Mark wants to find the mean repair cost of cars. He conducts research with a sample size of 120 cars, where the standard deviation of the cost of car repairs is $$56$. Using a 99% confidence level, what is the margin of error for the population mean? Round your answer to the nearest hundredth.

Question

Use the table below to find the critical value.
Confidence Level c Critical Value z
c
​

0.95 1.96
0.98 2.33
0.99 2.58

Answer 1

To calculate the margin of error for the population mean at a 99% confidence level, we can use the formula:

\[ \text{Margin of Error} = z \times \frac{\sigma}{\sqrt{n}} \]

Where:

• \( z \) is the critical value from the z-table for the desired confidence level,
• \( \sigma \) is the standard deviation,
• \( n \) is the sample size.

From the information provided:

• The critical value \( z \) for a 99% confidence level is 2.58.
• The standard deviation \( \sigma \) is 56.
• The sample size \( n \) is 120.

Now, we can plug in these values:

1. First, calculate the standard error (SE):

\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{56}{\sqrt{120}} \approx \frac{56}{10.95} \approx 5.11 \]

2. Now calculate the margin of error:

\[ \text{Margin of Error} = 2.58 \times 5.11 \approx 13.18 \]

Thus, the margin of error rounded to the nearest hundredth is:

\[ \boxed{13.18} \]