A simple random sample of 49 8th graders at a large suburban middle school indicated that 82% 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.

0.142

Step-by-step explanation:

From the question, we identify the following parameters;

n = 49

p = 82% = 82/100 = 0.82

alpha, α = 1-0.99 = 0.01

Zα/2 = Z_0.005 = 2.575

margin of error = Zα/2 * √( p(1-p)/n)

Margin of error = 2.575 * √(0.82)(1-0.82)/49

Margin of error =0.1416005 which is approximately 0.142

3 is greater then both of those numbers

i think it is 18

step-by-step explanation: