chapter 8 hypothesis testing answers

N ( 244 , 15 50 ) N ( 244 , 15 50 )

As the sample size increases, there will be less variability in the mean, so the interval size decreases.

X is the time in minutes it takes to complete the U.S. Census short form. X ¯ X ¯ is the mean time it took a sample of 200 people to complete the U.S. Census short form.

CI: (7.9441, 8.4559)

The level of confidence would decrease, because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases.

x ¯ x ¯ = 2.2

X ¯ X ¯ is the mean weight of a sample of 20 heads of lettuce.

EBM = 0.07 CI: (2.1264, 2.2736)

The interval is greater, because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals.

The confidence level would increase.

(24.52,36.28)

We are 95 percent confident that the true mean age for winter Foothill College students is between 24.52 and 36.28.

The error bound for the mean would decrease, because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean.

X is the number of hours a patient waits in the emergency room before being called back to be examined. X ¯ X ¯ is the mean wait time of 70 patients in the emergency room.

CI: (1.3808, 1.6192)

x ¯ x ¯ = 151
s x s x = 32
n – 1 = 107

X ¯ X ¯ is the mean number of hours spent watching television per month from a sample of 108 Americans.

CI: (142.92, 159.08)

(2.93, 3.59)

We are 95 percent confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.

The error bound would become EBM = 0.245. This error bound decreases, because as sample sizes increase, variability decreases, and we need less interval length to capture the true mean.

It would decrease, because the z -score would decrease, which would reduce the numerator and lower the number.

X is the number of successes where the woman makes the majority of the purchasing decisions for the household. P ′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.

CI: (0.5321, 0.6679)

EBM : 0.0679

X is the number of successes where an executive prefers a truck. P ′ is the percentage of executives sampled who prefer a truck.

CI: (0.19432, 0.33068)

EBM : 0.0707

The sampling error means that the true mean can be 2 percent above or below the sample mean.

P ′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election.

CI: (0.62735, 0.67265);

EBM: 0.02265

the number of girls, ages 8 to 12, in the 5 p.m. Monday night beginning ice-skating class

P ′ ~ N ( 0.8 , ( 0.8 ) ( 0.2 ) 80 ) P ′ ~ N ( 0.8 , ( 0.8 ) ( 0.2 ) 80 )

CI = (0.72171, 0.87829).

(0.72; 0.88)

With 92 percent confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72 percent and 88 percent.

The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error.

X is the height of a Swedish male, and is the mean height from a sample of 48 Swedish males.
Normal. We know the standard deviation for the population, and the sample size is greater than 30.
CI: (70.151, 71.49)
EBM = 0.849
The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.
x ¯ x ¯ = 23.6
X is the time needed to complete an individual tax form. X ¯ X ¯ is the mean time to complete tax forms from a sample of 100 customers.
N ( 23.6 , 7 100 ) N ( 23.6 , 7 100 ) because we know sigma.
(22.228, 24.972)
EBM = 1.372
It will need to change the sample size. The firm needs to determine what the confidence level should be and then apply the error bound formula to determine the necessary sample size.
The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.
According to the error bound formula, the firm needs to survey 206 people. Because we increase the confidence level, we need to increase either our error bound or the sample size.
X is the number of letters a single camper will send home. X ¯ X ¯ is the mean number of letters sent home from a sample of 20 campers.

N 7.9 ( 2.5 20 ) 7.9 ( 2.5 20 )

CI: (6.98, 8.82)
The error bound and confidence interval will decrease.
x ¯ x ¯ = $568,873
CL = 0.95, α = 1 – 0.95 = 0.05, z α 2 z α 2 = 1.96 EBM = z 0.025 σ n z 0.025 σ n = 1.96 909200 40 909200 40 = $281,764

Alternate solution:

Using the TI-83, 83+, 84, 84+ Calculator

Press STAT and arrow over to TESTS .
Arrow down to 7:ZInterval .
Press ENTER .
Arrow to Stats and press ENTER .
σ : 909,200
x ¯ x ¯ : 568,873
Arrow down to Calculate and press ENTER .
The confidence interval is ($287,114, $850,632).
Notice the small difference between the two solutions—these differences are simply due to rounding error in the hand calculations.
We estimate with 95 percent confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.

Use the formula for EBM , solved for n : n = z 2 σ 2 E B M 2 n = z 2 σ 2 E B M 2

From the statement of the problem, you know that σ = 2.5, and you need EBM = 1.

z = z 0.035 = 1.812.

(This is the value of z for which the area under the density curve to the right of z is 0.035.)

n = z 2 σ 2 E B M 2 = 1.812 2 2.5 2 1 2 ≈ 20.52 . n = z 2 σ 2 E B M 2 = 1.812 2 2.5 2 1 2 ≈ 20.52 .

You need to measure at least 21 male students to achieve your goal.

CI: (6244, 11,014)
It will become smaller.
x ¯ x ¯ = 2.51
s x s x = 0.318
The effective length of time for a tranquilizer
The mean effective length of time of tranquilizers from a sample of nine patients
We need to use a Student’s t -distribution, because we do not know the population standard deviation.
CI: (2.27, 2.76)
Check student's solution.
If we were to sample many groups of nine patients, 95 percent of the samples would contain the true population mean length of time.

x ¯ = $ 251 , 854.23 ; x ¯ = $ 251 , 854.23 ;

s = $ 521 , 130.41 . s = $ 521 , 130.41 .

Note that we are not given the population standard deviation, only the standard deviation of the sample.

There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29.

CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04.

α 2 = 0.02 t α 2 = t 0.02 α 2 = 0.02 t α 2 = t 0.02 = 2.150.

E B M = t α 2 ( s n ) = 2.150 ( 521 , 130.41 30 ) ~ $ 204 , 561.66 . E B M = t α 2 ( s n ) = 2.150 ( 521 , 130.41 30 ) ~ $ 204 , 561.66 .

x ¯ x ¯ - EBM = $251,854.23 - $204,561.66 = $47,292.57.

x ¯ x ¯ + EBM = $251,854.23 + $204,561.66 = $456,415.89.

We estimate with 96 percent confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.

Alternate Solution

The difference between solutions arises from rounding differences.

