Chapter 8.2, Problem 58E

(a)

Section 1:

To determine

To find: The $95\%$ confidence interval for the difference in proportions.

Answer to Problem 58E

Solution: The $95\%$ confidence interval for the difference is $\underset{\_}{\left(0.0806,0.3354\right)}$.

Explanation of Solution

Calculation: In order to test the difference in proportions for $95\%$ confidence interval, the number of successes as well as the number of failures should count at least 10. It is inferred form exercise 8.56 that the guidelines for using the $95\%$ confidence interval for large-sample method in this part are met. It is also provided that out of 69 (represented as $n_1$ ), about 40 customers (represented as $X_1$ ) left a tip for a server wearing a red shirt. The probability of a customer leaving a tip for the server wearing a red shirt is (represented as ${\widehat{p}}_1$ ) is calculated as:

$\begin{array}{c}{\widehat{p}}_1=\frac{X_1}{n_1}\\ =\frac{40}{69}\\ =0.58\end{array}$

Also, out of 349 customers (represented as $n_2$ ), about 130 male customers (represented as $X_2$ ) left a tip for a server wearing a different colored shirt. The probability of a customer leaving a tip for the server wearing a colored shirt is (represented as ${\widehat{p}}_2$ ) is calculated as:

$\begin{array}{c}{\widehat{p}}_2=\frac{X_2}{n_2}\\ =\frac{130}{349}\\ =0.372\end{array}$

The difference between the two sample proportions $\left({\widehat{p}}_1-{\widehat{p}}_2\right)$ is denoted by $D$ and is calculated as follows:

$\begin{array}{c}D={\widehat{p}}_1-{\widehat{p}}_2\\ =0.58-0.372\\ =0.208\end{array}$

The Standard Error of $D$ is calculated as follows:

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}\\ =\sqrt{\frac{0.58\left(1-0.58\right)}{69}+\frac{0.372\left(1-0.372\right)}{349}}\\ =0.065\end{array}$

For $95\%$ confidence level, the $z\text{-value}$ is $z=1.96$. Thus, the margin of error $\left(m\right)$ is obtained as:

$\begin{array}{c}m=z\times {\text{SE}}_D\\ =\left(1.96\right)\times \left(0.065\right)\\ =0.1274\end{array}$

Hence, the margin of error is about $12.74\%$.

The $95\%$ confidence interval is calculated as:

$\begin{array}{c}D\pm m=0.208\pm 0.1274\\ =\left(0.0806,0.3354\right)\end{array}$

Hence, it can be said that with $95\%$ confidence level, the difference between the proportions lie between the values 0.0806 and 0.3354.

Section 2:

To determine

To find: The $95\%$ confidence interval for the difference in proportions.

Answer to Problem 58E

Solution: The $95\%$ confidence interval for the difference is $\left(-0.10,0.2421\right)$.

Explanation of Solution

In order to test the difference in proportions for $95\%$ confidence interval, the number of successes as well as the number of failures should count at least 10. It is inferred form exercise 8.56 that the guidelines for using the $95\%$ confidence interval for large-sample method in this part are not met. Hence, the method that can be used for this problem is ‘Plus four Confidence Interval’.

It is provided that out of 40 runners (represented as $n_1$ ), 9 runners (represented as $X_1$ ) were satisfied or very satisfied with the first routine. The plus four estimate of the population proportion for runners who are satisfied or very satisfied after following the first routine is represented as ${\tilde{p}}_1$ and is calculated as:

$\begin{array}{c}{\tilde{p}}_1=\frac{X_1+1}{n_1+2}\\ =\frac{9+1}{40+2}\\ =0.238\end{array}$

Also, it is provided that out of 40 runners (represented as $n_2$ ), 6 runners (represented as $X_2$ ) were satisfied or very satisfied with the second routine. The plus four estimate of the population proportion for runners who are satisfied or very satisfied after following the second routine is represented as ${\tilde{p}}_2$ and is calculated as:

$\begin{array}{c}{\tilde{p}}_2=\frac{X_2+1}{n_2+2}\\ =\frac{6+1}{40+2}\\ =0.167\end{array}$

The difference between the two sample proportions $\left({\tilde{p}}_1-{\tilde{p}}_2\right)$ is denoted by $D$ and is calculated as follows:

$\begin{array}{c}D={\tilde{p}}_1-{\tilde{p}}_2\\ =0.238-0.167\\ =0.071\end{array}$

The Standard Error of $D$ is calculated as follows:

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}\\ =\sqrt{\frac{0.238\left(1-0.238\right)}{40+2}+\frac{0.167\left(1-0.167\right)}{40+2}}\\ =0.0873\end{array}$

For $95\%$ confidence level, the $z\text{-value}$ is $z=1.96$. Thus, the margin of error $\left(m\right)$ is obtained as:

$\begin{array}{c}m=z\times {\text{SE}}_D\\ =\left(1.96\right)\times \left(0.0873\right)\\ =0.1711\end{array}$

Hence, the margin of error is about $17.11\%$.

The $95\%$ confidence interval is calculated as:

$\begin{array}{c}D\pm m=0.071\pm 0.1711\\ =\left(-0.10,0.2421\right)\end{array}$

Hence, it can be said that with $95\%$ confidence level, the difference between the proportions lie between the values 0.10 and 0.2421.

Section 3:

To determine

To explain: The concept of confidence interval based on above Sections 1 and 2.

Answer to Problem 58E

Solution: The difference in the proportion of male customers who tip a server wearing red colored shirt and the proportion of male customers who tip a server wearing a different colored shirt than red, with $95\%$ surety lies between the intervals $\left(0.0806,0.3354\right)$. Also, the difference in the proportion of runners who are satisfied with the first routine and the proportion of runners who are satisfied with the second routine, with $95\%$ surety having a margin of error $17.11\%$ lies between the intervals $\left(-0.10,0.2421\right)$.

Explanation of Solution

Given: In sections 1 and 2 above, the $95\%$ confidence interval for part (a) is $\left(0.0806,0.3354\right)$ and for part (b) is $\left(-0.10,0.2421\right)$.

Explanation: The confidence interval shows the bracket of values which are an estimate that the value of the parameter will lie between these values. The confidence interval obtained for part (a) is $\left(0.0806,0.3354\right)$ which implies that the difference in the proportion of male customers who tip a server wearing red colored shirt and the proportion of male customers who tip a server wearing a different colored shirt than red, with $95\%$ surety having a margin of error $12.74\%$ lies between the intervals $\left(0.0806,0.3354\right)$. The confidence interval obtained for part (b) is $\left(-0.10,0.2421\right)$ which implies that the difference in the proportion of runners who are satisfied with the first routine and the proportion of runners who are satisfied with the second routine, with $95\%$ surety having a margin of error $17.11\%$ lies between the intervals $\left(-0.10,0.2421\right)$.