Testing Claims About Proportions. In Exercises 9–32, 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. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Tennis Instant Replay The Hawk-Eye electronic system is used in tennis for displaying an instant replay that shows whether a ball is in bounds or out of bounds so players can challenge calls made by referees. In a recent U.S. Open, singles players made 879 challenges and 231 of them were successful, with the call overturned. Use a 0.01 significance level to test the claim that fewer than 1/ 3 of the challenges are successful. What do the results suggest about the ability of players to see calls better than referees?

Out of 879 challenges made by single players, a total of 231 are overturned. It is claimed that less than of $\frac{1}{3}$challenges are successful.

The null hypothesis is written as follows.

The proportion of successful challenges is equal to $\frac{1}{3}$.

$H_0:p=0.333$

The alternative hypothesis is written as follows.

The proportion of successful challenges is less than $\frac{1}{3}$.

$H_1:p<0.333$

The test is left-tailed.

The sample size is equal to n=879.

The sample proportion of successful challenges is computed below.

$$\begin{gathered} \hat{p}=\frac{\text{Number}\text{of}\text{successful}\text{challenges}}{\text{Sample}\text{Size}} \\ =\frac{231}{879} \\ =0.263 \end{gathered}$$

The population proportion of successful challenges is equal to 0.333.

The value of the test statistic is computed below.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.263-0.333}{\sqrt{\frac{0.333\left(1-0.333\right)}{879}}} \\ =-4.404 \end{gathered}$$

Thus, z=-4.404.

Referring to the standard normal table, the critical value of z at $\alpha =0.01$for a left-tailed test is equal to -2.3263.

Referring to the standard normal table, the p-value for the test statistic value of -4.404 is equal to 0.00003.

As the p-value is less than 0.01, the null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of successful challenges is less than $\frac{\mathbf{1}}{\mathbf{3}}$.

About 26.3% of the challenges raised are overturned, which doesn’t seem to be a very high percentage. Thus, the ability of referees to make the correct decision is evident.