Since an instant replay system for tennis was introduced at a major tournament, men challenged 1420 referee calls, with the result that 416 of the calls were overturned. Women challenged 739 referee calls, and 215 of the calls were overturned. Use a 0.05 significance level to test the claim that men and women have equal success in challenging calls. Complete parts (a) through (c) below.

a. Test the claim using a hypothesis test.

Consider the first sample to be the sample of male tennis players who challenged referee calls and the second sample to be the sample of female tennis players who challenged referee calls. What are the null and alternative hypotheses for the hypothesis test?

A. [tex]\( H_0: p_1 = p_2 \)[/tex]

B. [tex]\( H_0: p_1 \neq p_2 \)[/tex] [tex]\( H_1: p_1 \ \textgreater \ p_2 \)[/tex] [tex]\( H_1: p_1 = p_2 \)[/tex]

C. [tex]\( H_0: p_1 = p_2 \)[/tex] [tex]\( H_1: p_1 \neq p_2 \)[/tex]

D. [tex]\( H_0: p_1 = p_2 \)[/tex] [tex]\( H_1: p_1 \ \textless \ p_2 \)[/tex]

E. [tex]\( H_0: p_1 \geq p_2 \)[/tex] [tex]\( H_1: p_1 \neq p_2 \)[/tex]

F. [tex]\( H_0: p_1 \leq p_2 \)[/tex] [tex]\( H_1: p_1 \neq p_2 \)[/tex]

Identify the test statistic.

[tex]\[ z = \square \][/tex]

(Round to two decimal places as needed.)



Answer :

To address the question, we need to go through the steps of the hypothesis testing method for comparing two proportions. Let's do this step-by-step.

### Step 1: State the Hypotheses

We are testing the claim that men and women have equal success in challenging calls. This sets up our hypotheses as follows:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\( p_1 = p_2 \)[/tex]
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): [tex]\( p_1 \neq p_2 \)[/tex]

From the provided options, the correct hypotheses setup is:

C. [tex]\( H_0: p_1 = p_2 \)[/tex] and [tex]\(H_1: p_1 \neq p_2\)[/tex]

### Step 2: Calculate the Sample Proportions

We have the following data:
- Men: [tex]\( n_1 = 1420 \)[/tex], [tex]\( x_1 = 416 \)[/tex]
- Women: [tex]\( n_2 = 739 \)[/tex], [tex]\( x_2 = 215 \)[/tex]

The sample proportions are calculated as:
[tex]\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{416}{1420} \approx 0.29296 \][/tex]
[tex]\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{215}{739} \approx 0.29093 \][/tex]

### Step 3: Calculate the Combined Proportion

The combined proportion [tex]\(\hat{p}\)[/tex] is calculated as:
[tex]\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{416 + 215}{1420 + 739} \approx 0.29226 \][/tex]

### Step 4: Calculate the Standard Error

The standard error (SE) for the difference in proportions is calculated using the combined proportion:
[tex]\[ SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \approx \sqrt{0.29226(1 - 0.29226) \left(\frac{1}{1420} + \frac{1}{739}\right)} \approx 0.02063 \][/tex]

### Step 5: Calculate the Test Statistic

The test statistic (z) for the difference in proportions is:
[tex]\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \approx \frac{0.29296 - 0.29093}{0.02063} \approx 0.098 \][/tex]

### Step 6: Determine the P-value

For a two-tailed test, the p-value is calculated as:
[tex]\[ p\text{-value} = 2 \times (1 - \Phi(|z|)) \][/tex]
Given our test statistic [tex]\( z \approx 0.098 \)[/tex], the corresponding p-value is approximately [tex]\( 0.9218 \)[/tex].

### Step 7: Conclusion

At a [tex]\( \alpha = 0.05 \)[/tex] significance level, we compare the p-value with [tex]\( \alpha \)[/tex]:
- If [tex]\( p\text{-value} < \alpha \)[/tex], we reject [tex]\( H_0 \)[/tex].
- If [tex]\( p\text{-value} \geq \alpha \)[/tex], we fail to reject [tex]\( H_0 \)[/tex].

Since [tex]\( p\text{-value} \approx 0.9218 \)[/tex] is much greater than [tex]\( 0.05 \)[/tex], we fail to reject the null hypothesis. This means that there is not enough evidence to suggest that the success rates of men and women in challenging referee calls are different.

### Summary

- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\( p_1 = p_2 \)[/tex]
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): [tex]\( p_1 \neq p_2 \)[/tex]
- Test Statistic ([tex]\(z\)[/tex]): [tex]\(\boxed{0.10}\)[/tex]

Thus, we conclude that the success rates for men and women in challenging calls are statistically equal in this context.