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 $\mu_d=\mu_1-\mu_2$.

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 solution to find the greatest common divisor of two integers n1 and n2 is as follows: First find d to be the minimum of n1 and n2, and then check whether d, d - 1, d - 2, ..., 2, or 1 is a divisor for both n1 and n2 in this order. The first such common divisor is the greatest common divisor for n1 and n2. Write the program.

Suppose two independent random samples of sizes $n_1=9$ and $n_2=7$ that have been taken from two normally distributed populations having variances $\sigma_1^2$ and $\sigma_2^2$ give sample variances of $s_1^2=100$ and $s_2^2=20$.

Test $H_0: \sigma_1^2=\sigma_2^2$ versus $H_a: \sigma_1^2 > \sigma_2^2$ with $\alpha=.05$. What do you conclude?

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?

Suppose we have taken independent, random samples of sizes $n_1=7$ and $n_2=7$ from two normally distributed populations having means $\mu_1$ and $\mu_2$, and suppose we obtain $\bar{x}_1=240, \bar{x}_2=$ $210, s_1=5$, and $s_2=6$.

Use critical values to test the null hypothesis $H_0: \mu_1-\mu_2 \leq 20$ versus the alternative hypothesis $H_a: \mu_1-\mu_2>20$ by setting $\alpha$ equal to $.10, .05, .01$, and $.001$. How much evidence is there that the difference between $\mu_1$ and $\mu_2$ exceeds $20$?

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 $\$ 22$.

Set up the null and alternative hypotheses needed to test whether there is a difference between the population mean dollar amounts owed by customers with delinquent charge accounts in Ohio and Illinois.

In the book Business Research Methods, Donald R. Cooper and C. William Emory ($1995$) discuss a manager who wishes to compare the effectiveness of two methods for training new salespeople. The authors describe the situation as follows:

The company selects $22$ sales trainees who are randomly divided into two experimental groups-one receives type $A$ and the other type $B$ training. The salespeople are then assigned and managed without regard to the training they have received. At the year's end, the manager reviews the performances of salespeople in these groups and finds the following results:

$$\begin{array}{lcc} & \text { A Group } & \text { B Group } \\ \text { Average Weekly Sales } & \$ 1,500 & \$ 1,300 \\ \text { Standard Deviation } & 225 & 251\end{array}$$

Use the equal variances procedure to calculate a $95$ percent confidence interval for the difference between the mean weekly sales obtained when type $A$ training is used and the mean weekly sales obtained when type $B$ training is used. Interpret this interval.

In the book Business Research Methods, Donald R. Cooper and C. William Emory ($1995$) discuss a manager who wishes to compare the effectiveness of two methods for training new salespeople. The authors describe the situation as follows:

The company selects $22$ sales trainees who are randomly divided into two experimental groups-one receives type $A$ and the other type $B$ training. The salespeople are then assigned and managed without regard to the training they have received. At the year's end, the manager reviews the performances of salespeople in these groups and finds the following results:

$$\begin{array}{lcc} & \text { A Group } & \text { B Group } \\ \text { Average Weekly Sales } & \$ 1,500 & \$ 1,300 \\ \text { Standard Deviation } & 225 & 251\end{array}$$

Set up the null and alternative hypotheses needed to attempt to establish that type $A$ training results in higher mean weekly sales than does type $B$ training.

In the book Essentials of Marketing Research, William R. Dillon, Thomas J. Madden, and Neil H. Firtle $(1993)$ present preexposure and postexposure attitude scores from an advertising study involving $10$ respondents. The data for the experiment are given in Table $10.3$. Assuming that the differences between pairs of postexposure and preexposure scores are normally distributed:

Table $10.3$

$$\begin{array}{llll} \text{Subject} & \text{Preexposure Attitudes}\ (A_{1}) & \text{Preexposure Attitudes}\ (A_{2}) & \text{Attitude Change}\ (d_{i}) \\ 1 &50& 53 &3 \\ 2 &25& 27 &2 \\ 3 &30 &38 &8\\ 4 &50 &55 &5\\ 5 &60 &61 &1\\ 6 &80 &85 &5\\ 7 &45 &45 &0\\ 8 &30 &31 &1\\ 9 &65 &72 &7\\ 10 &70& 78 &8\\ \end{array}$$

Estimate the minimum difference between the mean postexposure attitude score and the mean preexposure attitude score. Justify your answer.

In the book Essentials of Marketing Research, William R. Dillon, Thomas J. Madden, and Neil H. Firtle $(1993)$ present preexposure and postexposure attitude scores from an advertising study involving $10$ respondents. The data for the experiment are given in Table $10.3$. Assuming that the differences between pairs of postexposure and preexposure scores are normally distributed:

Table $10.3$

$$\begin{array}{llll} \text{Subject} & \text{Preexposure Attitudes}\ (A_{1}) & \text{Preexposure Attitudes}\ (A_{2}) & \text{Attitude Change}\ (d_{i}) \\ 1 &50& 53 &3 \\ 2 &25& 27 &2 \\ 3 &30 &38 &8\\ 4 &50 &55 &5\\ 5 &60 &61 &1\\ 6 &80 &85 &5\\ 7 &45 &45 &0\\ 8 &30 &31 &1\\ 9 &65 &72 &7\\ 10 &70& 78 &8\\ \end{array}$$

Set up the null and alternative hypotheses needed to attempt to establish that the advertisement increases the mean attitude score (that is, that the mean postexposure attitude score is higher than the mean preexposure attitude score).

The manufacturer of the ColorSmart-$5000$ television set claims that $95$ percent of its sets last at least five years without needing a single repair. In order to test this claim, a consumer group randomly selects $400$ consumers who have owned a ColorSmart-$5000$ television set for five years. Of these $400$ consumers, $316$ say that their ColorSmart-$5000$ television sets did not need repair, while $84$ say that their ColorSmart-$5000$ television sets did need at least one repair.

Use critical values and the previously given sample information to test the hypotheses you set up in part $a$ by setting $\alpha$ equal to $.10$, $.05$, $.01$, and $.001$. How much evidence is there that the manufacturer’s claim is false?

The manufacturer of the ColorSmart-$5000$ television set claims that $95$ percent of its sets last at least five years without needing a single repair. In order to test this claim, a consumer group randomly selects $400$ consumers who have owned a ColorSmart-$5000$ television set for five years. Of these $400$ consumers, $316$ say that their ColorSmart-$5000$ television sets did not need repair, while $84$ say that their ColorSmart-$5000$ television sets did need at least one repair.

Do you think the results of the consumer group’s survey have practical importance? Explain your opinion.

The manufacturer of the ColorSmart-$5000$ television set claims that $95$ percent of its sets last at least five years without needing a single repair. In order to test this claim, a consumer group randomly selects $400$ consumers who have owned a ColorSmart-$5000$ television set for five years. Of these $400$ consumers, $316$ say that their ColorSmart-$5000$ television sets did not need repair, while $84$ say that their ColorSmart-$5000$ television sets did need at least one repair.

Letting $p$ be the proportion of ColorSmart$-5000$ television sets that last five years without a single repair, set up the null and alternative hypotheses that the consumer group should use to attempt to show that the manufacturer’s claim is false.