Chapter 8.2, Problem 35E

a)

To determine

To findthe critical value z for 98% confidence interval for a proportion.

Answer to Problem 35E

The critical value z for 98% confidence interval for a proportion is 2.33

Explanation of Solution

Given:

  $\begin{array}{l}x=914\\ n=4579\end{array}$

Confidence level = 0.98

The confidence level = 0.98.

So, level of significance = 0.02

The z_c critical value = 2.33 …Using excel formula, =ABS(NORMSINV(0.02/2))

b)

To determine

To construct 98% confidence interval for a proportion.

Answer to Problem 35E

The 98% confidence interval for a proportion is 0.1859 < p < 0.2133

Explanation of Solution

Given:

  $\begin{array}{l}x=914\\ n=4579\end{array}$

Confidence level = 0.98

The z_c critical value = 2.33

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,

  $\widehat{p}=\frac{914}{4579}=0.1996$

The margin of error is,

  $\begin{array}{l}E=2.33\times \sqrt{\frac{0.1996(1-0.1996)}{4579}}\\ E=0.0137\end{array}$

The confidence interval is,

  $\begin{array}{l}(0.1996-0.0137,0.1996+0.0137)\\ (0.1859,0.2133)\end{array}$

Hence, the 98% confidence interval for population proportion is 0.1859 < p < 0.2133

c)

To determine

To interpret confidence interval.

Answer to Problem 35E

We are 98% confident that the true population proportion is between 0.1859 and 0.2133

Explanation of Solution

Given:

The 98% confidence interval for population proportion is 0.1859 < p < 0.2133

We are 98% confident that the true population proportion of all American adults who would report having earned money by selling something online in the previous year is between 0.1859 and 0.2133