Question

A simple random sample of 64 8th graders at a large suburban middle school indicated that 88% of them are involved with some type of after school activity. Find the 98% confidence interval that estimates the proportion of them that are involved in an after school activity.

Answer 1

Solution :

Given that,

n = 64

Point estimate = sample proportion = = 0.88=88%

1 - = 1-0.88=0.12

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.88*0.12) / 64)

= 0.091

A 98% confidence interval is ,

- E < p < + E

0.88-0.091< p <0.88+0.091

(0.789 , 0.971 )