Question

An automobile manufacturer would like to know the proportion of its customers that are not satisfied with the service provided by the local dealer. The customer relations department intends on surveying a random sample of customers to estimate this proportion within 1.5 percentage points at 99% confidence. Past studies suggest that the proportion of unsatisfied customers will be about 20%. Find the minimum number of customers that should be included in the survey.

Answer 1

Solution :

Given that,

= 20% = 0.20

1 - = 1 - 0.20 = 0.80

margin of error = E = 1.5% = 0.015

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

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.576 / 0.015)² * 0.20 * 0.80

= 4718.77

= 4719

The minimum number of customers that should be included in the survey = 4719