X is the number of unoccupied seats on a single flight. X ¯ X ¯ is the mean number of unoccupied seats from a sample of 225 flights.
We will use a Student’s t-distribution, because we do not know the population standard deviation.
CI: (11.12 , 12.08)
CI: (7.64, 9.36)
The sample should have been increased.
Answers will vary.
The sample size would need to be increased, because the critical value increases as the confidence level increases.

X = the number of people who believe that the president is doing an acceptable job;

P ′ = the proportion of people in a sample who believe that the president is doing an acceptable job.

N ( 0.61 , ( 0.61 ) ( 0.39 ) 1200 ) N ( 0.61 , ( 0.61 ) ( 0.39 ) 1200 )
CI: (0.59, 0.63)
Check student’s solution.
(0.72, 0.82)
(0.65, 0.76)
(0.60, 0.72)
Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families.
X = the number of adult Americans who believe that crime is the main problem; P′ = the proportion of adult Americans who believe that crime is the main problem.
Because we are estimating a proportion, that P′ = 0.2 and n = 1,000, the distribution we should use is N ( 0.2 , ( 0.2 ) ( 0.8 ) 1000 ) N ( 0.2 , ( 0.2 ) ( 0.8 ) 1000 ) .
CI: (0.18, 0.22)
One way to lower the sampling error is to increase the sample size.
The stated ± 3 percent represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3 percent. Thus, they estimate the percentage of adult Americans who the percentage of adult Americans who that crime is the main problem to be between 18 percent and 22 percent.
p′ = (0 .55 + 0 .49) 2 (0 .55 + 0 .49) 2 = 0.52; EBP = 0.55 – 0.52 = 0.03
No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.

STAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75.

Answer is (0.502, 0.538).

Yes, this interval does not fall below 0.50, so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75 percent confidence.

CL = 0.95; α = 1 – 0.95 = 0.05; α 2 α 2 = 0.025; z α 2 z α 2 = 1.96. Use p ′ = q ′ = 0.5.

n = z α 2 2 p ′ q ′ E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.05 2 = 384.16 . n = z α 2 2 p ′ q ′ E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.05 2 = 384.16 .

You need to interview at least 385 students to estimate the proportion to within 5 percent at 95 percent confidence.

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-statistics . Changes were made to the original material, including updates to art, structure, and other content updates.

Access for free at https://openstax.org/books/statistics/pages/1-introduction

Authors: Barbara Illowsky, Susan Dean
Publisher/website: OpenStax
Book title: Statistics
Publication date: Mar 27, 2020
Location: Houston, Texas
Book URL: https://openstax.org/books/statistics/pages/1-introduction
Section URL: https://openstax.org/books/statistics/pages/8-solutions

© Jan 23, 2024 Texas Education Agency (TEA). The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Introduction to Statistics

Chapter 8 hypothesis testing.

Learning Outcome

Perform hypotheses testing involving one population mean, one population proportion, and one population standard deviations/variance.

This chapter introduces the statistical method of hypothesis testing to test a given claim about a population parameter, such as proportion, mean, standard deviation, or variance. This method combines the concepts covered in the previous chapters, including sampling distribution, standard error, critical scores, and probability theory.

8.1 Hypothesis Testing

In statistics, a hypothesis is a claim or statement about a property of a population.

A hypothesis test (or test of significance) is a procedure for testing a claim about a property of a population.

The null hypothesis ( $H_0$ ) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value.

The alternative hypothesis ( $H_A$ ) is a statement that the parameter has a value that somehow differs from the null hypothesis.

Purpose of a Hypothesis Test

The purpose of a hypothesis test is to determine how plausible the null hypothesis is. At the start of a hypothesis test, we assume that the null hypothesis is true. Then we look at the evidence, which comes from data that have been collected. If the data strongly indicate that the null hypothesis is false, we abandon our assumption that it is true and believe the alternate hypothesis instead. This is referred to as rejecting the null hypothesis.

The evidence comes in the form of a test statistic. When the difference between the test statistic (such as, $z$ or $t$ scores) and the value in the null hypothesis is sufficiently large, we reject the null hypothesis.

\[ \fbox{Assume the null hypothesis is true} \xrightarrow{} \fbox{consider the evidence} \xrightarrow{} \fbox{decide whether to accept or reject the null hypothesis} \]

Probability of getting head from a single toss of coin, $p = 0.5$ .

Therefore, expected value of the number of heads from $20$ tosses = $10$ .

Suppose, on your first trial, you have tossed a coin $20$ times and seen $15$ heads, $\hat p = 0.75$ . On your second trial, you have tossed a coin $20$ times again and seen $12$ heads, $\hat p = 0.60$ .

Is the coin fair, or is it biased towards heads?

Null and Alternative Hypotheses

Null hypothesis $(H_0)$ : states that any deviation from what was expected is due to chance error (i.e. the coin is fair).

Alternative hypothesis $(H_A)$ : asserts that the observed deviation is too large to be explained by chance alone (i.e. the coin is biased towards heads).

\[ H_0: p = 0.5 \\ H_A: p > 0.5 \]

Now, what is the probability of $p \ge 0.75?$ What is the probability of $p \ge 0.60?$

From normal approximation of the sampling distribution of $\hat p$ ,

\[ \begin{align} p &= 0.5 \\ se &= \sqrt{(0.5)(0.5)/20} = 0.112 \\ z_1 &= (0.75 - 0.50)/0.112 = 2.236 \\ \\ z_2 &= (0.60 - 0.50)/0.112 = 0.893 \\ \\ \text {p-value} &= \begin{cases} P(z\ge 2.236) &= 0.0127 \\ P(z\ge 0.893) &= 0.1860 \end{cases} \end{align} \]

We see that the farther the test statistic is from the values specified by $H_0$ , the less likely the difference is due to chance - and the less plausible becomes. The question then is: How big should the difference be before we reject $H_0$ ?

To answer this question, we need methods that enable us to calculate just how plausible $H_0$ is. Hypothesis tests provide these methods taking into account things such as the size of the sample and the amount of spread in the distribution.

Interpretation of $\text{p-value}$

A $\text{p-value}$ is the probability of obtaining the observed effect (or larger) under a “null hypothesis”. Thus, a $\text{p-value}$ that is very small indicates that the observed effect is very unlikely to have arisen purely by chance, and therefore provides evidence against the null hypothesis.

