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.

Calculate a $95$ percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be $95$ percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor?

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.

In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspapers, radio, and so forth.

Compare the confidence intervals in parts $a$ and $b$. Does it appear that the mean percentage share of billing volume from spot television commercials for all U.S. advertising agencies is greater than the mean percentage share of billing volume from network television for all U.S. advertising agencies? Explain.

For decades, the Federal Trade Commission (FTC) resolved fair trade and advertising disputes through individual adjudications. In the 1960s, the FTC began promulgating rules that defined fair and unfair trade practices. In cases involving violations of these rules, the due process rights of participants were more limited and did not include cross-examination. Although anyone charged with violating a rule would receive a full adjudication, the legitimacy of the rule itself could not be challenged in the adjudication. Furthermore, a party charged with violating a rule was almost certain to lose the adjudication. Affected parties complained to a court, arguing that their rights before the FTC were unduly limited by the new rules. What would the court examine to determine whether to uphold the new rules?

Advertising researchers have developed a theory that states that commercials that appear in violent television shows are less likely to be remembered and will thus be less effective. After examining samples of viewers who watch violent and nonviolent programs and asking them a series of five questions about the commercials, the researchers produced the following probability distributions of the number of correct answers.

Calculate the mean and standard deviation of the number of correct answers among viewers of nonviolent television programs.

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 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 hypotheses needed to attempt to establish that the difference between the proportions of U.K. and U.S. ads using humor is more than $.05$ (five percentage points). Test these hypotheses by using a $p$-value and by setting $\alpha$ equal to $.10$, $.05$, $.01$, and $.001$. How much evidence is there that the difference between the proportions exceeds $.05?$

In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspapers, radio, and so forth.

Suppose that a random sample of $400$ U.S. advertising agencies gives an average percentage share of billing volume from network television equal to $7.46$ percent, and assume that the population standard deviation equals $1.42$ percent. Calculate a $95$ percent confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies.

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?

A marketing research firm wishes to compare the prices charged by two supermarket chains-Miller's and Albert's. The research firm, using a standardized one-week shopping plan (grocery list), makes identical purchases at $10$ of each chain's stores. The stores for each chain are randomly selected, and all purchases are made during a single week.

The shopping expenses obtained at the two chains, along with box plots of the expenses, are as follows:

$$\begin{array}{lllll}\text { Miller's } & & & & \\ \$ 119.25 & \$ 121.32 & \$ 122.34 & \$ 120.14 & \$ 122.19 \\ \$ 123.71 & \$ 121.72 & \$ 122.42 & \$ 123.63 & \$ 122.44\end{array}$$

$$\begin{array}{lllll}\text { Albert's } & & & & \\\$ 111.99 & \$ 114.88 & \$ 115.11 & \$ 117.02 & \$ 116.89 \\ \$ 116.62 & \$ 115.38 & \$ 114.40 & \$ 113.91 & \$ 111.87\end{array}$$

Because the stores in each sample are different stores in different chains, it is reasonable to assume that the samples are independent, and we assume that weekly expenses at each chain are normally distributed.

Figure $10.5$ gives the $p$-value for testing $H_0: \mu_M-\mu_A=$ 0 versus $H_a: \mu_M-\mu_A \neq 0$. Use the $p$-value to test $H_0$ versus $H_a$ by setting $\alpha$ equal to $.10$, $.05$, $.01$, and $.001$. How much evidence is there that the mean weekly expenses at Miller's and Albert's differ?

Figure $10.5$ Minitab Output of Testing the Equality of Mean Weekly Expenses at Miller’s and Albert’s Supermarket Chains Two-sample $\mathrm{T}$ for Millers vs Alberts

$$\begin{array}{lrrrr} & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { Millers } & 10 & 121.92 & 1.40 & 0.44 \\ \text { Alberts } & 10 & 114.81 & 1.84 & 0.58\end{array}$$

Difference $=\mathbf{m u}$ (Millers) $-\mathbf{m u}$ (Alberts)$\quad$ Estimate for difference: $7.10900$ $95 \%$ CI for difference: $(5.57350,8.64450)$ T-Test of diff $=0($ vs not $=):$ T-Value $=9.73 \quad \mathrm{P}$-Value $=0.000 \quad \mathrm{DF}=18$ Both use Pooled StDev $=1.6343$

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

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.

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?