In a recent tennis tournament, women playing singles matches used challenges on 138 calls made by the line judges. Among those challenges, 39 were found to be successful with the call overte 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: \( 23.1 \%

Ask by Christensen Parry. in the United States

Mar 20,2025

To construct a 99% confidence interval for the percentage of successful challenges, first calculate the sample proportion \( \hat{p} \): \[ \hat{p} = \frac{\text{successful challenges}}{\text{total challenges}} = \frac{39}{138} \approx 0.2826 \] Now, we can find the standard error (SE): \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.2826 \cdot (1 - 0.2826)}{138}} \approx 0.049 \] For a 99% confidence level, the z-score is approximately 2.576. Now calculate the margin of error (ME): \[ ME = z \cdot SE = 2.576 \cdot 0.049 \approx 0.126 \] Now, construct the confidence interval: \[ \hat{p} - ME < p < \hat{p} + ME \] \[ 0.2826 - 0.126 < p < 0.2826 + 0.126 \] \[ 0.1566 < p < 0.4086 \] Rounding to one decimal place as needed, the confidence interval is: **15.7% < p < 40.9%** For part (b), compare this interval to the men's successful challenges interval \( 23.1\% < p < 43.7\% \). The confidence intervals overlap (both include values between 23.1% and 40.9%). This suggests that there is no clear evidence that women's successful challenges are statistically significantly different from men's, implying that the success rates might be similar rather than one significantly more successful than the other.