It has been common practice to interpret a $\text{p-value}$ by examining whether it is smaller than particular threshold values or “significance level”. In particular, $\text{p-values}$ less than $5\%$ are often reported as “statistically significant”, and interpreted as being small enough to justify rejection of the null hypothesis. By definition, the significance level $\alpha$ is the probability of mistakenly rejecting the null hypothesis when it is true.

\[\textbf {Significance level } \alpha = P \textbf { (rejecting } H_0 \textbf { when } H_0 \textbf { is true)} \]

In common practice, $\alpha$ is set at $10\%, 5\%$ or $1\%$ .

In the coin toss example:

p-value = $1.27\%$ which is less than the $5\%$ significance level.

Therefore, the result is statistically significant.

Conclusion: The coin is biased towards heads.

Type I and Type II Errors

When testing a null hypothesis, sometimes the test comes to a wrong conclusion by rejecting it or failing to reject it. There are two kinds of errors: type I and type II errors.

$\textbf {Type I error} :$ The error of rejecting the null hypothesis when it is actually true.

$\alpha = P (\textbf{type I error}) = P (\text{rejecting } H_0 \text{ when } H_0 \text{ is true } )$

The probability of $\text {Type I}$ can be minimized by choosing a smaller $\alpha$ .

$\textbf {Type II error} :$ The error of failing to reject the null hypothesis when it is actually false.

$\beta = P (\textbf{type II error}) = P (\text{failing to reject } H_0 \text{ when } H_0 \text{ is false} )$

The probability of $\text {Type II}$ can be minimized by choosing a larger sample size $n$ .

In general, a $\text{Type I}$ error is more serious than a $\text{Type II}$ error. This is because a $\text{Type I}$ error results in a false conclusion, while a $\text{Type II}$ error results only in no conclusion. Ideally, we would like to minimize the probability of both errors. Unfortunately, with a fixed sample size, decreasing the probability of one type increases the probability of the other.

Hypothesis tests are often designed so that the probability of a $\text{Type I}$ error will be acceptably small, often $0.05$ or $0.01$ . This value is called the significance level of the test.

Significance Level

Notice that if $H_0$ is actually true, but $\hat p$ falls in the critical region, then a $\text{Type I}$ Error occurs. We begin a hypothesis test by setting the probability of a $\text{Type I}$ Error. This value is called the significance level and is denoted by $\alpha$ . The area of critical region is equal to $\alpha$ . The choice of $\alpha$ is determined by how strong we require the evidence against $H_0$ to be in order to reject it. The smaller the value of $\alpha$ , the stronger we require the evidence to be.

Critical Value Method

In a hypothesis test, the critical value(s) separates the critical region (where we reject the null hypothesis) from the values of the test statistic that do not lead to rejection of the null hypothesis.

With the critical value method of testing hypothesis, we make a decision by comparing the test statistic to the critical value(s).

Stating the Conclusion in a Hypothesis Test

If the null hypothesis is rejected, the conclusion of the hypothesis test is straightforward: We conclude that the alternate hypothesis, $H_0$ , is true.

If the null hypothesis is not rejected, we say that there is not enough evidence to conclude that the alternate hypothesis, $H_a$ , is true. This is not saying the null hypothesis is true. What we are saying is that the null hypothesis might be true.

One-sided and two-sided tests

If the researchers are only interested in showing an increase or a decrease, but not both, use a one-sided test . If the researchers would be interested in any difference from the null value - an increase or decrease - then the test should be two-sided .

After observing data, it is tempting to turn a two-sided test into a one-sided test. Hypotheses must be set up before observing the data. If they are not, the test must be two-sided.

Steps of a Formal Test of Hypothesis

Follow these seven steps when carrying out a hypothesis test.

State the name of the test being used.
Verify conditions to ensure the standard error estimate is reasonable and the point estimate follows the appropriate distribution and is unbiased.
Write the hypotheses and set them up in mathematical notation.
Identify the significance level $\alpha$ .
Calculate the test statistics (e.g. $z$ ), using an appropriate point estimate of the paramater of interest and its standard error. \[\text{test statistics} = \frac{\text{point estimate - null value}}{\text{SE of estimate}}\]
Find the $\text{p-value}$ , compare it to $\alpha$ , and state whether to reject or not reject the null hypothesis.
Write your conclusion in context.

8.2 Power of a Hypothesis Test (Optional)

The power of a hypothesis test is the probability $1-\beta$ of rejecting a false null hypothesis. The value of the power is computed by using a particular significance level $\alpha$ and a particular value of the population parameter that is an alternative to the value of assumed true in the null hypothesis.

In practice, statistical studies are commonly designed with a statistical power of at least $80%$ .

Post-hoc Power Calculation for One Study Group vs. Population

\[ H_0 : p = P_0 \\ H_A : p \ne P_0 \\ \] \[ \begin{align} P_0 &= \text{proportion of population} \\ P_1 &= \text{proportion observed from the data (an alternative population)} \\ N &= \text{sample size} \\ \alpha &= \text{probability of type I error} \\ \beta &= \text{probability of type II error} \\ z &= \text{critical z score for a given } \alpha \text { or } \beta \end{align} \]

Suppose, $P_1$ is an alternative to the value assumed in $H_0$ .

Under $H_0$ ,

\[ P'_0 = P_0 + z_{1-\alpha/2} \cdot \sqrt{\dfrac{P_0Q_0}{N}} \]

Under $H_A$ ,

\[ \therefore z_{\beta} = \dfrac{P'_0 - P_1}{\sqrt{\dfrac{P_1Q_1}{N}}} = \dfrac{\Bigg( P_0 + z_{1-\alpha/2} \cdot \sqrt{\dfrac{P_0Q_0}{N}} \Bigg) - P_1}{\sqrt{\dfrac{P_1Q_1}{N}}} \\ \\ P(\textbf{Type II error}) = \beta = \Phi \left \{ \dfrac{\Bigg( P_0 + z_{1-\alpha/2} \cdot \sqrt{\dfrac{P_0Q_0}{N}} \Bigg) - P_1}{\sqrt{\dfrac{P_1Q_1}{N}}} \right \} \\ \]

\[ \begin{align} \text{where,} \\ Q_0 &= 1 - P_0 \\ Q_1 &= 1 - P_1 \\ \Phi &= \text{cumulative normal distribution function} \end{align} \]

Example: Calculate statistical power for various alternative hypotheses.

