In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of $400$ television ads in the United Kingdom reveals that $142$ use humor, while a random sample of $500$ television ads in the United States reveals that $122$ use humor.

Test the hypotheses you set up in part $a$ by using critical values and by setting $\alpha$ equal to $.10$, $.05$, $.01$, and $.001$. How much evidence is there that the proportions of U.K. and U.S. ads using humor are different?

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.

Do the confidence intervals you computed in parts $a$ and $b$ suggest that a greater percentage of U.K. ads use humor? Explain. How might an ad agency use this information?

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.

Suppose that a random sample of $400$ television ads in the United Kingdom reveals that $142$ of these ads use humor. Find a point estimate of and a $95$ percent confidence interval for the proportion of all U.K. television ads that use humor.

Air traffic controllers have the crucial task of ensuring that aircraft don't collide. To do this, they must quickly discern when two planes are about to enter the same air space at the same time. They are aided by video display panels that track the aircraft in their sector and alert the controller when two flight paths are about to converge. The display panel currently in use has a mean "alert time" of 15 seconds. (The alert time is the time elapsing between the instant when two aircraft enter into a collision course and when a controller initiates a call to reroute the planes.) According to Ralph Rudd, a supervisor of air traffic controllers at the Greater Cincinnati International Airport, a new display panel has been developed that uses artificial intelligence to project a plane's current flight path into the future. This new panel provides air traffic controllers with an earlier warning that a collision is likely. It is hoped that the mean "alert time," $\mu$, for the new panel is less than 8 seconds. In order to test the new panel, 15 randomly selected air traffic controllers are trained to use the panel and their alert times for a simulated collision course are recorded. The sample alert times (in seconds) are: $7.2,7.5,8.0,6.8,7.2,8.4,5.3,7.3,7.6,7.1,9.4,6.4,7.9,6.2,8.7$. Using the fact that $\bar{x}=7.4$ and $s=1.026$, find a 95 percent confidence interval for the population mean alert time, $\mu$, for the new panel.

Air traffic controllers have the crucial task of ensuring that aircraft don't collide. To do this, they must quickly discern when two planes are about to enter the same air space at the same time. They are aided by video display panels that track the aircraft in their sector and alert the controller when two flight paths are about to converge. The display panel currently in use has a mean "alert time" of 15 seconds. (The alert time is the time elapsing between the instant when two aircraft enter into a collision course and when a controller initiates a call to reroute the planes.) According to Ralph Rudd, a supervisor of air traffic controllers at the Greater Cincinnati International Airport, a new display panel has been developed that uses artificial intelligence to project a plane's current flight path into the future. This new panel provides air traffic controllers with an earlier warning that a collision is likely. It is hoped that the mean "alert time," $\mu$, for the new panel is less than 8 seconds. In order to test the new panel, 15 randomly selected air traffic controllers are trained to use the panel and their alert times for a simulated collision course are recorded. The sample alert times (in seconds) are: $7.2,7.5,8.0,6.8,7.2,8.4,5.3,7.3,7.6,7.1,9.4,6.4,7.9,6.2,8.7$. Can we be 95 percent confident that $\mu$ is less than 8 seconds?

In an article in Accounting and Business Research, Carslaw and Kaplan investigate factors that influence "audit delay" for firms in New Zealand. Audit delay, which is defined to be the length of time (in days) from a company's financial year-end to the date of the auditor's report, has been found to affect the market reaction to the report. This is because late reports often seem to be associated with lower returns and early reports often seem to be associated with higher returns.

Carslaw and Kaplan investigated audit delay for two kinds of public companies-owner-controlled and manager-controlled companies. Here a company is considered to be owner-controlled if $30$ percent or more of the common stock is controlled by a single outside investor (an investor not part of the management group or board of directors). Otherwise, a company is considered manager controlled. It was felt that the type of control influences audit delay. To quote Carslaw and Kaplan:

Large external investors, having an acute need for timely information, may be expected to pressure the company and auditor to start and to complete the audit as rapidly as practicable.

Suppose that a random sample of $100$ public owner-controlled companies in New Zealand is found to give a mean audit delay of $82.6$ days, and assume that the population standard deviation equals $33$ days. Calculate a $95$ percent confidence interval for the population mean audit delay for all public owner-controlled companies in New Zealand.

Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter $2005$), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of $n = 8$ students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table $10.5$. Assume that the population of all possible paired differences is normally distributed.

