In the book Cases in Finance, Nunnally and Plath present a case in which the estimated percentage of uncollectible accounts varies with the age of the account. Here the age of an unpaid account is the number of days elapsed since the invoice date.

An accountant believes that the percentage of accounts that will be uncollectible increases as the ages of the accounts increase. To test this theory, the accountant randomly selects independent samples of $500$ accounts with ages between $31$ and $60$ days and $500$ accounts with ages between $61$ and $90$ days from the accounts receivable ledger dated one year ago. When the sampled accounts are examined, it is found that $10$ of the $500$ accounts with ages between $31$ and $60$ days were eventually classified as uncollectible, while $27$ of the $500$ accounts with ages between $61$ and $90$ days were eventually classified as uncollectible. Let $p_1$ be the proportion of accounts with ages between $31$ and $60$ days that will be uncollectible, and let $p_2$ be the proportion of accounts with ages between $61$ and $90$ days that will be uncollectible.

Use the Minitab output below to determine how much evidence there is that we should reject $H_0: p_1-p_2=0$ in favor of $H_a: p_1-p_2 \neq 0$.

$$\begin{array}{lcccl} \text {Sample } & \mathrm{X} & \mathrm{N} & \text { Sample } \mathrm{p} & \\ 1 \text { (31 to } 60 \text { days) } & 10 & 500 & 0.020000 & \text { Difference }=\mathrm{p}(1)-\mathrm{p}(2) \\ 2 \text { (61 to 90 days } & 27 & 500 & 0.054000 & \text { Estimate for difference: }-0.034 \\ 95 \% \text { CI for difference: }(-0.0573036,-0.0106964) & \\ \text {Test for difference }=0 \text { (vs not }=0): \quad Z=-2.85 & \text { P-Value }=0.004 \end{array}$$

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 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.

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 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?

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 $110$?

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 $H_0: \mu_d \leq 100$ versus $H_a: \mu_d>100$ by setting $\alpha$ equal to $.05$ and $.01$. How much evidence is there that $\mu_d=\mu_1-\mu_2$ exceeds $100$?

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 $5.$

Calculate $95$ percent and $99$ percent confidence intervals for $\mu_d=\mu_1-\mu_2$.

A large discount chain compares the performance of its credit managers in Ohio and Illinois by comparing the mean dollar amounts owed by customers with delinquent charge accounts in these two states. Here a small mean dollar amount owed is desirable because it indicates that bad credit risks are not being extended large amounts of credit. Two independent, random samples of delinquent accounts are selected from the populations of delinquent accounts in Ohio and Illinois, respectively. The first sample, which consists of $10$ randomly selected delinquent accounts in Ohio, gives a mean dollar amount of $\$ 524$ with a standard deviation of $\$ 68$. The second sample, which consists of $20$ randomly selected delinquent accounts in Illinois, gives a mean dollar amount of $\$ 473$ with a standard deviation of $\$ 22$.

Assuming that the normality assumption holds, calculate a $95$ percent confidence interval for the difference between the mean dollar amounts owed in Ohio and Illinois. Based on this interval, do you think that these mean dollar amounts differ in a practically important way?