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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

Expert verified

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

01

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.

02

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.

03

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.

04

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.

05

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.

06

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.

07

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.

08

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鈥檚 gender.

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