Question

1.

Identify the test statistic.
Z=_____(Round to two decimal places as needed.)

Identify the P-value.
P=_____(Round to three decimal places as needed.)

What is the conclusion based on the hypothesis test?

The P-value is (1)_____ the significance level of α=0.05, so (2)_____ the null hypothesis. There (3)_____ 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% or 99% 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 (4)_____ 0, there (5) _____ appear to be a significant difference between the two proportions. There (6)_____ evidence to warrant rejection of the claim that men and women have equal success in challenging calls

(1) less than
greater than

(2) fail to reject
reject

(3) is sufficient
is not sufficient

(4) include
do not include

(5) does
does not

(6) is sufficient
is not sufficient

Homework: Section 9.1 Homework Save Score: 0 of 1 pt 5 of 10 (4 complete) HW Score: 12.37%, 1.24 of 10 pts 9.1.10-T Question Help Since an instant replay system for tennis was introduced at a major tournament, men challenged 1432 referee calls, with the result that 415 of the calls were overturned. Women challenged 739 referee calls, and 214 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? C. Ho: P1 SP2 H4: P1 *P2 O A. Ho: P1 *P2 H:P1 = P2 OD. Ho: P1 = P2 H:P1 P2 O B. Ho: P1 = P2 H:P1>P2 E. Ho: P1 = P2 H:P1P2 OF. Ho: P1 P2 HEP1 *P2 Identify the test statistic. ZE (Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer. ? 5 parts Clear All Check Answer remaining

2.

3.

Answer 1

1. Minitab output:

Test and CI for Two Proportions

Sample X N Sample p
Men 415 1432 0.289804
Women 214 739 0.289581

Difference = p (Men) - p (Women)
Estimate for difference: 0.000223955
95% CI for difference: (-0.0400441, 0.0404920)
Test for difference = 0 (vs not = 0): Z = 0.01 P-Value = 0.991

Fisher's exact test: P-Value = 1.000

Z=0.01
P-value=0.991
The P-value is greater than the significance level of α=0.05, 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 95% confidence interval is -0.040 <(P₁-P₂)<0.040

Because the confidence interval limits include 0, 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

Problem 2:

(a) Minitab output:

Test and CI for Two Proportions

Sample X N Sample p
Oxygen 108 148 0.729730
Placebo 26 154    0.168831

Difference = p (Oxygen) - p (Placebo)
Estimate for difference: 0.560899
99% lower bound for difference: 0.450702
Test for difference = 0 (vs > 0): Z = 9.81 P-Value = 0.000

Fisher's exact test: P-Value = 0.000

Z=9.81

P-value=0.000

The P-value is less than the significance level of $\alpha$ =0.01, so reject the null hypothesis. There is sufficient evidence to support the claim that the cure rate with oxygen treatment is higher than the cure rate for those gives a placebo.

(b) Minitab output:

Test and CI for Two Proportions

Sample X N Sample p
Oxygen 108 148 0.729730
Placebo 26 154 0.168831

Difference = p (Oxygen) - p (Placebo)
Estimate for difference: 0.560899
98% CI for difference: (0.450702, 0.671095)
Test for difference = 0 (vs not = 0): Z = 9.81 P-Value = 0.000

Fisher's exact test: P-Value = 0.000

The 98% confidence interval is 0.451 <(P₁-P₂)<0.671

Because the confidence interval limits do not include 0, it appears that the two cure rates are significantly different. Because the confidence interval limits include positive values, it appears that the cure rate is higer for oxygen treatment than for placebo.

c. Option A.

3. Minitab output:

Test and CI for Two Proportions

Sample X N Sample p
Male 18 244 0.073770
Female 65 495 0.131313

Difference = p (Male) - p (Female)
Estimate for difference: -0.0575426
95% upper bound for difference: -0.0203792
Test for difference = 0 (vs < 0): Z = -2.33 P-Value = 0.010

Fisher's exact test: P-Value = 0.012

Z=-2.33

P-value=0.010

The P-value is less than the significance level of $\alpha$ =0.05, so reject the null hypothesis. There is sufficient evidence to support the claim that rate of left handedness among males is less than among females.

b. Minitab output:

Test and CI for Two Proportions

Sample X N Sample p
Male 18 244 0.073770
Female 65 495 0.131313

Difference = p (Male) - p (Female)
Estimate for difference: -0.0575426
90% CI for difference: (-0.0947061, -0.0203792)
Test for difference = 0 (vs not = 0): Z = -2.33 P-Value = 0.020

Fisher's exact test: P-Value = 0.019

The 90% confidence interval is -0.095 <(P₁-P₂)<-0.020

Because the confidence interval limits do not include 0, it appears that two rates of left handedness are significantly different. There is sufficient evidence to support the claim that rate of left handedness among males is less than among females.