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Sagot :
First, let's begin by defining the null and alternative hypotheses.
We are testing the claim that men and women have equal success in challenging calls. Therefore, we will compare the proportions of overturned calls for men ([tex]\(p_1\)[/tex]) and women ([tex]\(p_2\)[/tex]).
The null hypothesis ([tex]\(H_0\)[/tex]) will state that the proportions are equal, and the alternative hypothesis ([tex]\(H_1\)[/tex]) will state that the proportions are not equal.
So, the hypotheses can be stated as:
- Null hypothesis: [tex]\(H_0: p_1 = p_2\)[/tex]
- Alternative hypothesis: [tex]\(H_1: p_1 \neq p_2\)[/tex]
Among the given options, the correct one is:
- [tex]\(B. H_0: p_1 = p_2\)[/tex]
Next, let's identify the test statistic for the hypothesis test.
Given the data:
- Men challenged 1416 referee calls, and 426 of these calls were overturned.
- Women challenged 755 referee calls, and 225 of these calls were overturned.
Let's denote:
- The sample proportion of overturned calls for men as [tex]\(p_1\)[/tex].
- The sample proportion of overturned calls for women as [tex]\(p_2\)[/tex].
The calculations lead to the following results:
- [tex]\(p_1 = \frac{426}{1416} = 0.3008474576271186\)[/tex]
- [tex]\(p_2 = \frac{225}{755} = 0.2980132450331126\)[/tex]
The combined proportion ([tex]\(p_{combined}\)[/tex]) is:
[tex]\[ p_{combined} = \frac{426 + 225}{1416 + 755} = \frac{651}{2171} = 0.2998618148318747 \][/tex]
The standard error (SE) of the difference between two proportions is calculated using:
[tex]\[ SE = \sqrt{ \frac{p_{combined} (1 - p_{combined})}{n_1} + \frac{p_{combined} (1 - p_{combined})}{n_2} } \][/tex]
Where [tex]\(n_1\)[/tex] is the number of challenges by men (1416) and [tex]\(n_2\)[/tex] is the number of challenges by women (755).
[tex]\[ SE = \sqrt{ 0.2998618148318747 \times (1 - 0.2998618148318747) \times \left(\frac{1}{1416} + \frac{1}{755}\right)} = 0.02064796588746282 \][/tex]
The test statistic ([tex]\(z\)[/tex]) is then given by:
[tex]\[ z = \frac{p_1 - p_2}{SE} = \frac{0.3008474576271186 - 0.2980132450331126}{0.02064796588746282} = 0.13726352559149235 \][/tex]
Therefore, the test statistic is:
[tex]\[ z = 0.14 \][/tex]
(Rounded to two decimal places as required).
We are testing the claim that men and women have equal success in challenging calls. Therefore, we will compare the proportions of overturned calls for men ([tex]\(p_1\)[/tex]) and women ([tex]\(p_2\)[/tex]).
The null hypothesis ([tex]\(H_0\)[/tex]) will state that the proportions are equal, and the alternative hypothesis ([tex]\(H_1\)[/tex]) will state that the proportions are not equal.
So, the hypotheses can be stated as:
- Null hypothesis: [tex]\(H_0: p_1 = p_2\)[/tex]
- Alternative hypothesis: [tex]\(H_1: p_1 \neq p_2\)[/tex]
Among the given options, the correct one is:
- [tex]\(B. H_0: p_1 = p_2\)[/tex]
Next, let's identify the test statistic for the hypothesis test.
Given the data:
- Men challenged 1416 referee calls, and 426 of these calls were overturned.
- Women challenged 755 referee calls, and 225 of these calls were overturned.
Let's denote:
- The sample proportion of overturned calls for men as [tex]\(p_1\)[/tex].
- The sample proportion of overturned calls for women as [tex]\(p_2\)[/tex].
The calculations lead to the following results:
- [tex]\(p_1 = \frac{426}{1416} = 0.3008474576271186\)[/tex]
- [tex]\(p_2 = \frac{225}{755} = 0.2980132450331126\)[/tex]
The combined proportion ([tex]\(p_{combined}\)[/tex]) is:
[tex]\[ p_{combined} = \frac{426 + 225}{1416 + 755} = \frac{651}{2171} = 0.2998618148318747 \][/tex]
The standard error (SE) of the difference between two proportions is calculated using:
[tex]\[ SE = \sqrt{ \frac{p_{combined} (1 - p_{combined})}{n_1} + \frac{p_{combined} (1 - p_{combined})}{n_2} } \][/tex]
Where [tex]\(n_1\)[/tex] is the number of challenges by men (1416) and [tex]\(n_2\)[/tex] is the number of challenges by women (755).
[tex]\[ SE = \sqrt{ 0.2998618148318747 \times (1 - 0.2998618148318747) \times \left(\frac{1}{1416} + \frac{1}{755}\right)} = 0.02064796588746282 \][/tex]
The test statistic ([tex]\(z\)[/tex]) is then given by:
[tex]\[ z = \frac{p_1 - p_2}{SE} = \frac{0.3008474576271186 - 0.2980132450331126}{0.02064796588746282} = 0.13726352559149235 \][/tex]
Therefore, the test statistic is:
[tex]\[ z = 0.14 \][/tex]
(Rounded to two decimal places as required).
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