Question

A random sample of 200 computers chips is obtained from one factory and 4% are found to be defective. Construct and interpret a 99% confidence interval for the percentage of all computer chips from that factory that are not defective.

Answer 1

Solution :

Given that,

n = 200

Point estimate = sample proportion = =4%=0.04

1 -   = 1- 0.04 =0.96

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 ( Using z table )

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

= 2.576* (((0.04*0.96) /200 )

E = 0.036

A 99% confidence interval proportion p is ,

- E < p < + E

0.04-0.036 < p < 0.04+0.036

0.004< p < 0.076