Asimple 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.

(0.7478, 0.9522)

Step-by-step explanation:

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

, where n= sample size.

= sample proportion.

= 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

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

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

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

the probability is 0