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 99.5% confidence interval for the proportion who are not satisfied.

(a) Past studies suggest that this proportion will be about 0.2. 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.)(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Answer 1

a) A sample size of 5615 is needed.

b) 0.012

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

99.5% confidence level

So
`\alpha = 0.005`, z is the value of Z that has a pvalue of
`1 - (0.005)/(2) = 0.9975`, so
`Z = 2.81`.

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that
`\pi = 0.2`

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.015 = 2.81\sqrt{(0.2*0.8)/(n)}`

`0.015√(n) = 2.81√(0.2*0.8)`

`√(n) = (2.81√(0.2*0.8))/(0.015)`

`(√(n))^(2) = ((2.81√(0.2*0.8))/(0.015))^(2)`

`n = 5615`

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now
`\pi = 0.12, n = 5615`.

We have to find M.

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`M = 2.81\sqrt{(0.12*0.88)/(5615)}`

`M = 0.012`