Chapter 8.2, Problem 36E

a)

To determine

To find the critical value z for 96% confidence interval for a proportion.

Answer to Problem 36E

The critical value z for 96% confidence interval for a proportion is 2.04

Explanation of Solution

Given:

  $\begin{array}{l}x=19\\ n=172\end{array}$

Confidence level = 0.96

The confidence level = 0.98.

So, level of significance = a=0.04

The z_c=z _a/2 critical value = 2.04 …Using excel formula, =ABS(NORMSINV(0.04/2))

b)

To determine

To construct 96% confidence interval for a proportion.

Answer to Problem 36E

The 96% confidence interval for a proportion is 0.0617< p < 0.1593

Explanation of Solution

Given:

  $\begin{array}{l}x=19\\ n=172\end{array}$

Confidence level = 0.96

The z_c critical value = 2.04

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Margin of error:

  $E=Z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

The confidence interval:

  $(\widehat{p}-E,\widehat{p}+E)$

The sample proportion is,

  $\widehat{p}=\frac{19}{172}=0.1105$

The margin of error is,

  $\begin{array}{l}E=2.04\times \sqrt{\frac{0.1105(1-0.1105)}{172}}\\ E=0.0488\end{array}$

The confidence interval is,

  $\begin{array}{l}(0.1105-0.0488,0.1105+0.0488)\\ (0.0617,0.1593)\end{array}$

Hence, the 96% confidence interval for population proportion is 0.0617 < p < 0.1593

c)

To determine

To interpret confidence interval.

Answer to Problem 36E

We are 96% confident that the true population proportion is between 0.0617 and 0.1593

Explanation of Solution

Given:

The 96% confidence interval for population proportion is 0.0617 < p < 0.1593

We are 96% confident that the true population proportion of all undergraduate at this university who would go to the professor is between 0.0617 and 0.1593