\[ Suppose, \begin{cases} H_0: p = 0.5 \\ H_A: p \ne 0.5 \\ P(\text{type I error}) = \alpha = 0.05 \\ \text{Critical z score, } z_{1 - \alpha/2} = 1.96 \\ N = 14 \\ \end{cases} \]

\[ \begin{array}{r|r|r} P_1 & \Phi(z_{\beta}) = \beta & 1-\beta \\ \hline 0.6 & \Phi(1.2367) = 0.8919 & 0.1081 \\ 0.7 & \Phi(0.5055) = 0.6934 & 0.3066 \\ 0.8 & \Phi(-0.3562) = 0.3608 & 0.6392 \\ 0.9 & \Phi(-1.7222) = 0.0425 & 0.9575 \\ \hline \end{array} \]

Example: Sample size calculation to achieve power (when $P_0$ and $P_1$ are known)

\[ \begin{align} z_{\beta} = \Phi^{-1}(\beta) &= \dfrac{\Bigg( P_0 + z_{1-\alpha/2} \cdot \sqrt{\dfrac{P_0Q_0}{N}} \Bigg) - P_1}{\sqrt{\dfrac{P_1Q_1}{N}}} \\ \\ z_{1-\alpha/2} &= 1.96 \\ 1- \beta &= 0.8 \\ \\ \Phi^{-1}(0.2) = -0.84 &= \dfrac{\Bigg( 0.5 + 1.96 \cdot \sqrt{\dfrac{(0.5) (0.5)}{N}} \Bigg) - 0.9}{\sqrt{\dfrac{(0.9)(0.1)}{N}}} \\ \implies N &= \Bigg( \dfrac{1.96\sqrt{0.25} + 0.84\sqrt{0.09}}{0.4} \Bigg)^2 \\ &\approx 10 \end{align} \]

Example: Sample size calculation to achieve power (when $P_0$ and $P_1$ are unknown)

\[ N = \Bigg( \dfrac{z_{1-\alpha/2} + z_{1-\beta}}{ES} \Bigg)^2 \\ \] where,

\[ \text{effect size, } ES = \dfrac{|P_1-P_0|}{\sqrt{P_0Q_0}} \]

Statistical power and design of experiment: When designing an experiment, it is essential to determine the minimize sample size that would be needed to detect an acceptable difference between the true value of the population parameter and what is observed from the data. A $5\%$ significance level $(\alpha)$ and a statistical power of at least $80\%$ are common requirements for determining that a hypothesis test is effective.

8.3 Inference for a Single Proportion

Conduct a formal hypothesis test of a claim about a population proportion $p$ .

Requirements:

The sample observations are simple random sample.
The trials are independent with two possible outcomes.
The sampling distribution for $\hat p$ , taken from a sample of size $n$ from a population with a true proportion $p$ , is nearly normal when the sample observations are independent and we expect to see at least $10$ successes and $10$ failures in our sample, i.e. $np \ge 10$ and $n(1-p) \ge 10$ . This is called the success-failure condition. If the conditions are met, then the sampling distribution of $\hat p$ is nearly normal with mean $\mu_{\hat p} = p$ and standard deviation $\sigma_{\hat p} = \sqrt {\dfrac{p(1-p)}{n}}$ .

\[ \begin{cases} n = \text {sample size } \\ \hat p = \dfrac{x}{n} \text { (sample proportion) } \\ p = \text {population proportion } \\ q = 1 - p \\ \end{cases} \]

Test Statistic for Testing a Claim About a Proportion

\[ z = \dfrac{\hat p - p}{\sqrt {\dfrac{pq}{n}} } \]

The DMV claims that $80\%$ of all drivers pass the driving test. In a survey of $90$ teens, only $61$ passed. Is there evidence that teen pass rates are significantly below $80\%?$

Let’s say, $p$ is the true population proportion.

\[ \begin{align} \text {One-tailed test} &:\\ H_0&: p = 0.80 \\ H_A&: p < 0.80 \end{align} \]

Verify success-failure condition: \[ \begin{align} np \ge 10 \rightarrow 90 \times 0.80 \ge 10 \\ n(1-p) \ge 10 \rightarrow 90 \times (1-0.80) \ge 10 \end{align} \]

Therefore, the conditions for a normal model are met.

Now, \[ \begin{align} \hat p &= \frac {61}{90} = 0.678 \\ \\ SE(\hat p) &= \sqrt \frac{pq}{n} = \sqrt \frac{(0.80)(0.20)}{90} = 0.042 \\ \\ z &= \frac {0.678-0.80}{0.042} = -2.90 \\ \\ p\text{-value} &= P(z < -2.90) = 0.002 < 0.05 \end{align} \]

Hence, we reject $H_0$ . Teen pass rate is significantly below population pass rate.

Under natural conditions, $51.7\%$ of births are male. In Punjab India’s hospital $56.9\%$ of the $550$ births were male. Is there evidence that the proportion of male births is significantly different for this hospital?

\[ \begin{align} \text {Two-tailed test} &:\\ H_0&: p = 0.517 \\ H_A&: p \ne 0.517 \end{align} \]

Verify success-failure condition: \[ \begin{align} np \ge 10 \rightarrow 550 \times 0.517 \ge 10 \\ n(1-p) \ge 10 \rightarrow 550 \times (1-0.517) \ge 10 \end{align} \]

\[ \begin{align} \hat p &= 0.569 \\ \\ SE(\hat p) &= \sqrt \frac{pq}{n} = \sqrt \frac{(0.517)(1-0.517)}{550} = 0.0213 \\ \\ z &= \frac {0.569-0.517}{0.0213} = 2.44 \\ \\ p\text{-value} &= 2 \times P(z > 2.44) = 2 \times 0.0073 = 0.0146 < 0.05 \end{align} \]

Hence, we reject $H_0$ . Male birth rate is significantly higher at the hospital than the natural birth rate.

8.4 $z \text {-test}$ | Testing Hypothesis About $\mu$ with $\sigma$ Known

The null hypothesis claims about a population mean $\mu$ .

\[ \begin{cases} \mu_{\bar x} = \text {population mean } \\ \sigma = \text {population standard deviation } \\ n = \text {size of the sample drawn from the population } \\ \bar x = \text{sample mean} \\ \end{cases} \]

The sample is simple random sample.
The population is normally distributed or $n > 30$ .

