Question

A researcher records the repair cost for 27 randomly selected refrigerators. A sample mean of $97.89 and standard deviation of $17.53 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal.

Step 1 of 2 :

Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Solution :

Given that,

t /2,df = 1.706

Margin of error = E = t/2,df * (s /n)

= 1.706 * (17.53 / 27)

Margin of error = E = 5.755

The 90% confidence interval estimate of the population mean is,

- E < < + E

97.89 - 5.755 < < 97.89 + 5.755

92.135 < < 103.645

(92.135 , 103.645)