Question

Since an instant replay system for tennis was introduced at a major​ tournament, men challenged

1413

referee​ calls, with the result that

416

of the calls were overturned. Women challenged

776

referee​ calls, and

225

of the calls were overturned. Use a

0.01

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.

H0​:

p1≥p2

H1​:

p1≠p2

B.

H0​:

p1=p2

H1​:

p1>p2

C.

H0​:

p1=p2

H1​:

p1<p2

D.

H0​:

p1=p2

H1​:

p1≠p2

E.

H0​:

p1≠p2

H1​:

p1=p2

F.

H0​:

p1≤p2

H1​:

p1≠p2

Identify the test statistic.

z=nothing

​(Round to two decimal places as​ needed.)

Identify the​ P-value.

​P-value=nothing

​(Round to three decimal places as​ needed.)

What is the conclusion based on the hypothesis​ test?

The​ P-value is

▼

greater than

less than

the significance level of

α=0.01​,

so

▼

reject

fail to reject

the null hypothesis. There

▼

is not sufficient

is sufficient

evidence to warrant rejection of the claim that women and men have equal success in challenging calls.

b. Test the claim by constructing an appropriate confidence interval.

The

99​%

confidence interval is

nothing<p1−p2<nothing.

​(Round to three decimal places as​ needed.)

What is the conclusion based on the confidence​ interval?

Because the confidence interval limits

▼

include

do not include

​0, there

▼

does not

does

appear to be a significant difference between the two proportions. There

▼

is sufficient

is not sufficient

evidence to warrant rejection of the claim that men and women have equal success in challenging calls.

c. Based on the​ results, does it appear that men and women may have equal success in challenging​ calls?

A.

The confidence interval suggests that there is a significant difference between the success of men and women in challenging calls. It is reasonable to speculate that men have more success.

B.

The confidence interval suggests that there is no significant difference between the success of men and women in challenging calls.

C.

The confidence interval suggests that there is a significant difference between the success of men and women in challenging calls. It is reasonable to speculate that women have more success.

D.

There is not enough information to reach a conclusion.

Answer 1

Let p1 be proportion of male tennis players who challenged referee calls.

and p2 be proportion of female tennis players who challenged referee calls.

Here we have to test that

Null hypothesis:

Ho: P1 = P2

Alternative hypothesis: $H_1:p_1\neq p_2$

$n_1=1413,x_1=416$

n2 = 776,22 776,22 = 225

Sample proportions:

(Round to 4 decimal)

11 416 Ôi = 0.2944 ni 1413

$\hat{p}_2=\frac{x_2}{n_2}=\frac{225}{776}=0.2899$    (Round to 4 decimal)

Pooled proportion:

$\bar{p}=\frac{x_1+x_2}{n_1+n_2}$

$\bar{p}=\frac{416+225}{1413+776}$

$\bar{p}=\frac{641}{2189}$

$\bar{p}=0.2928$   (Round to 4 decimal)

Test statistic:

$z=\frac{\hat{p}_1-\hat{p}_2}{SE(\hat{p}_1-\hat{p}_2)}$

where

$SE(\hat{p}_1-\hat{p}_2)=\sqrt{\frac{\bar{p}*(1-\bar{p})}{n_1}+\frac{\bar{p}*(1-\bar{p})}{n_2}}$

$SE(\hat{p}_1-\hat{p}_2)=\sqrt{\frac{0.2928*(1-0.2928)}{1413}+\frac{0.2928*(1-0.2928)}{776}}$

$SE(\hat{p}_1-\hat{p}_2)=\sqrt{\frac{0.207068}{1413}+\frac{0.207068}{776}}$

$SE(\hat{p}_1-\hat{p}_2)=\sqrt{0.000147+0.000267}$

$SE(\hat{p}_1-\hat{p}_2)=\sqrt{0.000413}$

$SE(\hat{p}_1-\hat{p}_2)=0.0203$ (Round to 4 decimal)

$z=\frac{0.2944-0.2899}{0.0203}$

$z=\frac{0.0045}{0.0203}$

z = 0.22 (Round to 2 decimal)

Test statistic = z = 0.22

P value:

Test is two tailed test.

P value = 2 * P(z > 0.22)

= 2 * [1 - P(z < 0.22)]

= 2 * [1 - 0.5871] (From statistical table of z values)

= 2 * 0.4129

= 0.826 (Round to 3 decimal)

P value = 0.826

Level of significance = $\alpha$ = 0.01

The P value is greater than the significance level of  $\alpha$ = 0.01

So we fail to reject the null hypothesis.

There is not sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls.

99% confidence interval is

$\hat{p}_1-\hat{p}_2-z_c*SE(\hat{p}_1-\hat{p}_2)<p_1-p_2<\hat{p}_1-\hat{p}_2+z_c*SE(\hat{p}_1-\hat{p}_2)$

where zc is z critical value for (1+c)/2 = (1+0.99)/2 = 0.995

zc = 2.58 (From statistical table of z values)

$(0.2944-0.2899)-2.58*0.0203<p_1-p_2<(0.2944-0.2899)+2.58*0.0203$

$0.0045-0.052374<p_1-p_2<0.0045+0.052374$

$-0.048<p_1-p_2<0.057$ (Round to 3 decimal)

99% confidence interval is (-0.048, 0.057)

Because the confidence interval limits include 0, there does not appear to be a significant difference between the two proportions.

There is not sufficient evidence to warrant rejection of the claim that men and women have equal success in challenging calls.

The confidence interval suggests that there is no significant difference between the success of men and women in challenging calls.