Test Statistic for Testing a Claim About a Mean

\[ z = \dfrac{\bar x - \mu_{\bar x}}{\dfrac{\sigma}{\sqrt n}} \]

The American Automobile Association reported that the mean price of a gallon of regular gasoline in the city of Los Angeles in July 2019 was $\$4.07$ . A recently taken simple random sample of $50$ gas stations had an average price of $\$4.02$ . Assume that the standard deviation of prices is $\$0.15$ . An economist is interested in determining whether the mean price is less than $\$4.07$ . Perform a hypothesis test at the $\alpha = 0.05$ level of significance.

\[ \begin{align} H_0&: \mu = 4.07 \\ H_A&: \mu < 4.07 \\ \\ \bar x &= 4.02 \\ \sigma &= 0.15 \\ SE(\bar x) &= 0.15/\sqrt {50} = 0.0212 \\ \\ z &= (4.02 - 4.07)/0.0212 = -2.36 \\ p-value &= 0.09\% < 5\% \end{align} \]

Therefore, we reject $H_0$ at the $\alpha = 0.05$ level. We conclude that the mean price of a gallon of regular gasoline in Los Angeles is less than $\$4.07$ .

8.5 $t \text {-test}$ | Testing Hypothesis About $\mu$ with $\sigma$ Not Known

\[ \begin{cases} \mu_{\bar x} = \text {population mean } \\ s = \text {sample standard deviation } \\ n = \text {size of the sample drawn from the population } \\ \bar x = \text{sample mean} \\ \end{cases} \]

\[ t_{n-1} = \dfrac{\bar x - \mu_{\bar x}}{\dfrac{s}{\sqrt n}} \]

Average weight of a mice population of a particular breed and age is $30 \text{ gm}$ . Weights recorded from a random sample of $5$ mice from that population are ${31.8, 30.9, 34.2, 32.1, 28.8}.$ Test whether the sample mean is significantly greater than the population mean.

\[ \begin{align} H_0&: \mu = 30 \\ H_A&: \mu > 30 \\ \\ \bar x &= 31.56 \\ s &= 1.9604 \\ SE(\bar x) &= 1.9604/\sqrt 5 = 0.8767 \\ \\ t &= (31.56 - 30)/0.8767 = 1.779 \\ df &= (5 -1) = 4 \\ p-value &= 7.5\% > 5\% \end{align} \]

Conclusion: $H_0$ cannot be rejected. The sample mean is not significantly greater than the population mean.

EPA recommended mirex screening is 0.08 ppm. A study of a sample of 150 salmon found an average mirex concentration of 0.0913 ppm with a std. deviation of 0.0495 ppm. Are farmed salmon contaminated beyond the permitted EPA level? Also, find a $95\%$ confidence interval for the mirex concentration in salmon.

\[ \begin{align} H_0&: \mu = 0.08 \\ H_A&: \mu > 0.08 \\ \\ \bar x &= 0.0913 \\ s &= 0.0495 \\ SE(\bar x) &= 0.0495/\sqrt {150} = 0.0040 \\ \\ t_{149} &= \dfrac{\bar x - \mu}{SE(\bar x)} = \dfrac{(0.0913 - 0.08)}{0.0040} = 2.795 \\ df &= (150 -1) = 149 \\ p-value &= P(t_{149}>2.795)= 0.29\% < 5\% \end{align} \]

Conclusion: Reject $H_0$ . The sample mean mirex level significantly higher that the EPA screening level.

8.6 $\chi^2 \text{-test}$ | Testing Hypothesis About a Variance

Caution: The method of this section applies only for samples drawn from a normal distribution. If the distribution differs even slightly from normal, this method should not be used.

Listed below are the heights (cm) for the simple random sample of female supermodels. Use a $0.01$ significance level to test the claim that supermodels have heights with a standard deviation that is less than $\sigma=7.5 \text { cm}$ for the population of women. Does it appear that heights of supermodels vary less than heights of women from the population?

\[ \text{178, 177, 176, 174, 175, 178, 175, 178} \\ \text{178, 177, 180, 176, 180, 178, 180, 176} \\ s^2 = 3.4 \]

From $\chi^2$ table,

\[ \text {The critical value of } \chi^2 = 5.229 \text { at } \alpha = 0.01. \\ \] Hence, we reject $H_0$ .

Confidence Interval Calculation:

\[ \sqrt{ \dfrac{(n-1)s^2}{\chi_R^2} } < \sigma < \sqrt{ \dfrac{(n-1)s^2}{\chi_L^2} } \\ \sqrt{ \dfrac{(16-1)3.4}{30.578} } < \sigma < \sqrt{ \dfrac{(16-1)3.4}{5.229} } \\ 1.3 \text{ cm } < \sigma < 3.1 \text { cm } \]

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

8.1 Introduction to Hypothesis Testing

This is a picture of a Dalmation dog covered in black spots. He is wearing a red color, appears to be in a nature setting, and there is a spout of water from a water fountain in the foreground.

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

A statistician will make a decision about whether these claims are true or false. This process is called hypothesis testing . A hypothesis test involves collecting data from a sample and evaluating the data. From the evidence provided by the sample data, the statistician makes a decision as to whether or not there is sufficient evidence to reject or not reject the null hypothesis.

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

Hypothesis testing consists of two contradictory hypotheses, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will:

Set up two contradictory hypotheses. Only one of these hypotheses is true and the hypothesis test will determine which of the hypothesis is most likely true.
Collect sample data. (In homework problems, the data or summary statistics will be given to you.)
Determine the correct distribution to perform the hypothesis test.
Analyze the sample data by performing calculations that ultimately will allow you to reject or not reject the null hypothesis.
Make a decision and write a meaningful conclusion.

Attribution

“Chapter 9 Introduction” in Introductory Statistics by OpenStax is licensed under a Creative Commons Attribution 4.0 International License.

school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Download Page (PDF)
Download Full Book (PDF)
Periodic Table
Physics Constants
Scientific Calculator
Reference & Cite
Tools expand_more
Readability

selected template will load here

This action is not available.

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

Last updated
Save as PDF
Page ID 1146

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

9.1: Introduction

9.2: null and alternative hypotheses.

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

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

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

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

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

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

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

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

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

9.3: Outcomes and the Type I and Type II Errors

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

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

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

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

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

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

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

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

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

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

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

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

9.4: Distribution Needed for Hypothesis Testing

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

9.5: Rare Events, the Sample, Decision and Conclusion

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

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

9.6: Additional Information and Full Hypothesis Test Examples

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

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

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

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

So we won't ever see them cry.

