Question

The table below shows results since 2006 of challenged referee calls in the U.S. Open. Use a 0.05 significance level to test the claim that the gender of the tennis player is independent of whether the call is overturned. Do players of either gender appear to be better at challenging calls?

	Was the Challenge to the Call Successful?
	Yes	No
Men	161	376
Women	68	152

Short answer

The gender of the tennis player is independent of whether the call is overturned. Thus, none of the genders is better to overturn the calls.

Step-by-step solution

Given information

The data for the successfulresults who challenged referee calls in the U.S. Open and the gender is provided.

The level of significance is 0.05.

Compute the expected frequencies

Formula forexpected frequencyis,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The observed frequencies along with row and column totals is,

	Yes	No	Row Total
Men	161	376	537
Women	68	152	220
Column Total	229	528	757

Theexpected frequency tableis represented as,

	Yes	No
Men	162.4478	374.5522
Women	66.5522	153.4478

The expected values are larger than 5. Assuming the players are randomly selected, the requirements of the chi-square test are satisfied.

State the null and alternate hypothesis

To test if the successful call is independent of the gender of the player, the hypotheses are formulated as follows,

\({H_0}:\)The gender of the tennis player is independent of whether the call is overturned.

\({H_1}:\)The gender of the tennis player is dependent on whether the call is overturned.

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {161 - 162.4478} \right)}^2}}}{{162.4478}} + \frac{{{{\left( {376 - 374.5522} \right)}^2}}}{{374.5522}} + ... + \frac{{{{\left( {152 - 153.4478} \right)}^2}}}{{153.4478}}\\ = 0.0637\\ \approx 0.064\end{aligned}\]

Therefore, the value of the test statistic is 0.064.

Compute the degrees of freedom

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {2 - 1} \right)\\ = 1\end{aligned}\)

Therefore, the degrees of freedom are 1.

Compute the critical value

From chi-square table, the critical value for the row corresponding to 1 degree of freedom and at 0.05 level of significance is 3.841.

The p-value is computed as 0.801.

State the decision

Since the critical (3.841) is greater than the value of the test statistic (0.064). In this case, the null hypothesis fails to be rejected.

Therefore, the decision is to fail to reject the null hypothesis.

State the conclusion

There issufficient evidencethatthe gender of the tennis player is independentof the overturning of the call.

Thus, it can be concluded that neither of the genders appears to be better as the overturning of the call is not dependent on the player’s gender.