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 c
Let p be the population proportion of them that are involved in an after school activity.
As per given , we have
n= 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)