I myself don't think she's correct,

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

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

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

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

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

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

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

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

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

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

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

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

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

Price per box of Mac and Cheese:

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

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

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

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

THE CHARACTERS (in order of appearance):

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

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

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

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

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

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

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

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

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

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

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

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

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

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

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

(Curtain falls)

"Untitled," by Stephen Chen

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

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

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

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

"Japanese Girls’ Names"

by Kumi Furuichi

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

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

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

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

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

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

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

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

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

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

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

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

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

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

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

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

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

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

After conducting the test, your decision and conclusion are

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

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

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

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

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

Instructions: For the following ten exercises,

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

State the null and alternate hypothesis.

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

State $\alpha$.

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

9.7: Hypothesis Testing of a Single Mean and Single Proportion

Snapsolve any problem by taking a picture. Try it in the Numerade app?

Elementary Statistics

Mario f. triola, hypothesis testing - all with video answers.

Basics of Hypothesis Testing

A package label includes a claim that the mean weight of the M&Ms is 0.8535 g, and another package label includes the claim that the mean amount of aspirin in Bayer tablets is 325 mg. Which has more serious implications: rejection of the M&M claim or rejection of the aspirin claim? Is it wise to use the same significance level for hypothesis tests of both claims?

Data Set 20 in Appendix B includes sample weights of the M&Ms referenced in Exercise 1. We could use methods of Chapter 7 for making an estimate, or we could use those values to test some claim. What is the difference between estimating and hypothesis testing?

A formal hypothesis test is to be conducted using the claim that the mean body temperature is equal to 98.6 ° F. a. What is the null hypothesis, and how is it denoted? b. What is the alternative hypothesis, and how is it denoted? c. What are the possible conclusions that can be made about the null hypothesis? d. Is it possible to conclude that “there is sufficient evidence to support the claim that the mean body temperature is equal to 98.6 ° F ”?

When the clinical trial of the XSORT method of gender selection is completed, a formal hypothesis test will be conducted with the alternative hypothesis of p > 0.5 , which corresponds to the claim that the XSORT method increases the likelihood of having a girl, so that the proportion of girls is greater than 0.5. If you are responsible for developing the XSORT method and you want to show its effectiveness, which of the following P values would you prefer: 0.999, 0.5, 0.95, 0.05, 0.01, 0.001? Why?

Do the following: a. Express the original claim in symbolic form. b. Identify the null and alternative hypotheses. Claim: 20% of adults smoke. A recent Gallup survey of 1016 randomly selected adults showed that 21% of the respondents smoke.

Do the following: a. Express the original claim in symbolic form. b. Identify the null and alternative hypotheses. Claim: When parents use the XSORT method of gender selection, the proportion of baby girls is greater than 0.5. The latest actual results show that among 945 babies born to couples using the XSORT method of gender selection, 879 were girls.

Do the following: a. Express the original claim in symbolic form. b. Identify the null and alternative hypotheses. Claim: The mean pulse rate (in beats per minute) of adult females is 76 or lower. For the random sample of adult females in Data Set 1 from Appendix B, the mean pulse rate is 77.5.

Do the following: a. Express the original claim in symbolic form. b. Identify the null and alternative hypotheses. Claim: The standard deviation of pulse rates of adult women is at least 50. For the random sample of adult females in Data Set 1 from Appendix B, the pulse rates have a standard deviation of 11.6.

Refer to the exercise identified. Using only the rare event rule, make subjective estimates to determine whether results are likely, then state a conclusion about the original claim. For example, if the claim is that a coin favors heads and sample results consist of 11 heads in 20 flips, conclude that there is not sufficient evidence to support the claim that the coin favors heads (because it is easy to get 11 heads in 20 flips by chance with a fair coin). Exercise 5

Find the value of the test statistic. (Refer to Table 82 to select the correct expression for evaluating the test statistic.) Claim: Three fourths of all adults believe that it is important to be involved in their communities. Based on a USA Today/Gallup poll of 1021 randomly selected adults, 89% believe that it is important to be involved in their communities.

Find the value of the test statistic. (Refer to Table 82 to select the correct expression for evaluating the test statistic.) Claim: Among those who file tax returns, less than one half file them through an accountant or other tax professional. A Follows survey of 1002 people who file tax returns showed that 48% of them file through an accountant or other tax professional.

Find the value of the test statistic. (Refer to Table 82 to select the correct expression for evaluating the test statistic.) Claim: For adult females, the standard deviation of their white blood cell counts is equal to 5.00. The random sample of 40 adult females in Data Set 1 from Appendix B has white blood cell counts with a standard deviation of 2.28.

Find the value of the test statistic. (Refer to Table 82 to select the correct expression for evaluating the test statistic.) Claim: For adult females, the mean of their white blood cell counts is equal to 8.00. The random sample of 40 adult females in Data Set 1 from Appendix B has white blood cell counts with a mean of 7.15 and a standard deviation of 2.28. (The population standard deviation ? is not known.)

Assume that the significance level is ? = 0.05 ; use the given statement and find the P value and critical values. (See Figure 84.) The test statistic of z = 2.00 is obtained when testing the claim that p > 0.5 .

Assume that the significance level is ? = 0.05 ; use the given statement and find the P value and critical values. (See Figure 84.) The test statistic of z = ? 2.00 is obtained when testing the claim that p < 0.5 .

Assume that the significance level is ? = 0.05 ; use the given statement and find the P value and critical values. (See Figure 84.) The test statistic of z = ? 1.75 is obtained when testing the claim that p = 1 / 3 .

Assume that the significance level is ? = 0.05 ; use the given statement and find the P value and critical values. (See Figure 84.) The test statistic of z = 1.50 is obtained when testing the claim that p ? 0.25 .

