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’s gender.

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Most popular questions from this chapter

Bias in Clinical Trials? Researchers investigated the issue of race and equality of access to clinical trials. The following table shows the population distribution and the numbers of participants in clinical trials involving lung cancer (based on data from “Participation in Cancer Clinical Trials,” by Murthy, Krumholz, and Gross, Journal of the American Medical Association, Vol. 291, No. 22). Use a 0.01 significance level to test the claim that the distribution of clinical trial participants fits well with the population distribution. Is there a race/ethnic group that appears to be very underrepresented?

Race/ethnicity

White

non-Hispanic

Hispanic

Black

Asian/

Pacific

Islander

American Indian/

Alaskan Native

Distribution of

Population

75.6%

9.1%

10.8%

3.8%

0.7%

Number in Lung

Cancer Clinical Trials

3855

60

316

54

12

Benford’s Law. According to Benford’s law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. In Exercises 21–24, test for goodness-of-fit with the distribution described by Benford’s law.

Leading Digits

Benford's Law: Distributuon of leading digits

1

30.10%

2

17.60%

3

12.50%

4

9.70%

5

7.90%

6

6.70%

7

5.80%

8

5.10%

9

4.60%

Author’s Computer Files The author recorded the leading digits of the sizes of the electronic document files for the current edition of this book. The leading digits have frequencies of 55, 25, 17, 24, 18, 12, 12, 3, and 4 (corresponding to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively). Using a 0.05 significance level, test for goodness-of-fit with Benford’s law.

Is the hypothesis test described in Exercise 1 right tailed, left-tailed, or two-tailed? Explain your choice.

Weather-Related Deaths For a recent year, the numbers of weather-related U.S. deaths for each month were 28, 17, 12, 24, 88, 61, 104, 32, 20, 13, 26, 25 (listed in order beginning with January). Use a 0.01 significance level to test the claim that weather-related deaths occur in the different months with the same frequency. Provide an explanation for the result.

The accompanying table is from a study conducted

with the stated objective of addressing cell phone safety by understanding why we use a particular ear for cell phone use. (See “Hemispheric Dominance and Cell Phone Use,” by Seidman, Siegel, Shah, and Bowyer, JAMA Otolaryngology—Head & Neck Surgery,Vol. 139, No. 5.)

The goal was to determine whether the ear choice is associated with auditory or language brain hemispheric dominance. Assume that we want to test the claim that handedness and cell phone ear preference are independent of each other.

a. Use the data in the table to find the expected value for the cell that has an observed frequency of 3. Round the result to three decimal places.

b. What does the expected value indicate about the requirements for the hypothesis test?

Right Ear

Left Ear

No Preference

Right-Handed

436

166

40

Left-Handed

16

50

3

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