Question

find a 99% interval for the population proportion if a sample of 200 had a sample proportion of 42%

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = 0.42

1 - = 1 - 0.42 = 0.58

Z/2 = 2.576

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

= 2.576 * (((0.42 * 0.58) / 200)

Margin of error = E = 0.090

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.42 - 0.090 < p < 0.42 + 0.090

0.330 < p < 0.510

The 99% confidence interval for the population proportion p is : 0.330 , 0.510