Assume that the significance level is ? = 0.05 ; use the given statement and find the P value and critical values. (See Figure 84.) With $\mathrm{H} 1: \mathrm{p} \neq 0.25,$ the test statistic is $z=-1.23$

Assume that the significance level is ? = 0.05 ; use the given statement and find the P value and critical values. (See Figure 84.) With $\mathrm{H} 1: \mathrm{p} \neq 2 / 3,$ the test statistic is $z=2.50$

Assume that the significance level is ? = 0.05 ; use the given statement and find the P value and critical values. (See Figure 84.) With $\mathrm{H} 1: \mathrm{p}<0.6,$ the test statistic is $z=-3.00$

Assume that the significance level is ? = 0.05 ; use the given statement and find the P value and critical values. (See Figure 84.) With $\mathrm{H} 1: \mathrm{p}>7 / 8,$ the test statistic is $z=2.88$

Assume a significance level of ? = 0.05 and use the given information for the following: a. State a conclusion about the null hypothesis. (Reject H 0 or fail to reject H 0 . ) b. Without using technical terms, state a final conclusion that addresses the original claim. Original claim: The percentage of blue M&Ms is greater than 5%. The hypothesis test results in a P-value of 0.0010.

Assume a significance level of ? = 0.05 and use the given information for the following: a. State a conclusion about the null hypothesis. (Reject H 0 or fail to reject H 0 . ) b. Without using technical terms, state a final conclusion that addresses the original claim. Original claim: Fewer than 20% of M&M candies are green. The hypothesis test results in a P-value of 0.0721.

Assume a significance level of ? = 0.05 and use the given information for the following: a. State a conclusion about the null hypothesis. (Reject H 0 or fail to reject H 0 . ) b. Without using technical terms, state a final conclusion that addresses the original claim. Original claim: Women have heights with a mean equal to 160.00 cm. The hypothesis test results in a Pvalue of 0.0614.

Assume a significance level of ? = 0.05 and use the given information for the following: a. State a conclusion about the null hypothesis. (Reject H 0 or fail to reject H 0 . ) b. Without using technical terms, state a final conclusion that addresses the original claim. Original claim: Women have heights with a standard deviation equal to 5.00 cm. The hypothesis test results in a Pvalue of 0.0055.

Use the given information to answer the following: a. Identify the null hypothesis and the alternative hypothesis. b. What is the value of ? ? c. What is the sampling distribution of the sample statistic? d. Is the test twotailed, lefttailed, or righttailed? e. What is the value of the test statistic? f. What is the Pvalue? g. What is the critical value? h. What is the area of the critical region? A 0.01 significance level is used for a hypothesis test of the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is greater than 0.5. Assume that sample data consist of 55 girls born in 100 births, so the sample statistic of 0.55 results in a z score that is 1.00 standard deviation above 0.

Use the given information to answer the following: a. Identify the null hypothesis and the alternative hypothesis. b. What is the value of ? ? c. What is the sampling distribution of the sample statistic? d. Is the test twotailed, lefttailed, or righttailed? e. What is the value of the test statistic? f. What is the Pvalue? g. What is the critical value? h. What is the area of the critical region? A 0.05 significance level is used for a hypothesis test of the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is different from 0.5. Assume that sample data consist of 55 girls born in 100 births, so the sample statistic of 0.55 results in a z score that is 1.00 standard deviation above 0.

Identify expressions that identify the type I error and the type II error that correspond to the given claim. (Although conclusions are usually expressed in verbal form, the answers here can be expressed with statements that include symbolic expressions such as p = 0.1. ) The proportion of people who write with their left hand is equal to 0.1.

Identify expressions that identify the type I error and the type II error that correspond to the given claim. (Although conclusions are usually expressed in verbal form, the answers here can be expressed with statements that include symbolic expressions such as p = 0.1. ) With $\mathrm{H} 1: \mathrm{p}>7 / 8,$ the test statistic is $z=2.88$

Chantix tablets are used as an aid to help people stop smoking. In a clinical trial, 129 subjects were treated with Chantix twice a day for 12 weeks, and 16 subjects experienced abdominal pain (based on data from Pfizer, Inc.). If someone claims that more than 8% of Chantix users experience abdominal pain, that claim is supported with a hypothesis test conducted with a 0.05 significance level. Using 0.18 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test.

Consider a hypothesis test of the claim that the MicroSort method of gender selection is effective in increasing the likelihood of having a baby girl, so that the claim is p > 0.5 . Assume that a significance level of ? = 0.05 is used, and the sample is a simple random sample of size n = 64. a. Assuming that the true population proportion is 0.65, find the power of the test, which is the probability of rejecting the null hypothesis when it is false. (Hint: With a 0.05 significance level, the critical value is z = 1.645 , so any test statistic in the right tail of the accompanying top graph is in the rejection region where the claim is supported. Find the sample proportion p ^ in the top graph, and use it to find the power shown in the bottom graph.) b. Explain why the redshaded region of the bottom graph represents the power of the test. (FIGURE CAN'T COPY)

Researchers plan to conduct a test of a gender-selection method. They plan to use the alternative hypothesis of H 1 : p > 0.5 and a significance level of ? = 0.05 . Find the sample size required to achieve at least 80% power in detecting an increase in p from 0.5 to 0.55. (This is a very difficult exercise. Hint: See Exercise 36.)

IMAGES

CHAPTER 8: Hypothesis Testing
PPT
Chapter 8 Hypothesis Testing I
Chapter 8 Hypothesis Testing
PPT
Chapter 8

VIDEO

Chapter 8 Hypothesis testing
Testing Of Hypothesis L-3
Chapter 8 Introduction to Hypothesis Testing
[Chapter 8] Hypothesis Testing Part 1
Math 14 8.2.18-T Identify the null hypothesis, alternative hypothesis, test statistic
regression analysis simplify complex data relationships coursera weekly challenges 4 answers

