Question

since an instant replay system for tennis was introduced at a major tournament, men challenged 1394 referee calls, with the result that 417 of the calls were overturned. women challenged 779 referee calls, and 216 of the calls were overturned. use a 0.05 significance level to test the claim that men and women have equal success in challenging calls. complete parts (a) through (c) below.

a. test the claim using a hypothesis test
consider the first sample to be the sample of male tennis players who challenged referee calls and the second sample to be the sample of female tennis players who challenged referee calls. what are the null and alternative hypotheses for the hypothesis test?

a. ( h_0: p_1 geq p_2 ) ( h_1: p_1
eq p_2 )

b. ( h_0: p_1 = p_2 ) ( h_1: p_1 > p_2 )

c. ( h_0: p_1
eq p_2 ) ( h_1: p_1 = p_2 )

d. ( h_0: p_1 = p_2 ) ( h_1: p_1
eq p_2 )

e. ( h_0: p_1 leq p_2 ) ( h_1: p_1
eq p_2 )

f. ( h_0: p_1 = p_2 ) ( h_1: p_1 < p_2 )

identify the test statistic

( z = ) (round to two decimal places as needed )

Explanation:

Step1: Identify the claim

The claim is that men and women have equal success in challenging calls, so \( p_1 = p_2 \) (where \( p_1 \) is the proportion of men's successful challenges and \( p_2 \) is for women). The null hypothesis \( H_0 \) usually contains the equality, and the alternative hypothesis \( H_1 \) is the complement of the claim if testing a two - tailed hypothesis (since we are testing for equality, the alternative is that they are not equal).

So \( H_0:p_1 = p_2 \) and \( H_1:p_1
eq p_2 \), which corresponds to option D.

Step2: Calculate the sample proportions

For men: \( n_1 = 1394 \), \( x_1 = 417 \), so \( \hat{p}_1=\frac{x_1}{n_1}=\frac{417}{1394}\approx0.2992 \)

For women: \( n_2 = 779 \), \( x_2 = 216 \), so \( \hat{p}_2=\frac{x_2}{n_2}=\frac{216}{779}\approx0.2773 \)

The pooled proportion \( \hat{p}=\frac{x_1 + x_2}{n_1 + n_2}=\frac{417+216}{1394 + 779}=\frac{633}{2173}\approx0.2913 \)

Step3: Calculate the test statistic \( z \)

The formula for the z - statistic for two - proportion test is:

\( z=\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}} \)

Substitute the values:

\( \hat{p}_1-\hat{p}_2=0.2992 - 0.2773 = 0.0219 \)

\( 1-\hat{p}=1 - 0.2913 = 0.7087 \)

\( \frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{1394}+\frac{1}{779}\approx\frac{779+1394}{1394\times779}=\frac{2173}{1086926}\approx0.001999 \)

\( \sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{0.2913\times0.7087\times0.001999} \)

\( 0.2913\times0.7087\approx0.2065 \)

\( 0.2065\times0.001999\approx0.000413 \)

\( \sqrt{0.000413}\approx0.0203 \)

Then \( z=\frac{0.0219}{0.0203}\approx1.08 \)

Answer:

The null and alternative hypotheses are D. \( H_0:p_1 = p_2 \), \( H_1:p_1
eq p_2 \) and the test statistic \( z\approx1.08 \)