Question

A sample of 200 computer chips is obtained and 4% are found to be defective. Construct a 90% confidence interval for the percentage of all computer chips that are not defective.

Answer 1

We have

Sample size n = 200

sample proportion of computer chips that are not defective p^ = 0.96

The 90% confidence interval for the percentage of all computer chips that are not defective is

p^ - E < p < p^ + E

Where E = Z_{​​​​​​a/2}* sqrt [ p^ *(1- p^)/n]

For a = 0.10 , Z_{​​​​​​a/2} = Z_{​​​​​​0.05} = 1.645

E = 1.645* sqrt [ 0.96*0.04/200] = 0.0228

0.96 - 0.023 < p < 0.96 + 0.023

0.94 < p < 0.98

We are 90% confident that percentage of all computer chips that are not defective is lies between 94% to 98%