Question

Since an instant replay system for tennis was introduced at a major tournament, men challenged 1433 referee calls, with the result that 417 of the calls were overturned. Women challenged 758 referee calls, and 222 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. 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? O A. Ho: *P2 HP, = P2 OD. Ho: P = P2 H:P, > P2 OB. Ho: P, SP2 HD *P2 O E. HOP, = P2 H:P, <P2 O C. Hop, 2P2 HP, P2 OF. Ho: P = P2 H:D P2 Identify the test statistic. zan (Round to two decimal places as needed.) Identify the P-value P-value=D (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? the null hypothesis. There evidence to warrant rejection of the claim that women and The P-value is the significance level of a = 0.01, SO men have equal success in challenging calls. b. Test the claim by constructing an appropriate confidence interval. The 99% confidence interval is <0,-P2) - (Round to three decimal places as needed.) What is the conclusion based on the confidence interval? Because the confidence interval limits , there appear to be a significant difference between the two proportions. There evidence to warrant

Options for "The P-value is" Greater than. Less than....... so reject or fail to reject...... There is sufficient or nonsufficient evidence

he Because the confidence interval limit

include

do not include

does

does not

appear to be a significant difference between the two proportions. There

is not sufficient

is sufficient

evidence to warrant rejection of the claim that men and

Answer 1

For sample 1, we have that the sample size is N1 = 1433, the number of favorable cases is X1 = 417, so then the sample proportion is i = = = 0.291 For sample 2, we have that the sample size is N2 = 758, the number of favorable cases is X2 = 222, so then the sample proportion is p2 = * = 222 = 0.2929 The value of the pooled proportion is computed as p= $.t. " Ni+Ng 417 +222_02916 1433+758 = 0.2916 Also, the given significance level is a = 0.01 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p1 = P2 Ha: px = P2 This corresponds to a two-tailed test, for which a z-test for two population proportions needs to be conducted. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a two-tailed test is %c = 2.58. The rejection region for this two-tailed testis R= {2:21 > 2.58} (3) Test Statistics The Z-statistic is computed as follows: 2 = 1 - 2 P(1 - P(1/ni +1/n2) 0.291 - 0.2929 0.2916 . (1 - 0.2916)(1/1433 + 1/758) = -0.092 (4) Decision about the null hypothesis Since it is observed that 2 = 0.092 < c = 2.58, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.9267, and since p = 0.9267 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion pı is different than P2, at the 0.01 significance level. Confidence Interval The 99% confidence interval for P1 - P2 is: -0.054 < P1 - P2 < 0.051.

a) Option F represent correct hypothesis.

Z = -0. 092

P = 0.927

The p value is greater than significance level alpha, so fail to reject the null hypothesis. There is no evidence to warrant the rejection of the claim.....

b) 99% CI is (-0.054, 0.051)

Confidence interval limits include zero, there is doesn't apper to be significant difference bet two proportion there is not sufficient evidence to Warrent....

Option b is correct.

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