Table $10.5$ Weekly Study Time Data for Students Who Perform Well on the MidTerm

Use the p-value to test the hypotheses at the $.10$, $.05$, and $.01$ levels of significance. How much evidence is there against the null hypothesis?

Paired T-Test and CI: StudyBefore, StudyAfter

$$\begin{array}{lrrrr} \text{Paired T for} & & \text{StudyBefore}&\text{- StudyAfter}\\ & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { StudyBefore } & 8 & 15.6250 & 1.9955 & 0.7055 \\ \text { StudyAfter } & 8 & 11.5000 & 3.4226 & 1.2101 \\ \text { Difference } & 8 & 4.12500 & 2.99702 & 1.05961\end{array}$$

$95 \%$ CI for mean difference: $(1.61943,6.63057)$ $T$-Test of mean difference $=0$ (vs not $=0$ ): $\mathrm{T}$-Value $=3.89 \quad \mathrm{P}$-Value $=\mathbf{0 . 0 0 6}$

Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter $2005$), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of $n = 8$ students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table $10.5$. Assume that the population of all possible paired differences is normally distributed.

Table $10.5$ Weekly Study Time Data for Students Who Perform Well on the MidTerm

The Minitab output for the paired differences test is presented below. Use the output and critical values to test the hypotheses at the $.10$, $.05$, and $.01$ levels of significance. Has the population mean study time changed?

Paired T-Test and CI: StudyBefore, StudyAfter

$$\begin{array}{lrrrr} \text{Paired T for} & & \text{StudyBefore}&\text{- StudyAfter}\\ & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { StudyBefore } & 8 & 15.6250 & 1.9955 & 0.7055 \\ \text { StudyAfter } & 8 & 11.5000 & 3.4226 & 1.2101 \\ \text { Difference } & 8 & 4.12500 & 2.99702 & 1.05961\end{array}$$

$95 \%$ CI for mean difference: $(1.61943,6.63057)$ $T$-Test of mean difference $=0$ (vs not $=0$ ): $\mathrm{T}$-Value $=3.89 \quad \mathrm{P}$-Value $=\mathbf{0 . 0 0 6}$

Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter $2005$), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of $n = 8$ students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table $10.5$. Assume that the population of all possible paired differences is normally distributed.

Table $10.5$ Weekly Study Time Data for Students Who Perform Well on the MidTerm

Set up the null and alternative hypotheses to test whether there is a difference in the population mean study time before and after the midterm exam.

National Paper Company must purchase a new machine for producing cardboard boxes. The company must choose between two machines. The machines produce boxes of equal quality, so the company will choose the machine that produces (on average) the most boxes. It is known that there are substantial differences in the abilities of the company's machine operators. Therefore, National Paper has decided to compare the machines using a paired difference experiment. Suppose that eight randomly selected machine operators produce boxes for one hour using machine 1 and for one hour using machine 2, with the following results: Ds BoxYield

Estimate the minimum and maximum differences between the mean outputs of the two machines. Justify your answer.

National Paper Company must purchase a new machine for producing cardboard boxes. The company must choose between two machines. The machines produce boxes of equal quality, so the company will choose the machine that produces (on average) the most boxes. It is known that there are substantial differences in the abilities of the company's machine operators. Therefore, National Paper has decided to compare the machines using a paired difference experiment. Suppose that eight randomly selected machine operators produce boxes for one hour using machine 1 and for one hour using machine 2, with the following results: Ds BoxYield

Assuming normality, perform a hypothesis test to determine whether there is a difference between the mean hourly outputs of the two machines. Use $\alpha=.05$

Suppose a sample of $11$ paired differences that has been randomly selected from a normally distributed population of paired differences yields a sample mean of $103.5$ and a sample standard deviation of $5.$

Test the null hypothesis $H_0: \mu_d \geq 110$ versus $H_a: \mu_d<110$ by setting $\alpha$ equal to $.05$ and $.01$. How much evidence is there that $\mu_d=\mu_1-\mu_2$ is less than $110$?

A population has a mean $\mu$=100 and a standard deviation $\sigma$=15. Find the mean and standard deviation of a sampling distribution of sample means with the given sample size n.

n=100

In a study of performance ratings of ex-smokers, a random sample of 34 ex-smokers had a mean rating of 2.21 and a sample standard deviation of 2.21. For an independent random sample of 86 long-term ex-smokers, the mean rating was 1.47 and the sample standard deviation was 1.69. Find the lowest level of significance at which the null hypothesis of equality of the two population means can be rejected against a two-sided alternative.