Questions: In a recent tennis tournament, women playing singles matches used challenges on 136 calls made by the line judges. Among those challenges, 33 were found to be successful with the call overturned. a. Construct a 99% confidence interval for the percentage of successful challenges. b. Compare the results from part (a) to this 99% confidence interval for the percentage of successful challenges made by the men playing singles matches: 18.4%<p<38.3%. Does it appear that either gender is more successful than the other? a. Construct a 99% confidence interval. %<p<% (Round to one decimal place as needed.)

Solution Steps

The sample proportion of successful challenges is calculated as follows:

\[ \hat{p} = \frac{\text{successful challenges}}{\text{total challenges}} = \frac{33}{136} \approx 0.2426 \]

To construct a \(99\%\) confidence interval for the proportion of successful challenges, we use the formula:

\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Where:

• \(\hat{p} \approx 0.243\)
• \(z \approx 2.576\) (for \(99\%\) confidence level)
• \(n = 136\)

Calculating the margin of error:

\[ \text{Margin of Error} = 2.576 \cdot \sqrt{\frac{0.243(1 - 0.243)}{136}} \approx 0.094 \]

Thus, the confidence interval is:

\[ (0.243 - 0.094, 0.243 + 0.094) \approx (0.148, 0.337) \]

Converting this to percentage form gives:

\[ 14.8\% < p < 33.7\% \]

The \(99\%\) confidence interval for the percentage of successful challenges made by men is given as:

\[ 18.4\% < p < 38.3\% \]

To determine if either gender is more successful, we compare the intervals:

• Women's confidence interval: \(14.8\% < p < 33.7\%\)
• Men's confidence interval: \(18.4\% < p < 38.3\%\)

Since the upper limit of the women's interval (\(33.7\%\)) is less than the lower limit of the men's interval (\(18.4\%\)), there is no overlap. Therefore, we conclude that:

There is no clear evidence that one gender is more successful than the other.

Final Answer