Questions: A recent study by the American Automobile Dealers Association surveyed a random sample of 17 dealers. The data revealed a mean amount of profit per car sold was 190, with a standard deviation of 124. Develop a 99% confidence interval for the population mean of profit per car. Note: Round your answers to 2 decimal places. The confidence interval is between and

Solution Steps

We have the following data from the study:

• Sample mean (\(\bar{x}\)) = 190
• Sample standard deviation (\(s\)) = 124
• Sample size (\(n\)) = 17
• Confidence level = 99%

For a 99% confidence level and a sample size of 17, we need to find the critical value \(t\) from the t-distribution. The degrees of freedom (\(df\)) is given by \(n - 1 = 16\). The critical value \(t\) for \(df = 16\) at a 99% confidence level is approximately \(2.92\).

The standard error (SE) is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{124}{\sqrt{17}} \approx 30.16 \]

The margin of error (ME) is calculated as: \[ ME = t \cdot SE = 2.92 \cdot 30.16 \approx 88.06 \]

The confidence interval is given by: \[ \bar{x} \pm ME = 190 \pm 88.06 \] Calculating the lower and upper bounds:

• Lower bound: \(190 - 88.06 \approx 101.94\)
• Upper bound: \(190 + 88.06 \approx 278.06\)

Final Answer