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Sagot :
To address the question, we need to go through the steps of the hypothesis testing method for comparing two proportions. Let's do this step-by-step.
### Step 1: State the Hypotheses
We are testing the claim that men and women have equal success in challenging calls. This sets up our hypotheses as follows:
- 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]
From the provided options, the correct hypotheses setup is:
C. [tex]\( H_0: p_1 = p_2 \)[/tex] and [tex]\(H_1: p_1 \neq p_2\)[/tex]
### Step 2: Calculate the Sample Proportions
We have the following data:
- Men: [tex]\( n_1 = 1420 \)[/tex], [tex]\( x_1 = 416 \)[/tex]
- Women: [tex]\( n_2 = 739 \)[/tex], [tex]\( x_2 = 215 \)[/tex]
The sample proportions are calculated as:
[tex]\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{416}{1420} \approx 0.29296 \][/tex]
[tex]\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{215}{739} \approx 0.29093 \][/tex]
### Step 3: Calculate the Combined Proportion
The combined proportion [tex]\(\hat{p}\)[/tex] is calculated as:
[tex]\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{416 + 215}{1420 + 739} \approx 0.29226 \][/tex]
### Step 4: Calculate the Standard Error
The standard error (SE) for the difference in proportions is calculated using the combined proportion:
[tex]\[ SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \approx \sqrt{0.29226(1 - 0.29226) \left(\frac{1}{1420} + \frac{1}{739}\right)} \approx 0.02063 \][/tex]
### Step 5: Calculate the Test Statistic
The test statistic (z) for the difference in proportions is:
[tex]\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \approx \frac{0.29296 - 0.29093}{0.02063} \approx 0.098 \][/tex]
### Step 6: Determine the P-value
For a two-tailed test, the p-value is calculated as:
[tex]\[ p\text{-value} = 2 \times (1 - \Phi(|z|)) \][/tex]
Given our test statistic [tex]\( z \approx 0.098 \)[/tex], the corresponding p-value is approximately [tex]\( 0.9218 \)[/tex].
### Step 7: Conclusion
At a [tex]\( \alpha = 0.05 \)[/tex] significance level, we compare the p-value with [tex]\( \alpha \)[/tex]:
- If [tex]\( p\text{-value} < \alpha \)[/tex], we reject [tex]\( H_0 \)[/tex].
- If [tex]\( p\text{-value} \geq \alpha \)[/tex], we fail to reject [tex]\( H_0 \)[/tex].
Since [tex]\( p\text{-value} \approx 0.9218 \)[/tex] is much greater than [tex]\( 0.05 \)[/tex], we fail to reject the null hypothesis. This means that there is not enough evidence to suggest that the success rates of men and women in challenging referee calls are different.
### Summary
- 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]
- Test Statistic ([tex]\(z\)[/tex]): [tex]\(\boxed{0.10}\)[/tex]
Thus, we conclude that the success rates for men and women in challenging calls are statistically equal in this context.
### Step 1: State the Hypotheses
We are testing the claim that men and women have equal success in challenging calls. This sets up our hypotheses as follows:
- 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]
From the provided options, the correct hypotheses setup is:
C. [tex]\( H_0: p_1 = p_2 \)[/tex] and [tex]\(H_1: p_1 \neq p_2\)[/tex]
### Step 2: Calculate the Sample Proportions
We have the following data:
- Men: [tex]\( n_1 = 1420 \)[/tex], [tex]\( x_1 = 416 \)[/tex]
- Women: [tex]\( n_2 = 739 \)[/tex], [tex]\( x_2 = 215 \)[/tex]
The sample proportions are calculated as:
[tex]\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{416}{1420} \approx 0.29296 \][/tex]
[tex]\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{215}{739} \approx 0.29093 \][/tex]
### Step 3: Calculate the Combined Proportion
The combined proportion [tex]\(\hat{p}\)[/tex] is calculated as:
[tex]\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{416 + 215}{1420 + 739} \approx 0.29226 \][/tex]
### Step 4: Calculate the Standard Error
The standard error (SE) for the difference in proportions is calculated using the combined proportion:
[tex]\[ SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \approx \sqrt{0.29226(1 - 0.29226) \left(\frac{1}{1420} + \frac{1}{739}\right)} \approx 0.02063 \][/tex]
### Step 5: Calculate the Test Statistic
The test statistic (z) for the difference in proportions is:
[tex]\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \approx \frac{0.29296 - 0.29093}{0.02063} \approx 0.098 \][/tex]
### Step 6: Determine the P-value
For a two-tailed test, the p-value is calculated as:
[tex]\[ p\text{-value} = 2 \times (1 - \Phi(|z|)) \][/tex]
Given our test statistic [tex]\( z \approx 0.098 \)[/tex], the corresponding p-value is approximately [tex]\( 0.9218 \)[/tex].
### Step 7: Conclusion
At a [tex]\( \alpha = 0.05 \)[/tex] significance level, we compare the p-value with [tex]\( \alpha \)[/tex]:
- If [tex]\( p\text{-value} < \alpha \)[/tex], we reject [tex]\( H_0 \)[/tex].
- If [tex]\( p\text{-value} \geq \alpha \)[/tex], we fail to reject [tex]\( H_0 \)[/tex].
Since [tex]\( p\text{-value} \approx 0.9218 \)[/tex] is much greater than [tex]\( 0.05 \)[/tex], we fail to reject the null hypothesis. This means that there is not enough evidence to suggest that the success rates of men and women in challenging referee calls are different.
### Summary
- 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]
- Test Statistic ([tex]\(z\)[/tex]): [tex]\(\boxed{0.10}\)[/tex]
Thus, we conclude that the success rates for men and women in challenging calls are statistically equal in this context.
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