Question

In a random sample of 55 refrigerators, the mean repair cost was $120.00 and the population standard deviation is $15.10. Construct a 90% confidence interval for the population mean repair cost. Interpret the results Construct a 90% confidence interval for the population mean repair cost. The 90% confidence interval is Interpret you results. Choose the correct answer below. (Round to two decimal places as needed.) OA. With 90% confidence, it can be said that the confidence interval contains the sample mean repair cost. OB. With 90% confidence, it can be said that the confidence interval contains the true mean repair cost OC. The confidence interval contains 90% of the mean repair costs

Answer 1

Answer: To construct a 90% confidence interval for the population mean repair cost, we'll use the formula for the confidence interval:

Confidence Interval = Sample Mean ± (Critical Value * (Population Standard Deviation / √Sample Size))

Given:

Sample Mean = $120.00

Population Standard Deviation = $15.10

Sample Size = 55

Step 1: Find the critical value for a 90% confidence level.

Since the sample size is relatively large (n = 55), we can use the z-distribution table. For a 90% confidence level, the critical value is approximately 1.645.

Step 2: Calculate the confidence interval.

Confidence Interval = $120.00 ± (1.645 * ($15.10 / √55))

Confidence Interval ≈ $120.00 ± ($1.645 * $2.03454)

Confidence Interval ≈ $120.00 ± $3.34

Confidence Interval ≈ ($116.66, $123.34)

Interpretation:

With 90% confidence, we can say that the true population mean repair cost is between $116.66 and $123.34.

The correct answer is:

OB. With 90% confidence, it can be said that the confidence interval contains the true mean repair cost.