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.

Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.

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.

Suppose a random sample of $500$ television ads in the United States reveals that $122$ of these ads use humor. Find a point estimate of and a $95$ percent confidence interval for the proportion of all U.S. 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 facts that the sample mean and sample standard deviation are $7.4$ and $1.026$, respectively,

(1) Find a $95$ percent confidence interval for the population mean alert time, $\mu$, for the new panel. Assume normality.

(2) Can we be $95$ percent confident that $\mu$ is less than $8$ seconds? Explain.

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?

The mean and the standard deviation of the sample of $40$ trash bag breaking strengths in the table are $50.575$ and $1.6438$, respectively.

Assuming that the population of all individual trash bag breaking strengths is normally distributed, estimate tolerance intervals of the form $[\mu \pm z_{\alpha/2}\sigma]$ that contain (1) $95$ percent of all individual trash bag breaking strengths and (2) $98.8$ percent of all individual trash bag breaking strengths.

Consider the trash bag problem. Suppose that an independent laboratory has tested trash bags and has found that no $30$-gallon bags that are currently on the market have a mean breaking strength of $50$ pounds or more. On the basis of these results, the producer of the new, improved trash bag feels sure that its $30$-gallon bag will be the strongest such bag on the market if the new trash bag's mean breaking strength can be shown to be at least $50$ pounds. The mean of the sample of $40$ trash bag breaking strengths in Table $1.10$ is $50.575$. If we let $\mu$ denote the mean of the breaking strengths of all possible trash bags of the new type and assume that the population standard deviation equals $1.65$:

Based on your answers to parts $b$ and $c$, how convinced are you that the new $30$-gallon trash bag is the strongest such bag on the market?

The mean and the standard deviation of the sample of $40$ trash bag breaking strengths are $\bar{x}=50.575$ and $s=1.6438$.

What does the histogram in the figure before say about whether the Empirical Rule should be used to describe the trash bag breaking strengths?

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.

Use the confidence intervals you computed in parts $a$ and $b$ to compare the mean audit delay for all public owner-controlled companies versus that of all public manager-controlled companies. How do the means compare? Explain.

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 $=\mathbf{0 . 0 0 6}$

When the number of trials, $n$, is large, binomial probability tables may not be available. Furthermore, if a computer is not available, hand calculations will be tedious. As an alternative, the Poisson distribution can be used to approximate the binomial distribution when $n$ is large and $p$ is small. Here the mean of the Poisson distribution is taken to be $\mu=n p$. That is, when $n$ is large and $p$ is small, we can use the Poisson formula with $\mu=n p$ to calculate binomial probabilities, and we will obtain results close to those we would obtain by using the binomial formula. A common rule is to use this approximation when $n / p \geq 500$.

To illustrate this approximation, in the movie Coma, a young female intern at a Boston hospital was very upset when her friend, a young nurse, went into a coma during routine anesthesia at the hospital. Upon investigation, she found that $10$ of the last $30,000$ healthy patients at the hospital had gone into comas during routine anesthesias. When she confronted the hospital administrator with this fact and the fact that the national average was $6$ out of $100,000$ healthy patients going into comas during routine anesthesias, the administrator replied that $10$ out of $30,000$ was still quite small and thus not that unusual.

Given the hospital’s record and part a, what conclusion would you draw about the hospital’s medical practices regarding anesthesia?

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.