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in a random sample of 55 refrigerators, the mean repair cost was $142.00 and the standard deviation was $17.50. A 90% confidence interval for the population mean repair cost is (138.12,145.88). Change the sample size to n=110. Construct a 90% confidence interval for the population mean repair cost. Which confidence interval is wider? Explain

The 90% confidence interval is (       ,    )

Solution

Given a=1-0.9=0.1, Z(0.05)=1.645 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =142 -1.645*17.5/sqrt(110) =139.2552

So the upper bound is

xbar +Z*s/vn =142 +1.645*17.5/sqrt(110) =144.7448

confidence interval for n=55 is wider