Question

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.

Answer 1

To calculate the 90% confidence interval for the mean repair cost of refrigerators, a t-distribution is used due to the sample size being less than 30 and the population assumed to be normal. The calculation includes finding the critical t-value, then using the sample mean, standard deviation, and sample size to find the margin of error.

Step-by-step explanation:

The researcher has recorded the repair cost for 27 randomly selected refrigerators and calculated a sample mean and standard deviation. To determine the 90% confidence interval for the mean repair cost of refrigerators, we will use the formula for the confidence interval which is mean ± (critical value) × (standard deviation / √ sample size). As the sample size is less than 30 and we are assuming the population to be normal, we'll use the t-distribution to find the critical value.

First, let's identify the degrees of freedom (df), which is the sample size minus one. Here, df = 27 - 1 = 26. Next, we look up the t-score for a 90% confidence level with 26 degrees of freedom. For a two-tailed test, this value will be approximately 1.706. Now we can plug the values into the confidence interval formula:

Confidence Interval = 60.52 ± 1.706 × (23.29 / √ 27)

Finally, calculate the margins of error and add to subtract from the sample mean to get the confidence interval:

Margin of Error = 1.706 × (23.29 / √ 27) × 1.706 × (23.29 / 5.196)

Margin of Error = 1.706 × 4.4819

Margin of Error = 7.65 (rounded to two decimal places)

So the confidence interval is:

52.87 to 68.17 dollars

Therefore, the researcher can be 90% confident that the population mean repair cost lies between $52.87 and $68.17.

Answer 2

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