Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 324 with 211 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.80% C.I. =

Answer 1

Sample size: 324

Number of successes: 211

Confidence level: 80%

The sample proportion of positive results is:

$\bar{p}=\frac{\text{ Number of successes}}{\text{ Sample size}}=\frac{211}{324}\approx0.65123$

Now, we calculate the significance level α:

$\alpha=\frac{1-0.8}{2}=0.1$

The corresponding z-score is (we look at tables of Z-distribution):

$Z_{\alpha}=Z_{0.1}\approx1.28155$

Finally, we apply the formula for the confidence interval (n is the sample size):

$\begin{gathered} 80\text{\% }C.I.=0.65123\pm\sqrt{\frac{\bar{0.65123}(1-0.65123)}{324}} \\ \\ \therefore80\text{\% }C.I.=(0.617,0.685) \end{gathered}$