Question

Suppose a preliminary sample of 200 parts produced by a machine showed that 7% of them are defective. How large a sample should the company select so that the 95% confidence interval for p is within .02 of the population proportion?

Answer 1

Solution :

Given that,

= 0.07

1 - = 1 - 0.07 = 0.93

margin of error = E = 0.02

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.02)2 * 0.07 * 0.93

= 625.22

Sample size =626