The accompanying table shows results of challenged referee calls in a major tennis tournament. Use a 0.05 significance level to test the claim that the gender of the tennis player is independent of whether a call is overturned.
Upper H 0​: The gender of the tennis player is independent of whether a call is overturned.
Upper H 1​: The gender of the tennis player is not independent of whether a call is overturned. what is the test statistic and p-value _ Successful_Challenge; Unsuccessful_Challenge
Men 247 ;978
Women 290 ;971

To test the claim that the gender of the tennis player is independent of whether a call is overturned, we can use the Chi-Square Test of Independence. Here are the steps:

1. **Set up the hypotheses:**
   - \(H_0\): The gender of the tennis player is independent of whether a call is overturned.
   - \(H_1\): The gender of the tennis player is not independent of whether a call is overturned.

2. **Create the contingency table:**

   \[
   \begin{array}{|c|c|c|c|}
   \hline
   & \text{Successful Challenge} & \text{Unsuccessful Challenge} & \text{Total} \\
   \hline
   \text{Men} & 247 & 978 & 1225 \\
   \text{Women} & 290 & 971 & 1261 \\
   \hline
   \text{Total} & 537 & 1949 & 2486 \\
   \hline
   \end{array}
   \]

3. **Calculate the expected frequencies:**

   \[
   E_{ij} = \frac{(row\ total \times column\ total)}{grand\ total}
   \]

   \[
   E_{11} = \frac{1225 \times 537}{2486} = 264.79
   \]

   \[
   E_{12} = \frac{1225 \times 1949}{2486} = 960.21
   \]

   \[
   E_{21} = \frac{1261 \times 537}{2486} = 272.21
   \]

   \[
   E_{22} = \frac{1261 \times 1949}{2486} = 988.79
   \]

4. **Calculate the Chi-Square test statistic:**

   \[
   \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}
   \]

   \[
   \chi^2 = \frac{(247 - 264.79)^2}{264.79} + \frac{(978 - 960.21)^2}{960.21} + \frac{(290 - 272.21)^2}{272.21} + \frac{(971 - 988.79)^2}{988.79}
   \]

   \[
   \chi^2 = \frac{(-17.79)^2}{264.79} + \frac{(17.79)^2}{960.21} + \frac{(17.79)^2}{272.21} + \frac{(-17.79)^2}{988.79}
   \]

   \[
   \chi^2 = \frac{316.0841}{264.79} + \frac{316.0841}{960.21} + \frac{316.0841}{272.21} + \frac{316.0841}{988.79}
   \]

   \[
   \chi^2 = 1.193 + 0.329 + 1.161 + 0.319
   \]

   \[
   \chi^2 = 3.002
   \]

5. **Determine the degrees of freedom:**

   \[
   df = (r - 1)(c - 1) = (2-1)(2-1) = 1
   \]

6. **Find the p-value:**
   Using the Chi-Square distribution table or a calculator for \( \chi^2 = 3.002 \) with \( df = 1 \):

   \[
   p \text{-value} = 0.083
   \]

7. **Conclusion:**
   Compare the p-value with the significance level \( \alpha = 0.05 \):

   Since \( p = 0.083 > 0.05 \), we fail to reject the null hypothesis \( H_0 \).

Thus, there is not enough evidence to reject the claim that the gender of the tennis player is independent of whether a call is overturned.

Do you want details or have any questions?

Here are five questions you can ask:
1. How is the Chi-Square test of independence different from other Chi-Square tests?
2. Can you explain how to calculate expected frequencies in more detail?
3. What are some assumptions made when performing a Chi-Square test of independence?
4. How do degrees of freedom affect the Chi-Square test?
5. What are the implications if we reject the null hypothesis in this context?

**Tip:** When performing statistical tests, always ensure your sample size is sufficiently large to meet the assumptions of the test.

Explore the Chi-Square test of independence applied to a study on whether the gender of tennis players affects call overturns in a major tournament. Includes step-by-step calculations, test statistic, p-value, and conclusion.

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