Question

1. Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 159 with 40% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

C.I. =

Answer 1

Solution :

Given that,

n = 159

Point estimate = sample proportion = =0.40

1 - = 0.60

At 99.9% confidence level the z is ,

= 1 - 99.9% = 1 - 0.999 = 0.001

/ 2 = 0.001 / 2 = 0.0005

Z_/2 = Z_0.0005 = 3.291

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

= 3.291 * (((0.40*0.60) / 159)

= 0.128

A 99.9% confidence interval for population proportion p is ,

- E < p < + E

0.40 - 0.128 < p < 0.40 + 0.128

0.272 < p < 0.528

CI = ( 0.272 , 0.528 )