Question

Suppose that we randomly select the records of 100 cars sold in 2002 and check to see if they have ever had repair work done that was covered by the warranty. 35 of them had.

a. Give a 95% confidence interval for the proportion of all the cars we sold in 2002 that have had work done under warranty.

b. If we wanted to have a margin of error in estimating the proportion of cars having warranty work done to be 0.1 with a confidence level of 90% and no prior knowledge of the true proportion, how many records should we inspect?

Answer 1

a) n = number of cars selected randomly = 100

x = number of cars have ever had repair work done that was covered by the warranty = 35

Sample proportion:

Confidence level = c = 0.95

z critical value for (1+c)/2 = (1+0.95)/2 = 0.975

zc = 1.96 (From statistical table of z values)

95% confidence interval for the proportion of all the cars we sold in 2002 that have had work done under warranty is

(Round to 3 decimal)

95% confidence interval for the proportion of all the cars we sold in 2002 that have had work done under warranty is (0.257, 0.443)

b)

Margin of error = e = 0.1

Confidence level = c = 0.90

z critical value for (1+c)/2 = (1+0.90)/2 = 0.95

zc = 1.645 (From statistical table of z values, average of 1.64 and 1.65,(1.64+1.65)/2 = 1.645)

Consider p = 0.5 because  no prior knowledge of the true proportion.

Sample size (n) :

n = 67.6506

n = 68 (Round to nearest integer)

We should inspect 68 records.