In: Math
Since an instant replay system for tennis was introduced at a major tournament, men challenged 1389 referee calls, with the result that 427 of the calls were overturned. Women challenged 779 referee calls, and 227 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? A. Upper H 0: p 1equalsp 2 Upper H 1: p 1not equalsp 2 Your answer is correct.B. Upper H 0: p 1equalsp 2 Upper H 1: p 1less thanp 2 C. Upper H 0: p 1less than or equalsp 2 Upper H 1: p 1not equalsp 2 D. Upper H 0: p 1equalsp 2 Upper H 1: p 1greater thanp 2 E. Upper H 0: p 1not equalsp 2 Upper H 1: p 1equalsp 2 F. Upper H 0: p 1greater than or equalsp 2 Upper H 1: p 1not equalsp 2 Identify the test statistic. zequals .78 . 78 (Round to two decimal places as needed.) Identify the P-value. P-valueequals .782218435 . 435 (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P-value is greater than the significance level of alphaequals0.01, so fail to reject the null hypothesis. There 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 99% confidence interval is -.199 negative . 199less thanleft parenthesis p 1 minus p 2 right parenthesisless than -.200 negative . 200. (Round to three decimal places as needed.)
Let be the true proportion of referee calls that were successfully challenged by men, and be the proportion of calls that were successfully challeged by women tennis players.
We want to test the that men and women have equal success in challenging calls. That is we want to test the claim that
The hypotheses are
ans: A
We have the following sample information
is the sample proportion of success among men
is the sample size of challenges made by men
is the sample proportion of success among women
is the sample size of challenges made by women
Since both the sample sizes are greater than 30, we will use normal distribution to model the difference between the sample proportions. Since the 2 samples are independent we will use 2 sample independent test for proportions
First we will estimate the overall proportion of success among the combined population of men and women
the estimated standard error of the difference between 2 proportions is
The test statistics is
ans: the test statistics is z=0.78
This is a 2 tailed test (because the alternative hypothesis has "not equal to")
The p-value is P(Z<-0.78) + P(Z>0.78)
Using the standard normal tables we get
P(Z>0.78) = 1-P(Z<0.78) = 1-(0.5+0.2823) =0.2177
Due to the symmetry of normal distribution we have P(Z<-0.78) = P(Z>0.78)=0.2177
The p-value is 0.2177*2=0.435
Ans: P-value equals 0.435
We will reject the null hypothesis if the p-value is less than the significance level of .
Ans:The P-value is greater than the significance level of , so fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls.
b) 99% confidence interval has a level of significance
the critical value of z is . This can also be written as
Using the standard normal tables we get that for z=2.58, we get P(Z<2.58) = 0.5+0.4951=0.995
The 99% confidence interval is
ans: 99% confidence interval is -0.037 and 0.069
The hypothesized difference between the proportions is zero (from the null hypothesis). Since the interval contains zero, we do not reject the null hypothesis.
There is not sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls.
part c is not provided in the question