Question

Since an instant replay system for tennis was introduced a a major tournament, men challenged 1438 referee calls, with the result that 422 of the calls were overturned. Women challenged 740 referee calls, and 212 of the calls were overturned. Use a .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? Identify the test statistic. Z= (round to 2 decimal places as needed) Identify the P-value. P= (round to 3 decimal places as needed) What is the conclusion based on the hypothesis test? The P-value is Less (than/greater) than the significance level of alpha= .05, so (fail to reject/reject) the null hypothesis. There (is sufficient/is not sufficient) 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 x<(p1-p2)

Answer 1

Part A

Z test for population proportions

H₀: p₁ = p₂ versus H_a: p₁ ≠ p₂

α = 0.05

WE are given

X1=422, N1=1438, X2=212, N2=740

P1=X1/N1=422/1438= 0.293463143

P2=X2/N2=212/740= 0.286486486

P=(X1+X2)/(N1+N2)=(422+212)/(1438+740) = 0.2911

Z = (P1-hat – P2-hat) / sqrt(p-hat*(1 – P-hat)*((1/N1) + (1/N2)))

Z = (0.293463143 – 0.286486486) / sqrt(0.2911*(1 – 0.2911)*((1/1438) + (1/740)))

Z = 0.3395

P-value = 0.7343

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that men and women have equal success in challenging calls.

Part B

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = (0.293463143 – 0.286486486) ± 1.96*sqrt[(0.293463143*(1 – 0.293463143)/1438) + (0.286486486*(1 – 0.286486486)/740)]

Confidence interval = (0.293463143 – 0.286486486) ± 0.0402

Confidence interval = 0.006976657 ± 0.0402

Lower limit = 0.006976657 - 0.0402 =-0.033223343

Upper limit = 0.006976657 + 0.0402 = 0.047176657

Confidence interval = (0.0332, 0.0472)