Questions: 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. Determine the null and alternative hypotheses. A. H0 : The gender of the tennis player is not independent of whether a call is overturned. H1 : The gender of the tennis player is independent of whether a call is overturned. B. H0 : Male tennis players are not more successful in overturning calls than female players. H1 : Male tennis players are more successful in overturning calls than female players. C. H0 : The gender of the tennis player is independent of whether a call is overturned. H1 : The gender of the tennis player is not independent of whether a call is overturned. D. H0 : Male tennis players are more successful in overturning calls than female players. H1 : Male tennis players are not more successful in overturning calls than female players. Determine the test statistic. x^2= (Round to three decimal places as needed.) Was the Challenge to the Call Successful? Yes No Men 412 607 Women 497 837

Solution Steps

We are testing the claim that the gender of the tennis player is independent of whether a call is overturned. The hypotheses are:

• Null Hypothesis (\(H_0\)): The gender of the tennis player is independent of whether a call is overturned.
• Alternative Hypothesis (\(H_1\)): The gender of the tennis player is not independent of whether a call is overturned.

The expected frequency for each cell in a contingency table is calculated using the formula:

\[ E = \frac{R_i \times C_j}{N} \]

where \(R_i\) is the total for row \(i\), \(C_j\) is the total for column \(j\), and \(N\) is the grand total.

• For cell (1, 1): \(E = \frac{1019 \times 909}{2353} = 393.655\)
• For cell (1, 2): \(E = \frac{1019 \times 1444}{2353} = 625.345\)
• For cell (2, 1): \(E = \frac{1334 \times 909}{2353} = 515.345\)
• For cell (2, 2): \(E = \frac{1334 \times 1444}{2353} = 818.655\)

The expected frequencies are:

\[ \begin{bmatrix} 393.655 & 625.345 \\ 515.345 & 818.655 \end{bmatrix} \]

The Chi-Square test statistic is calculated using:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

where \(O\) is the observed frequency and \(E\) is the expected frequency.

• For cell (1, 1): \(\frac{(412 - 393.655)^2}{393.655} = 0.855\)
• For cell (1, 2): \(\frac{(607 - 625.345)^2}{625.345} = 0.538\)
• For cell (2, 1): \(\frac{(497 - 515.345)^2}{515.345} = 0.653\)
• For cell (2, 2): \(\frac{(837 - 818.655)^2}{818.655} = 0.411\)

Sum of all cells: \(\chi^2 = 0.855 + 0.538 + 0.653 + 0.411 = 2.325\)

For a Chi-Square distribution with 1 degree of freedom at \(\alpha = 0.05\), the critical value is:

\[ \chi^2_{(0.05, 1)} = 3.841 \]

The p-value associated with the test statistic \(\chi^2 = 2.325\) is:

\[ P(\chi^2 > 2.325) = 0.127 \]

Since the test statistic \(\chi^2 = 2.325\) is less than the critical value \(3.841\) and the p-value \(0.127\) is greater than \(\alpha = 0.05\), we fail to reject the null hypothesis. There is not enough evidence to conclude that the gender of the tennis player is not independent of whether a call is overturned.

Final Answer