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Since an instant replay system for tennis was introduced at a major tournament, men challenged 1407 referee calls, with the result that 414 of the calls were overturned. Women challenged 770 referee calls, and 224 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.
[tex]\[
\begin{array}{r}
H_0: p_1 = p_2 \\
H_1: p_1 \neq p_2
\end{array}
\][/tex]

Identify the test statistic.
[tex]\[
z = 0.16
\][/tex]
(Round to two decimal places as needed.)

Identify the P-value.
[tex]\[
\text{P-value} = 0.87
\][/tex]
(Round to three decimal places as needed.)

B.
[tex]\[
\begin{array}{l}
H_0: p_1 = p_2 \\
H_1: p_1 \ \textless \ p_2
\end{array}
\][/tex]

C.
[tex]\[
\begin{array}{l}
H_0: p_1 = p_2 \\
H_1: p_1 \neq p_2
\end{array}
\][/tex]

D.
[tex]\[
\begin{aligned}
H_0: p_1 = p_2 \\
H_1: p_1 \ \textgreater \ p_2
\end{aligned}
\][/tex]

E.
[tex]\[
\begin{array}{l}
H_0: p_1 = p_2 \\
H_1: p_1 \neq p_2
\end{array}
\][/tex]

F.
[tex]\[
\begin{array}{l}
H_0: p_1 \geq p_2 \\
H_1: p_1 \ \textless \ p_2
\end{array}
\][/tex]

What is the conclusion based on the hypothesis test?
The P-value is greater than the significance level of [tex]\alpha = 0.01[/tex], 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.

Test the claim by constructing an appropriate confidence interval.
The [tex]99\%[/tex] confidence interval is [tex]\square \ \textless \ (p_1 - p_2) \ \textless \ \square[/tex]
(Round to three decimal places as needed.)



Answer :

GinnyAnswer
Let's break down the process of solving this problem step-by-step:

### Step 1: Establish Hypotheses
We need to determine which hypothesis is appropriate for our test of whether men and women have equal success rates in overturning referee calls.

The appropriate hypotheses for testing equality are:
- Null Hypothesis ([tex]\( H_0 \)[/tex]): [tex]\( p_1 = p_2 \)[/tex]
- Alternative Hypothesis ([tex]\( H_1 \)[/tex]): [tex]\( p_1 \neq p_2 \)[/tex]

So the correct hypotheses are:
[tex]\[ H_0: p_1 = p_2 \][/tex]
[tex]\[ H_1: p_1 \neq p_2 \][/tex]

### Step 2: Calculate Proportions
From the data:
- Men: [tex]\( 1407 \)[/tex] total challenges, [tex]\( 414 \)[/tex] overturned.
- Women: [tex]\( 770 \)[/tex] total challenges, [tex]\( 224 \)[/tex] overturned.

Calculate the sample proportions:
[tex]\[ \hat{p}_1 = \frac{414}{1407} = 0.294 \][/tex]
[tex]\[ \hat{p}_2 = \frac{224}{770} = 0.291 \][/tex]

### Step 3: Calculate the Pooled Proportion
The pooled proportion ([tex]\( \hat{p} \)[/tex]) combines both groups:
[tex]\[ \hat{p} = \frac{414 + 224}{1407 + 770} = 0.293 \][/tex]

### Step 4: Calculate the Standard Error
Using the pooled proportion, calculate the standard error:
[tex]\[ \text{SE} = \sqrt{\hat{p} \cdot (1 - \hat{p}) \left(\frac{1}{1407} + \frac{1}{770}\right)} = 0.0204 \][/tex]

### Step 5: Calculate the Test Statistic
The Z-test statistic for the difference between proportions is given by:
[tex]\[ Z = \frac{\hat{p}_1 - \hat{p}_2}{\text{SE}} = \frac{0.294 - 0.291}{0.0204} = 0.163 \][/tex]
The test statistic is [tex]\( Z = 0.16 \)[/tex] (rounded to two decimal places).

### Step 6: Calculate the P-value
With the Z-value, we find the P-value:
[tex]\[ \text{P-value} = 2 \cdot (1 - \text{norm.cdf}(|Z|)) \][/tex]
For [tex]\( Z = 0.16 \)[/tex], the P-value is approximately [tex]\( 0.87 \)[/tex] (rounded to three decimal places).

### Conclusion Based on Hypothesis Test
Since the P-value ([tex]\( 0.87 \)[/tex]) is greater than the significance level ([tex]\( \alpha = 0.01 \)[/tex]), we fail to reject the null hypothesis. This means there is not sufficient evidence to conclude that men and women have different success rates in challenging calls.

### Step 7: Construct a Confidence Interval
Next, we construct a [tex]\( 99\% \)[/tex] confidence interval for [tex]\( \hat{p}_1 - \hat{p}_2 \)[/tex].

The margin of error is:
[tex]\[ \text{Margin of Error} = Z_{\alpha/2} \times \text{SE} \][/tex]
For a [tex]\( 99\% \)[/tex] confidence interval, [tex]\( Z_{\alpha/2} \approx 2.576 \)[/tex]:
[tex]\[ \text{Margin of Error} = 2.576 \times 0.0204 = 0.0526 \][/tex]

The confidence interval for the difference in proportions is:
[tex]\[ (\hat{p}_1 - \hat{p}_2) \pm \text{Margin of Error} \][/tex]
[tex]\[ 0.003 \pm 0.0526 \][/tex]

So the [tex]\( 99\% \)[/tex] confidence interval is:
[tex]\[ -0.049 \text{ to } 0.056 \][/tex]

### Final Answer
A. The test statistic is:
[tex]\[ Z = 0.16 \][/tex]

The P-value is:
[tex]\[ \text{P-value} = 0.87 \][/tex]

There is not sufficient evidence to reject the null hypothesis.

B. The [tex]\( 99\% \)[/tex] confidence interval is:
[tex]\[ -0.049 < (p_1 - p_2) < 0.056 \][/tex]

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