A supervisor records the repair cost for 14 randomly selected refrigerators. A sample mean of $79.20 and standard deviation of $10.41 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.

($74.623, $83.777)

The 90% confidence interval is = ($74.623, $83.777)

Critical value at 90% confidence = 1.645

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $79.20

Standard deviation r = $10.41

Number of samples n = 14

Confidence interval = 90%

Using the z table;

The critical value that should be used in constructing the confidence interval.

z (α=0.05) = 1.645

Critical value at 90% confidence z = 1.645

Substituting the values we have;

$79.20+/-1.645 ($10.42/√14)

$79.20+/-1.645 ($2.782189528308)

$79.20+/-$4.576701774067

$79.20+/-$4.577

($74.623, $83.777)

The 90% confidence interval is = ($74.623, $83.777)