Question

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a

Answer 1

Answer:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$

Step-by-step explanation:

Information given:

$n=131$ represent the sample size

$\hat p=0.81$ represent the estimated proportion

The confidence interval would be given by this formula

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

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.326$

And replacing into the confidence interval formula we got:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$