Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Tennis Challenges Since the Hawk-Eye instant replay system for tennis was introduced at the U.S. Open in 2006, men challenged 2441 referee calls, with the result that 1027 of the calls were overturned. Women challenged 1273 referee calls, and 509 of the calls were overturned. We want to use a 0.05 significance level to test the claim that men and women have equal success in challenging calls.

a. Test the claim using a hypothesis test.

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

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

Out of 2441 calls made by men, 1027 calls were overturned. Out of 1273 calls made by women, 509 calls were overturned. It is claimed that men and women have equal success in challenging calls.

Null hypothesis:Men and women have equal success in challenging calls.

$H_0:p_1=p_2$

Alternate hypothesis:Men and women do not have equal success in challenging calls.

$H_1:p_1\ne p_2$

Let $n_1$denote the sample size of the calls challenged by men and $n_2$denote the sample size of the calls challenged by women.

Here, $n_1=2441$and$n_2=1273$

Assume that $x_1$and $x_2$are the number of overturned calls made by men and women respectively.

Let${\hat{p}}_1$be the sample proportion of calls that were made by men and got overturned.

Thus,

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{1027}{2441} \\ =0.4207 \end{gathered}$$

$$\begin{gathered} {\hat{q}}_1=1-{\hat{p}}_1 \\ =0.5793 \end{gathered}$$

Let${\hat{p}}_2$be the sample proportion of calls that were made by women and got overturned.

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{509}{1273} \\ {\hat{p}}_2=0.3998 \end{gathered}$$

Thus,

$$\begin{gathered} {\hat{q}}_2=1-{\hat{p}}_2 \\ =0.6002 \end{gathered}$$

The value of the pooled sample proportion is equal to:

$$\begin{gathered} \overline{p}=\frac{x_1+x_2}{n_1+n_2} \\ =\frac{1027+509}{2441+1273} \\ =0.4136 \end{gathered}$$

Hence,

$$\begin{gathered} \overline{q}=1-\overline{p} \\ =1-0.4136 \\ =0.5864 \end{gathered}$$

The test statistic is equal to:

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}}} \\ =\frac{\left(0.4207-0.3998\right)-0}{\sqrt{\frac{\left(0.4136\right)\left(0.5864\right)}{2441}+\frac{\left(0.4136\right)\left(0.5864\right)}{1273}}} \\ =1.227 \end{gathered}$$

Referring to the standard normal distribution table, the critical values of z corresponding to $\alpha =0.05$for a two-tailed test are equal to -1.96 and 1.96.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.02199.

Here, the value of the test statistic lies between the two critical values.

Therefore, the null hypothesis is failed to reject.

a.

There is not sufficient evidence to reject the claim thatmen and women have equal success in challenging calls.

b.

If the level of significance for a two-tailed test is equal to 0.05, then the corresponding confidence level to construct the confidence interval is equal to 95%.

The confidence interval estimate has the following formula:

$\left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E$

Here, E is the margin of error.

The value of the margin of error is computed below:

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2{\hat{q}}_2}{n_2}} \\ =1.96\times \sqrt{\frac{\left(0.4207\right)\left(0.5793\right)}{2441}+\frac{\left(0.3998\right)\left(0.6002\right)}{1273}} \\ =0.0333 \end{gathered}$$

b.

Substituting the required values, the following confidence interval is obtained:

$$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-E<p_1-p_2<\left({\hat{p}}_1-{\hat{p}}_2\right)+E \\ (0.4207-0.3998)-0.0333<p_1-p_2<(0.4207-0.3998)+0.0333 \\ -0.0124<p_1-p_2<0.0542 \end{gathered}$$

Thus, the 95% confidence interval is equal to (-0.0124, 0.0542).

This confidence interval contains zero that means the difference in the proportions of overturned calls can be equal to 0.

Therefore, the confidence interval suggests that there is not sufficient evidence to reject the claim thatmen and women have equal success in challenging calls.

c.

The sample success ratein challenging callsfor men is equal to 42.07% and the sample success rate in challenging calls for women is 39.98%.

Therefore,it appears that men and women have equal success in challenging calls.