A researcher records the repair cost for 27 randomly selected refrigerators. A sample mean of $60.52 and standard deviation of $23.29 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal.

53.15≤x≤67.89

Step-by-step explanation:

The formula for calculating confidence interval is expressed as;

CI = xbar±(z×s/√n)

xbar is the sample mean

z is the z score at 90% CI

s is the standard deviation

n is the sample size

Given

xbar = $60.52

s = $23.29

n = 27

z = 1.645

Substitute into the formula

CI = xbar±(z×s/√n)

CI = 60.52±(1.645×23.29/√27)

CI = 60.52±(1.645×23.29/5.196)

CI = 60.52±(1.645×4.482)

CI = 60.52±7.373

CI = [60.52-7.373, 60.52+7.373]

CI = [53.15, 67.89]

Hence the 90% confidence interval for the mean repair cost for the refrigerators is 53.15≤x≤67.89

the answer to this question is 3/19

step-by-step explanation:

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