(1 pt) 3.0902 An automobile manufacturer would like to know what proportion of i
ID: 3130982 • Letter: #
Question
(1 pt) 3.0902
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 99.8% confidence interval for the proportion who are not satisfied.
(a) Past studies suggest that this proportion will be about 0.16. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.01.
(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, 19% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?
MoE:
I got 2444.6 for my sample size, which is wrong. I'm doing this based on other answers given to the same problem with just different numbers, yet still getting the wrong values. Please help!
Explanation / Answer
a.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.002 is = 3.09
Sample Proportion = 0.16
ME = 0.01
n = ( 3.09 / 0.01 )^2 * 0.16*0.84
= 12832.646 ~ 12833
b.
Margin of Error = Z a/2 Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Sample Size(n)=12833
Sample proportion =0.19
Z- Table at 0.002 is = 3.09
Margin of Error = Z a/2 * ( Sqrt ( (0.19*0.81) /12833) )
= 3.09* Sqrt(0)
=0.011
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