A simple random sample of 81 8th graders at a large suburban middle school indicated that 85% of them are involved with some type of after school activity. Find the 99% confidence interval that estimates the proportion of them that are involved in an after school activity.

Answer: (0.7478, 0.9522)Step-by-step explanation:We know that the confidence interval for population proportion is given by :-[tex]\hat{p}\pm z_{c}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex], where n= sample size.[tex]\hat{p}[/tex] = sample proportion.[tex]z_{c}[/tex] = Two -tailed z-value for confidence level of cLet p be the population proportion of them that are involved in an after school activity.As per given , we haven= 81[tex]\hat{p}=85\%=0.85[/tex]By using z-value table, Two -tailed z-value for 99% confidence[tex]=z_{c}=2.576[/tex]Then, the 99% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-[tex]0.85\pm (2.576)\sqrt{\dfrac{0.85(1-0.85)}{81}}\\\\0.85\pm0.1022\\\\=(0.85-0.1022,\ 0.85+0.1022)=(0.7478,\ 0.9522) [/tex]Hence, the 99% confidence interval that estimates the proportion of them that are involved in an after school activity= (0.7478, 0.9522)