Question

In a random sample of 60 refrigerators, the mean repair cost was $150.00 and the standard deviation was $15.50. Construct a 99% confidence interval for the population mean repair cost. Interpret the results.

Answer 1

The constructed 99% confidence interval for the population mean repair cost of refrigerators is from $144.85 to $155.15, and we can be 99% confident that the true population mean lies within this range.

Step-by-step explanation:

The student has asked to construct a 99% confidence interval for the population mean repair cost of refrigerators based on a sample mean of $150.00 and a standard deviation of $15.50 with a sample size of 60. To address this, we use the formula for a confidence interval which is given by:

Confidence Interval = Sample Mean ± (z-value * (Standard Deviation / √Sample Size))

First, we need to find the z-value that corresponds to a 99% confidence level. Looking up the z-value in the standard normal distribution table, we find that the z-value is approximately 2.576 for a 99% confidence interval.

Then we calculate the margin of error:

Margin of Error = 2.576 * ($15.50 / √60)
Margin of Error = 2.576 * ($15.50 / 7.746)
Margin of Error = 2.576 * 2.001
Margin of Error ≈ $5.15

Now we can construct the confidence interval:

Lower Limit = $150.00 - $5.15 = $144.85
Upper Limit = $150.00 + $5.15 = $155.15

The 99% confidence interval for the population mean repair cost is from $144.85 to $155.15. Interpreting the results, we can be 99% confident that the true population mean for the repair costs of refrigerators lies within this interval.