Question

Compute the 99% confidence interval estimate for the population proportion, p, based on a sample size of 100 when the sample proportion, p (overbar), is equal to 0.26

Answer 1

solution :

Given that,

n = 100

= 0.26

1 - = 1 - 0.26 = 0..76

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01/ 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576* (((0.26 *0.74) / 100) = 0.113

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.26 - 0.113 < p < 0.26 + 0.113

0.147< p < 0.373