Question

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 1

Answer: (0.7478, 0.9522)

Explanation:

We know that the confidence interval for population proportion is given by :-

`\hat{p}\pm z_(c)\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

, where n= sample size.

`\hat{p}` = sample proportion.

`z_(c)` = 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

`\hat{p}=85\%=0.85`

By using z-value table, Two -tailed z-value for 99% confidence

`=z_(c)=2.576`

Then, the 99% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

`0.85\pm (2.576)\sqrt{(0.85(1-0.85))/(81)}\\\\0.85\pm0.1022\\\\=(0.85-0.1022,\ 0.85+0.1022)=(0.7478,\ 0.9522)`

Hence, the 99% confidence interval that estimates the proportion of them that are involved in an after school activity= (0.7478, 0.9522)