Question

2.5758
An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by the local dealer. The customer relations department will survey a random sample of customers and compute a 99% confidence interval for the proportion who are not satisfied.

(a) Past studies suggest that this proportion will be about 0.19. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.
(You will need a critical value accurate to at least 4 decimal places.)
Sample size:

(b) Using the sample size above, when the sample is actually contacted, 18% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?
MoE:

Answer 1

Date. / A. Solution : Given that, E = 0.015 - P = 0.19 - 1-P = 0.81 At 99% confidence interval z id, d - 1-0.99 = 0.01 do 0.01 0.005 22 24 = 2.576 1. Sample size ne, ne ( 2012 ) xP (1-P) - 12.576 )x0.81(2-0.81) 0.015 / 2 4538.87 - 4539 Sample size n = 4539

(Date. B. Solution : Given that, n = 4539 Construct the 99% CI for p x = -6 1-0.99 = 0.01 de 0.01 0.005 2 2 Za = 2.576 .. Margin of essor f. 2x 2018 (1-0.18) 2 4539 - 2.5.76 / 0.18 X 0.82 . 4539 E = 0.0147