Chapter 7.3, Problem 41E

a.

To determine

Construct a 95% confidence interval for the proportion of tenth-graders who plan to attend college using Wilson’s method.

Answer to Problem 41E

The 95% confidence interval for the proportion of tenth-graders who plan to attend college using Wilson’s method is $\left(0.358,0.802\right)$.

Explanation of Solution

Calculation:

The given information is that,in a certain college 9 said that they planned to go to college after graduatingwhen 15 tenth-graders were asked.

Wilson’s interval:

For constructing a confidence interval the small-sample method is a simple approximation of very complicated interval that is, Wilson’s interval. Consider $\widehat{p}=\frac{x}{n}$.

Wilson’s confidence interval for p is given by,

$\frac{\widehat{p}+\frac{z_{\frac{\alpha }{2}}^2}{2n}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{z_{\frac{\alpha }{2}}^2}{4n^2}}}{1+\frac{z_{\frac{\alpha }{2}}^2}{n}}$

Point estimate:

The point estimate $\widehat{p}$ is obtained by the dividing the number of observations x with the sample size n.

$\widehat{p}=\frac{x}{n}$

Substitute x as 9 and 15 as n in the formula,

$\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{9}{15}\\ =0.6\end{array}$

Thus, the point estimate $\widehat{p}$ is 0.6.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $z_{\frac{\alpha }{2}}$ for 95% confidence level is observed as 1.96.

Now, substitute $\widehat{p}$ as 0.6, $z_{\frac{\alpha }{2}}$ as 1.96and n as 15 in the formula,

$\begin{array}{c}\text{Wilson's interval}=\frac{0.6+\frac{{1.96}^2}{2\left(15\right)}\pm 1.96\sqrt{\frac{0.6\left(1-0.6\right)}{15}+\frac{{1.96}^2}{4\left({15}^2\right)}}}{1+\frac{{1.96}^2}{15}}\\ =\frac{0.6+\frac{3.8416}{30}\pm 1.96\sqrt{\frac{0.6\left(0.4\right)}{15}+\frac{3.8416}{4\left(225\right)}}}{1+\frac{3.8416}{15}}\\ =\frac{0.6+0.1281\pm 1.96\sqrt{\frac{0.24}{15}+\frac{3.8416}{900}}}{1+0.2561}\\ =\frac{0.7281\pm 1.96\sqrt{0.0203}}{1.2561}\end{array}$

$\begin{array}{c}=\frac{0.7281\pm 0.2790}{1.2561}\\ =\left(\frac{0.7281-0.2790}{1.2561},\frac{0.7281+0.2790}{1.2561}\right)\\ =\left(0.3575,0.8018\right)\end{array}$

Thus, the 95% confidence interval for the proportion of tenth-graders who plan to attend college using Wilson’s method is $\left(0.358,0.802\right)$.

b.

To determine

Construct a 95% confidence interval for the proportion of tenth-graders who plan to attend college using small-sample method.

Answer to Problem 41E

The 95% confidence interval for the proportion of tenth-graders who plan to attend college using small-sample method is $0.357<p<0.801$.

Explanation of Solution

Calculation:

Constructing confidence intervals for a proportion with small samples:

If x represents the number of individuals in a sample of size n that has certain characteristic and p is the population proportion, then

The adjusted sample proportion is, $\tilde{p}=\frac{x+2}{n+4}$.

The confidence interval for p is, $\tilde{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{n+4}}<p<\tilde{p}+z_{\frac{\alpha }{2}}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{n+4}}$ or $\tilde{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{n+4}}$.

Substitute x as 9 and n as 15 in the formula of adjusted sample proportion,

$\begin{array}{c}\tilde{p}=\frac{x+2}{n+4}\\ =\frac{9+2}{15+4}\\ =\frac{11}{19}\\ =0.5789\end{array}$

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $z_{\frac{\alpha }{2}}$ for 95% confidence level is observed as 1.96

Now, substitute $\tilde{p}$ as 0.5789, $z_{\frac{\alpha }{2}}$ as 1.96 and n as 15 in the formula,

$\begin{array}{c}\text{Confidence interval}=\tilde{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{n+4}}<p<\tilde{p}+z_{\frac{\alpha }{2}}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{n+4}}\\ =\left(\begin{array}{l}0.5789-1.96\sqrt{\frac{0.5789\left(1-0.5789\right)}{15+4}}<p\\ <0.5789+1.96\sqrt{\frac{0.5789\left(1-0.5789\right)}{15+4}}\end{array}\right)\\ =0.5789-1.96\sqrt{\frac{0.2438}{19}}<p<0.5789+1.96\sqrt{\frac{0.2438}{19}}\\ =0.5789-1.96\left(0.1133\right)<p<0.5789+1.96\left(0.1133\right)\end{array}$

$\begin{array}{c}=0.5789-0.2220<p<0.5789+0.2220\\ =0.3569<p<0.8009\\ =0.357<p<0.801\end{array}$

Thus, the 95% confidence interval for the proportion of tenth-graders who plan to attend college using small-sample method is $0.357<p<0.801$.

c.

To determine

Construct a 95% confidence interval for the proportion of tenth-graders who plan to attend college using traditional method.

Answer to Problem 41E

The 95% confidence interval for the proportion of tenth-graders who plan to attend college using traditional method is $\left(0.352,0.848\right)$.

Explanation of Solution

Calculation:

Confidence interval:

Software procedure:

Step-by-step software procedure to obtain the confidence interval using MINITAB software is as follows,

• Choose Stat > Basic Statistics > 1-Proportion.
• Choose Summarized data.
• Enter Number of events as 9 and Number of trials as 15.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output using MINITAB software is as follows

Thus, the 95% confidence interval for the proportion of tenth-graders who plan to attend college using traditional method is $\left(0.352,0.848\right)$.