In a recent tennis tournament, women playing singles matches used challenges on 132 calls made by the line judges. Among those challenges, 31 were found to be successful with the call overturned.
a. Construct a $95 \%$ confidence interval for the percentage of successful challenges.
b. Compare the results from part (a) to this $95 \%$ confidence interval for the percentage of successful challenges made by the men playing singles matches: $20.5 \%
Solution
Step 1 ：Given that women playing singles matches used challenges on 132 calls made by the line judges. Among those challenges, 31 were found to be successful with the call overturned.
Step 2 ：We are asked to construct a 95% confidence interval for the percentage of successful challenges.
Step 3 ：The formula for the confidence interval of a proportion is given by: \[\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] where \(\hat{p}\) is the sample proportion, \(Z\) is the Z-score corresponding to the desired confidence level (for a 95% confidence level, \(Z=1.96\)), and \(n\) is the sample size.
Step 4 ：In this case, \(\hat{p}\) is the proportion of successful challenges, which is 31 out of 132, and \(n\) is the total number of challenges, which is 132.
Step 5 ：Substituting the given values into the formula, we get \(\hat{p} = 0.23484848484848486\), \(se = 0.03689611480869199\), \(ci_{lower} = 0.16253209982344857\), and \(ci_{upper} = 0.30716486987352115\).
Step 6 ：Thus, the 95% confidence interval for the percentage of successful challenges is \(\boxed{16.25 \% < p < 30.72 \%}\).
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Solution
Step 1 ：Given that women playing singles matches used challenges on 132 calls made by the line judges. Among those challenges, 31 were found to be successful with the call overturned.
Step 2 ：We are asked to construct a 95% confidence interval for the percentage of successful challenges.
Step 3 ：The formula for the confidence interval of a proportion is given by: \[\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] where \(\hat{p}\) is the sample proportion, \(Z\) is the Z-score corresponding to the desired confidence level (for a 95% confidence level, \(Z=1.96\)), and \(n\) is the sample size.
Step 4 ：In this case, \(\hat{p}\) is the proportion of successful challenges, which is 31 out of 132, and \(n\) is the total number of challenges, which is 132.
Step 5 ：Substituting the given values into the formula, we get \(\hat{p} = 0.23484848484848486\), \(se = 0.03689611480869199\), \(ci_{lower} = 0.16253209982344857\), and \(ci_{upper} = 0.30716486987352115\).
Step 6 ：Thus, the 95% confidence interval for the percentage of successful challenges is \(\boxed{16.25 \% < p < 30.72 \%}\).
