Question

Selling online According to a recent Pew Research Center report, many

American adults have made money by selling something online. In a random sample of $4579$American adults, $914$reported that they earned money by selling something online in the previous year. Assume the conditions for inference are met.

Determine the critical value $z*$ for a $98\%$ confidence interval for a proportion.

b. Construct a $98\%$ confidence interval for the proportion of all American adults who

would report having earned money by selling something online in the previous year.

c. Interpret the interval from part (b).

Short answer

a. The critical value for $98\%$confidence interval is $2.33$

b. The confidence interval for proportion is $0.1859<p<0.2133$

c. The confidence interval lie between$0.1859\text{and}0.2133$

Step-by-step solution

Given Information

It is given that $x=914$

$n=4579$

Confidence Level $=0.98$

Level of significance$=0.02$

Critical Value

Using EXCEL FORMULA$=\mathrm{ABS}\left(\mathrm{NORMSINV}\right(0.02/2\left)\right)$, critical value${\mathrm{Z}}_{\mathrm{C}}=2.33$

To construct 98% confidence interval for a proportion

Sample Proportion $\hat{p}=\frac{x}{n}$

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

Margin of Error $E=Z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$E=2.33\times \sqrt{\frac{0.1996(1-0.1996)}{4579}}$

$E=0.0137$

Confidence Interval is $(\hat{p}-E,\hat{p}+E)$

$(0.1996-0.0137,0.1996+0.0137)=(0.1859,0.2133)$

$98\%$confidence interval is$0.1859<p<0.2133$

Interpreting Confidence Interval

Based for given data and above calculations, there is $98\%$confidence that correct population proportion of American adults reported to earn money by selling online previous year lie between$0.1859\text{and}0.2133$.