Answer to Question #323925 in Statistics and Probability for rere

Question

We take a random sample of 60 households to estimate the percentage of all

homes in the Spring Town Village that have a refrigerator. It turns out that 49 of

the 60 homes in our sample have a refrigerator. Use 95% confidence level to

calculate the length of confidence interval estimation of population proportion of

households with refrigerator.

Answer 1

$\hat{p}=\frac{49}{60}=0.816667\\\left( \hat{p}-\frac{\sqrt{\hat{p}\left( 1-\hat{p} \right)}}{\sqrt{n}}z_{\frac{1+\gamma}{2}},\frac{\sqrt{\hat{p}\left( 1-\hat{p} \right)}}{\sqrt{n}}z_{\frac{1+\gamma}{2}} \right) \\Length:\\2\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}}z_{\frac{1+\gamma}{2}}=2\sqrt{\frac{0.816667\left( 1-0.816667 \right)}{60}}\cdot 1.960=0.195818$