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}}=(211)/(324)\approx0.65123`

Now, we calculate the significance level α:

`\alpha=(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}`