Question

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 90% confidence interval for the proportion who are not satisfied.

(a) Past studies suggest that this proportion will be about 0.21. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.025.
(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, 29% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Answer 1

a)

sample proportion ,   p̂ =    0.21                          
sampling error ,    E =   0.025                          
Confidence Level ,   CL=   0.90                          
                                  
alpha =   1-CL =   0.10                          
Z value =    Zα/2 =    1.6449    [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.6449 /   0.025   ) ² *   0.21   * ( 1 -   0.21   ) =    718.2
                                  
                                  
so,Sample Size required=       719                          

b)

Level of Significance,   α =    0.10   
Sample Size,   n =    719          
                  
Sample Proportion ,    p̂ =    0.29
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0169          
margin of error , E = Z*SE =    1.645   *   0.0169   =   0.0278