Question

A newspaper reporter asked an SRS of $100$residents in a large city for their opinion about the mayor’s job performance. Using the results from the sample, the $C\%$confidence interval for the proportion of all residents in the city who approve of the mayor’s job performance is $0.565\mathrm{to}0.695$. What is the value of $C$?

a. $82$

b. $86$

c. $90$

d. $95$

e. $99$

Short answer

Option (a) is correct.

Step-by-step solution

Given Information

It is given that $n=100$

$0.565<p<0.695$

Calculation of Confidence Level

Sample Proportion is calculated as

$\hat{p}=\frac{\text{Lower}\mathrm{limit}+\mathrm{Upperlimit}}{2}$

$\hat{p}=\frac{0.565+0.695}{2}=0.63$

Confidence Interval is given by

$\hat{p}\pm Z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$0.63+Z_{\alpha /2}\times \sqrt{\frac{0.63(1-0.63)}{100}}=0.695$

$Z_{\alpha /2}=\frac{0.065}{\sqrt{\frac{0.63(1-0.63)}{100}}}=1.35$

Hence, probability

$P(-1.35<Z<1.35)=P(Z<1.35)-P(Z<-1.35)$

$P(-1.35<Z<1.35)=0.9115-0.0885$

$P(-1.35<Z<1.35)=0.8230=82.30\%=82\%$

Confidence Level is $82\%$

Option (a) is correct.