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since an instant replay system for tennis was introduced at a major tou…

Question

since an instant replay system for tennis was introduced at a major tournament, men challenged 1432 referee calls, with the result that 419 of the calls were overturned. women challenged 769 referee calls, and 212 of the calls were overturned. use a 0.05 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. h₀: p₁ = p₂ h₁: p₁ > p₂ b. h₀: p₁ ≥ p₂ h₁: p₁ ≠ p₂ c. h₀: p₁ = p₂ h₁: p₁ < p₂ d. h₀: p₁ ≠ p₂ h₁: p₁ = p₂ e. h₀: p₁ = p₂ h₁: p₁ ≠ p₂ f. h₀: p₁ ≤ p₂ h₁: p₁ ≠ p₂ identify the test statistic. z = (round to two decimal places as needed.) identify the p - value. p - value = (round to three decimal places as needed.) what is the conclusion based on the hypothesis test? the p - value is the significance level of α = 0.05, so the null hypothesis. there evidence to warrant rejection of the claim that women and men have equal success in challenging calls. b. test the claim by constructing an appropriate confidence interval. the 95% confidence interval is < (p₁ - p₂) < (round to three decimal places as needed.) what is the conclusion based on the confidence interval? because the confidence interval limits 0, there appear to be a significant difference between the two proportions. there evidence to warrant rejection of the claim that men and women have equal success in challenging calls.

Explanation:

Step 1: Identify Hypotheses

The claim is that men and women have equal success in challenging calls, so the null hypothesis \( H_0 \) is \( p_1 = p_2 \). The alternative hypothesis \( H_1 \) (since we're testing equality) is \( p_1
eq p_2 \). So the correct option is D: \( H_0: p_1 = p_2 \), \( H_1: p_1
eq p_2 \).

Step 2: Calculate Proportions

For men: \( n_1 = 1432 \), \( x_1 = 419 \), so \( \hat{p}_1 = \frac{419}{1432} \approx 0.2926 \).
For women: \( n_2 = 769 \), \( x_2 = 212 \), so \( \hat{p}_2 = \frac{212}{769} \approx 0.2757 \).
Pooled proportion \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{419 + 212}{1432 + 769} = \frac{631}{2201} \approx 0.2867 \).

Step 3: Calculate Test Statistic

The formula for the z - statistic in two - proportion test is:

$$ z=\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}} $$

Substitute the values:

$$ LATEXBLOCK0 $$

Step 4: Calculate P - value

Since \( H_1: p_1
eq p_2 \), it is a two - tailed test. The P - value is \( 2\times P(Z > |z|) \). For \( z = 0.84 \), \( P(Z>0.84)=1 - P(Z\leq0.84) \). From standard normal table, \( P(Z\leq0.84) = 0.7995 \), so \( P(Z > 0.84)=1 - 0.7995 = 0.2005 \). Then P - value \(=2\times0.2005 = 0.401 \).

Step 5: Conclusion for Hypothesis Test

Since the P - value (\( 0.401 \)) is greater than \( \alpha=0.05 \), we fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have equal success in challenging calls.

Step 6: Confidence Interval for \( p_1 - p_2 \)

The formula for the 95% confidence interval for \( p_1 - p_2 \) is:

$$ (\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}

\( z_{\alpha/2}=z_{0.025}=1.96 \)
\( \frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}=\frac{0.2926\times(1 - 0.2926)}{1432}=\frac{0.2926\times0.7074}{1432}\approx\frac{0.2070}{1432}\approx0.0001446 \)
\( \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}=\frac{0.2757\times(1 - 0.2757)}{769}=\frac{0.2757\times0.7243}{769}\approx\frac{0.1997}{769}\approx0.0002597 \)
\( \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}=\sqrt{0.0001446 + 0.0002597}=\sqrt{0.0004043}\approx0.0201 \)
\( \hat{p}_1-\hat{p}_2 = 0.0169 \)
Lower limit: \( 0.0169-1.96\times0.0201=0.0169 - 0.0394=-0.0225 \)
Upper limit: \( 0.0169 + 1.96\times0.0201=0.0169+0.0394 = 0.0563 \)
Since 0 is included in the interval \((-0.0225,0.0563)\), there does not appear to be a significant difference between the two proportions. There is not sufficient evidence to warrant rejection of the claim that men and women have equal success in challenging calls.

Answer:

a. Hypotheses: D. \( H_0: p_1 = p_2 \), \( H_1: p_1
eq p_2 \); Test statistic \( z\approx0.84 \); P - value \( = 0.401 \); Conclusion: Fail to reject \( H_0 \), not sufficient evidence.
b. 95% confidence interval: \( - 0.023<(p_1 - p_2)<0.056 \) (rounded to three decimal places); Since 0 is in the interval, no significant difference, not sufficient evidence.