To test the claim that men and women have equal success rates in challenging calls, we'll perform a hypothesis test for the difference between two population proportions.

Part (a): Define the Hypotheses

Given the information provided:

1. Null Hypothesis (H0): The success rates for men and women are equal.
2. Alternative Hypothesis (H1): The success rates for men and women are not equal.

Therefore, the correct answer is:

• Option D: $H_0: P_1 = P_2$ and $H_1: P_1 \neq P_2$

Part (b): Calculate the Test Statistic

Step 1: Define the Proportions

1. For men:

   • Total challenges: $n_1 = 1432$
   • Successful challenges (overturned calls): $x_1 = 422$
   • Proportion of successful challenges: $\hat{p}_1 = \frac{x_1}{n_1} = \frac{422}{1432} \approx 0.2947$

2. For women:

   • Total challenges: $n_2 = 758$
   • Successful challenges (overturned calls): $x_2 = 225$
   • Proportion of successful challenges: $\hat{p}_2 = \frac{x_2}{n_2} = \frac{225}{758} \approx 0.2963$

Step 2: Calculate the Pooled Proportion

The pooled proportion $\hat{p}$ is calculated as follows: $\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{422 + 225}{1432 + 758} = \frac{647}{2190} \approx 0.2950$

Step 3: Calculate the Standard Error

The standard error $SE$ for the difference between two proportions is given by: $SE = \sqrt{\hat{p} \left( 1 - \hat{p} \right) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}$ Substituting the values: $SE = \sqrt{0.2950 \times (1 - 0.2950) \times \left( \frac{1}{1432} + \frac{1}{758} \right)} \approx 0.0262$

Step 4: Calculate the Z-Statistic

The Z-statistic for the difference in proportions is calculated by: $Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.2947 - 0.2963}{0.0262} \approx -0.06$

Part (c): Conclusion of the Hypothesis Test

Using a significance level of $\alpha = 0.05$:

• Since this is a two-tailed test, we look at the critical values for $Z$ at $\pm 1.96$.
• The calculated $Z$-value of $-0.06$ is within the range $-1.96 < Z < 1.96$, so we do not reject the null hypothesis.

Conclusion

There is not enough evidence at the 0.05 significance level to conclude that there is a difference in success rates between men and women in challenging calls.

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Do you have questions, or would you like further details?

Follow-Up Questions:

1. How would the result change if we used a one-tailed test instead?
2. What is the practical interpretation of failing to reject the null hypothesis in this context?
3. How would a larger sample size impact the significance of the test?
4. Could the result change if the success rates for both groups were higher overall?
5. What are other potential factors that could influence the success rate in challenging calls?

Tip: When comparing proportions, always check if the sample sizes are large enough for reliable results.