Question

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?

Short answer

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

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

Test statistic: -4.404

Critical value: -2.3263

P-value: 0.00003

The null hypothesis is rejected.

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

The ability of referees to make the correct decision is quite well as only 26.3% of the challenges raised are overturned, which does not seem to be a considerably high percentage.

Step-by-step solution

Given information

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.

Hypotheses

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.

Sample size, sample proportion, and population proportion

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.

Test statistic

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.

Critical value and p-value

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.

Conclusion of the test

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.