Question

A simple random sample of 81 8th graders at a large suburban middle school indicated that 87% of them are involved with some type of after school activity. Find the 90% confidence interval that estimates the proportion of them that are involved in an after school activity.

Answer 1

Solution :

Given that,

n = 81

   =87% = 0.87

1 - = 1 - 0.0.87 = 0.13

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.87 * 0.13) / 81) = 0.0615

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.87 - 0.0615< p < 0.87 + 0.0615

0.8085 < p < 0.9315

The 90% confidence interval for the population proportion p is : ( 0.8085, 0.9315)