COMMENTS

PDF Introduction to Hypothesis Testing
Answers: 1. The population mean; 2. False. Researchers select a sample from a population to learn more about ... CHAPTER 8: INTRODUCTION TO HYPOTHESIS TESTING 5 Step 2: Set the criteria for a decision. To set the criteria for a decision, we state the level of significance for a test. This is similar to the criterion that jurors use in a
Chapter 8: Hypothesis Testing
29 Multiple choice questions. Definition. state the hypotheses. The power of the decision-making process is. A measure of the size of an effect in a population is called. The one-sample z test is a hypothesis test used to test hypotheses. The first step to hypothesis testing requires that a researcher. 1 of 29.
Chapter 8: Hypothesis Testing Flashcards
Steps of a Hypothesis Test. (Two Tailed) Step 1: State the Hypothesis and select the alpha level. Step 2: Set the location criteria by locating the critical region. a) if the alpha level is .05 then look at the z-score table and find .025 proportion in the tail to find the corresponding z-score aka the critical region.
Chapter 8: Introduction to hypothesis testing
hypothesis testing involves set of logical procedures and rules that enable us to make general statements about a population when all we have is sample data. This logic is reflected in 4 steps: 1. state hypothesis and set alpha level 2. locate critical region 3. compute test statistic (in this case, z-score) for sample 4. make decision about H0 ...
PDF Lecture #8 Chapter 8: Hypothesis Testing 8-2 Basics of hypothesis
This chapter introduces another major topic of inferential statistics: testing claims (or hypothesis) made about population parameters. 8-2 Basics of hypothesis testing In this section, 1st we introduce the language of hypothesis testing, then we discuss the formal process of testing a hypothesis. A hypothesis is a statement or claim regarding ...
Ch. 8 Solutions
9.6 Hypothesis Testing of a Single Mean and Single Proportion; Key Terms; Chapter Review; Formula Review; Practice; Homework; References; Solutions; 10 Hypothesis Testing with Two Samples. Introduction; 10.1 Two Population Means with Unknown Standard Deviations; ... Answer is (0.502, 0.538). Yes, this interval does not fall below 0.50, so we ...
Chapter 8, Hypothesis Testing Video Solutions ...
Video answers for all textbook questions of chapter 8, Hypothesis Testing, Understandable Statistics, Concepts and Methods by Numerade ... Chapter 8 Hypothesis Testing - all with Video Answers. Educators. Section 1. Introduction to Statistical Tests 06:02 ...
Chapter 8 Hypothesis Testing
8.1 Hypothesis Testing. In statistics, a hypothesis is a claim or statement about a property of a population.. A hypothesis test (or test of significance) is a procedure for testing a claim about a property of a population.. The null hypothesis ($H_0$) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value.
8.1: Steps in Hypothesis Testing
Figure 8.1.1 8.1. 1: You can use a hypothesis test to decide if a dog breeder's claim that every Dalmatian has 35 spots is statistically sound. (Credit: Robert Neff) A statistician will make a decision about these claims. This process is called "hypothesis testing." A hypothesis test involves collecting data from a sample and evaluating the data.
8.1 Introduction to Hypothesis Testing
This process is called hypothesis testing . A hypothesis test involves collecting data from a sample and evaluating the data. From the evidence provided by the sample data, the statistician makes a decision as to whether or not there is sufficient evidence to reject or not reject the null hypothesis. In this chapter, you will conduct hypothesis ...
Chapter 8, Introduction to Hypothesis Testing Video Solutions
If there actually has been a 7-point increase in the average IQ during the past ten years, then find the power of the hypothesis test for each of the following. a. The researcher uses a two-tailed hypothesis test with $\alpha=.05$ to determine whether the data indicate a significant change in IQ over the past 10 years. b.
MINDTAP
Terms in this set (18) step 1. The 4 steps of a hypothesis test: _____ - statement of the hypothesis. step 2. The 4 steps of a hypothesis test: ____ - setting of the criteria for a decision. step 3. The 4 steps of a hypothesis test: ____ - collection of data and computation of sample statistics. step 4.
Chapter 8, Hypothesis Testing Video Solutions, Essentials of Statistics
Problem 10. Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value (s), conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise.
Solved Chapter 8: Introduction to Hypothesis Testing What
Math. Statistics and Probability. Statistics and Probability questions and answers. Chapter 8: Introduction to Hypothesis Testing What is a hypothesis test? What is the null hypothesis and what is the alternative hypothesis? Draw a distribution showing where sample means are likely to be obtained and where they are unlikely to be obtained.
PDF Triola Chapter 8
The General Idea of a Hypothesis Test To run a hypothesis test, there are a few general steps, which will be elaborated on later. 1. Based on the question you want to answer, formulate the test as a choice between two hypotheses (the null and alternative). 2. Find a test statistic whose distribution is known when the null hypothesis H 0 is true. 3.
Chapter 8 Introduction to Hypothesis Testing Mastery Training
Chapter 8 Introduction to Hypothesis Testing Mastery Training. The four steps of a hypothesis test: step 1. Click the card to flip 👆. statement of the hypothesis. Click the card to flip 👆. 1 / 18.
9.E: Hypothesis Testing with One Sample (Exercises)
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.
Chapter 8, Hypothesis Testing Video Solutions, Elementary Statistics
Video answers for all textbook questions of chapter 8, Hypothesis Testing, Elementary Statistics by Numerade ... Is it valid to use these sample results for testing the claim that the majority of the general population answers "yes"? Sheryl Ezze Numerade Educator 01:12. Problem 4 ...
Chapter 8: Hypothesis Testing
Chapter 8: Hypothesis Testing . Section 8.1 . Note: For all graphs provided, the P value is indicated by the shaded portion in the tails. 1. See text for definitions. Essays may include (a) A working hypothesis about the population parameter in question is called the null hypothesis. The value specified in the null hypothesis is often a historical value, a claim, or a production specification.
Chapter 8: Introduction to Hypothesis Testing Flashcards
Improve your grades and reach your goals with flashcards, practice tests and expert-written solutions today. ... For "Chapter 8: Introduction to Hypothesis Testing" Password. Enter Password. About us. About Quizlet; How Quizlet works; Careers; Advertise with us; Get the app; For students. Flashcards; Test; Learn;
Chapter 8, Hypothesis Testing Video Solutions, Essentials of ...
Estimates and Hypothesis Tests Data Set 3 "Body Temperatures" in Appendix B includes sample body temperatures. We could use methods of Chapter 7 for making an estimate, or we could use those values to test the common belief that the mean body temperature is $98.6^{\circ} \mathrm{F}$. What is the difference between estimating and hypothesis testing?
Chapter 8, Sampling Distributions and Hypothesis Testing Video
In the exercises in Chapter 2, we discussed a study of alkowances in fourth-grade children. We considered that study again in Chapter 4 , where you generated data that might have been found in such a study. a) Consider how you would go about testing the research hypothesis that boys receive more allowance than girls. What would be the null ...
Chapter 8, Hypothesis Testing Video Solutions, Elementary ...
a. State a conclusion about the null hypothesis. (Reject H 0 or fail to reject H 0 . b. Without using technical terms, state a final conclusion that addresses the original claim. Original claim: Women have heights with a standard deviation equal to 5.00 cm. The hypothesis test results in a Pvalue of 0.0055.