Math / Data & Statistics

QuestionSince an instant replay system for tennis was introduced at a major tournament, men challenged 1384 referee calls, with the result that 431 of the calls were overturned. Women challenged 762 referee calls, and 212 of the calls were overtumed. 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. $H_{0}: p_{1} \leqslant p_{2}$ $H_{1}: p_{1} \neq p_{2}$ D. $H_{0}: P_{1}=p_{2}$ $H_{1}: P_{1}>P_{2}$ B. $H_{0}: p_{1} \geq p_{2}$ $H_{1}: p_{1} \neq p_{2}$ E. $H_{0}: p_{1}=p_{2}$ $H_{1}: p_{1} \neq p_{2}$ C. $H_{0}: p_{1}=p_{2}$ $H_{1}: p_{1}<p_{2}$ F. $H_{0}: p_{1} \neq p_{2}$ $H_{1}: p_{1}=p_{2}$ Identify the test statistic. $z=$ $\square$ (Round to two decimal places as needed.)

Studdy Solution

STEP 1

1. We are conducting a hypothesis test to compare two proportions.
2. The significance level is $\alpha = 0.01$ .
3. The first sample is from male tennis players, and the second sample is from female tennis players.

STEP 2

1. Define the null and alternative hypotheses.
2. Calculate the test statistic.
3. Determine the conclusion of the hypothesis test.

STEP 3

Define the null and alternative hypotheses.
For this problem, we want to test if men and women have equal success in challenging calls. Therefore, the appropriate hypotheses are:
$H_0: p_1 = p_2$ $H_1: p_1 \neq p_2$
This corresponds to option E.

STEP 4

Calculate the test statistic.
First, calculate the sample proportions: - For men: $\hat{p}_1 = \frac{431}{1384}$ - For women: $\hat{p}_2 = \frac{212}{762}$
Next, calculate the pooled sample proportion: $\hat{p} = \frac{431 + 212}{1384 + 762}$
Calculate the standard error: $SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{1384} + \frac{1}{762}\right)}$
Finally, calculate the test statistic $z$ : $z = \frac{\hat{p}_1 - \hat{p}_2}{SE}$
Perform the calculations:
$\hat{p}_1 = \frac{431}{1384} \approx 0.3114$ $\hat{p}_2 = \frac{212}{762} \approx 0.2782$ $\hat{p} = \frac{431 + 212}{1384 + 762} = \frac{643}{2146} \approx 0.2996$ $SE = \sqrt{0.2996 \times (1 - 0.2996) \times \left(\frac{1}{1384} + \frac{1}{762}\right)} \approx 0.0231$ $z = \frac{0.3114 - 0.2782}{0.0231} \approx 1.44$

STEP 5

Determine the conclusion of the hypothesis test.
Since the calculated $z$ -value is approximately 1.44, we compare this to the critical $z$ -value for a two-tailed test at the 0.01 significance level, which is approximately $\pm 2.576$ .
Since $1.44$ is within the range $-2.576$ to $2.576$ , we fail to reject the null hypothesis.
The test statistic is:
$z = \boxed{1.44}$

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