject the null hypothesis and conclude that the mean waiting time is greater than 10 minutes.
Null and Alternative Hypotheses
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 o: $\overline{x}$ = 4.5, H a: \(\overline{x ...
7.3: The Research Hypothesis and the Null Hypothesis
This null hypothesis can be written as: H0: X¯ = μ H 0: X ¯ = μ. For most of this textbook, the null hypothesis is that the means of the two groups are similar. Much later, the null hypothesis will be that there is no relationship between the two groups. Either way, remember that a null hypothesis is always saying that nothing is different.
How to Formulate a Null Hypothesis (With Examples)
To distinguish it from other hypotheses, the null hypothesis is written as H 0 (which is read as "H-nought," "H-null," or "H-zero"). A significance test is used to determine the likelihood that the results supporting the null hypothesis are not due to chance. A confidence level of 95% or 99% is common. Keep in mind, even if the confidence level is high, there is still a small chance the ...
Hypothesis Testing
There are 5 main steps in hypothesis testing: State your research hypothesis as a null hypothesis and alternate hypothesis (H o) and (H a or H 1 ). Collect data in a way designed to test the hypothesis. Perform an appropriate statistical test. Decide whether to reject or fail to reject your null hypothesis.
Hypothesis Test for a Mean
The first step is to state the null hypothesis and an alternative hypothesis. Null hypothesis: μ >= 110. Alternative hypothesis: μ < 110. Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small. Formulate an analysis plan. For this analysis, the significance level is 0.01.
13.1 Understanding Null Hypothesis Testing
A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the p value. A low p value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A p value that is not low means that ...
Hypothesis Test: Difference in Means
The null hypothesis will be rejected if the difference between sample means is bigger than would be expected by chance. Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis. Analyze sample data. Using sample data, we compute the standard ...
8.6: Hypothesis Test of a Single Population Mean with Examples
The sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data. Calculate the $p$-value using the Student's $t$-distribution: ... If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.
Null Hypothesis
Here, the hypothesis test formulas are given below for reference. The formula for the null hypothesis is: H 0 : p = p 0. The formula for the alternative hypothesis is: H a = p >p 0, < p 0 ≠ p 0. The formula for the test static is: Remember that, p 0 is the null hypothesis and p - hat is the sample proportion.
9.E: Hypothesis Testing with One Sample (Exercises)
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: H0: ˉx = 4.5, Ha: ˉx > 4.5. H0: μ ≥ 4.5, Ha: μ < 4.5. H0: μ = 4.75, Ha: μ > 4.75.
9.1 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. H 0: The null hypothesis: It is a statement of no difference between the variables—they are not related. This can often be considered the status quo and as a result if you cannot accept the null it requires some action.
An Introduction to t Tests
The null hypothesis (H 0) is that the true difference between these group means is zero. The alternate hypothesis ... or use a one-sample t-test to compare the group mean to a standard value. If you are studying two groups, use a two-sample t-test. If you want to know only whether a difference exists, use a two-tailed test.
PDF Hypothesis Testing
Given Null Hypothesis: µ = k k is a value of the mean given µ is the population mean discussed throughout the worksheet ... Why do hypothesis testing? Sample mean may be di↵erent from the population mean. Type of Test to Apply: Rightailed T. µ>kuYo believe that µ is more than value stated in H 0. Left-Tailed. µ<kuYo believe that µ is ...
5 Tips for Interpreting P-Values Correctly in Hypothesis Testing
Here are five essential tips for ensuring the p-value from a hypothesis test is understood correctly. 1. Know What the P-value Represents. First, it is essential to understand what a p-value is. In hypothesis testing, the p-value is defined as the probability of observing your data, or data more extreme, if the null hypothesis is true.
Search
Our objective here is to test the null hypothesis that the population means are equal against the alternative hypothesis that means are not equal, as described in the …. \ (Y_i = X_ {1i}-X_ {2i}\) and \ (\mu_Y = \mu_1-\mu_2\) Testing the null hypothesis for the equality of the …. 13.3. Test for Relationship Between Canonical Variate Pairs.

Understanding Null Hypothesis Testing

The Purpose of Null Hypothesis Testing

The Logic of Null Hypothesis Testing

Role of Sample Size and Relationship Strength

Statistical Significance Versus Practical Significance

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What is The Null Hypothesis & When Do You Reject The Null Hypothesis

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Video transcript

Module 9: Hypothesis Testing With One Sample

Concept Review

Formula Review

Null and Alternative Hypotheses

Chapter Review

Formula Review

Null Hypothesis Examples

What Is the Null Hypothesis?

Examples of the Null Hypothesis

Other Types of Hypotheses

Key Takeaways

13.1 Understanding Null Hypothesis Testing

The Purpose of Null Hypothesis Testing

The Logic of Null Hypothesis Testing

The Misunderstood p Value

Role of Sample Size and Relationship Strength

Statistical Significance Versus Practical Significance

Key Takeaways

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Hypothesis Test: Difference Between Means

State the Hypotheses

Formulate an Analysis Plan

Analyze Sample Data

Interpret Results

Test Your Understanding

Margin Size

8.6: Hypothesis Test of a Single Population Mean with Examples

Steps for performing Hypothesis Test of a Single Population Mean

The following examples illustrate a left-, right-, and two-tailed test.

Exercise \(\PageIndex{1}\)

Example \(\PageIndex{2}\)

Exercise \(\PageIndex{2}\)

Example \(\PageIndex{3}\)

Exercise \(\PageIndex{3}\)

Full Hypothesis Test Examples

Exercise \(\PageIndex{4}\)

Example \(\PageIndex{5}\)

Null Hypothesis

Null Hypothesis Definition

Null Hypothesis Symbol

Null Hypothesis Principle

Null Hypothesis Formula

Types of Null Hypothesis

Null Hypothesis Rejection

How do you Find the Null Hypothesis?

When is Null Hypothesis Rejected?

Null Hypothesis and Alternative Hypothesis

Null Hypothesis Examples

Frequently Asked Questions on Null Hypothesis

What are the benefits of hypothesis testing?

When a null hypothesis is accepted and rejected?

Why is the null hypothesis important?

How to accept or reject the null hypothesis in the chi-square test?

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Margin Size

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