Question

A student records the repair cost for 13 randomly selected TVs. A sample mean of $72.19 and standard deviation of $15.88 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

Answer 1

(60.382 ; 83.998)

Explanation:

The confidence interval is given as :

Mean ± Margin of error

Mean = 72.19

The margin of error is : Tcritical * s/√n

n = 13 ; s = 15.88

Tcritical at 98%, df = 13-1 = 12 = 2.681

Margin of Error = 2.681 * 15.88/√13 = 11.808

Confidence interval = 72.19 ± 11.808

(60.382 ; 83.998)