Since an instant replay system for tennis was introduced at a major tournament, men challenged 1405 referee calls, with the result that 426 of the calls were overturned. Women challenged 758 referee calls, and 212 of the calls were overturned. Use a 0.01 significance level to test the claim that men and women have equal success in challenging calls. Complete parts (a) through (c) below.
\[
H_{1}: p_{1}< p_{2}
\]
\[
H_{1}: p_{1} \neq p_{2}
\]
\[
H_{1}: p_{1} \neq p_{2}
\]
Identify the test statistic.
\[
z=1.14
\]
(Round to two decimal places as needed.)
Jdentify the P-value.
\[
P \text {-value }=0.252
\]
(Round to three decimal places as needed
What is the conclusion based on the hypothesis test?
The P-value is greater than the significance level of $\alpha=0.01$, so fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls.
b. Test the claim by constructing an appropriate confidence interval
The $99 \%$ confidence interval is $\square< \left(p_{1}-p_{2}\right)< \square$. (Round to three decimal places as needed.)

Step 1 ：The problem is asking to test the claim that men and women have equal success in challenging calls in tennis using a 0.01 significance level. The null hypothesis is that the success rates are equal, and the alternative hypothesis is that they are not equal.

Step 2 ：The test statistic and P-value have already been calculated and provided. The test statistic is \(z = 1.14\) and the P-value is \(P\text{-value} = 0.252\).

Step 3 ：Based on the hypothesis test, the P-value is greater than the significance level of \(\alpha = 0.01\), so we fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls.

Step 4 ：Now, we need to construct a 99% confidence interval for the difference in proportions. The formula for a confidence interval for the difference in proportions is \((p_1 - p_2) \pm z \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\), where \(p_1\) and \(p_2\) are the sample proportions, \(n_1\) and \(n_2\) are the sample sizes, and \(z\) is the z-score corresponding to the desired level of confidence.

Step 5 ：In this case, \(p_1\) is the proportion of successful challenges by men, \(p_2\) is the proportion of successful challenges by women, \(n_1\) is the number of challenges by men, \(n_2\) is the number of challenges by women, and \(z\) is the z-score for a 99% confidence interval, which is approximately 2.576.

Step 6 ：Using these values, we calculate the 99% confidence interval for the difference in proportions to be \(-0.029 < (p_{1}-p_{2}) < 0.076\).

Step 7 ：\(\boxed{-0.029 < (p_{1}-p_{2}) < 0.076}\). This interval includes zero, which indicates that there is not a significant difference between the success rates of men and women in challenging calls in tennis at the 0.01 significance level.