Question

This Question: 1 pt 19 of 24 16 complete This Test: 24 pts poss Men A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normaly distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts Women P2 50 X 9765 F 0.84 0.737 a. Test the claim that men have a higher mean body temperature than women. What are the nul and alternative hypotheses? ОА но и на OBH: HE ODH осы и на The best statistics (Round to two decimal places as needed.) The Pris (Round to three decimal places as needed.) State the conclusion for the test. O A Reject the mullypothesis. There is suficient evidence to support the claim that men have a higher mean body temperature Than women OB. Reject the null hypothesis. There is not contevidence to support the claim that man have a higher mean bodylomerwure than women Oc. Fall to reject the nul hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women Click to select your answers

Answer 1

Answer: A study was done on body temperatures of men and women. The result are shown in table.

Solution:

The null and alternative hypothesis:

Ho: μ1 = μ2

Ha: μ1 > μ2

The option B is correct answer.

Test statistic t:

t = (x̄1 - x̄2)/√(s1^2/n1 + s2^2/n2)

t = (97.65 - 97.42)/√(0.84^2/11 + 0.73^2/59)

t = 0.23 / 0.2705

t = 0.8503

Test statistic t = 0.85

P-value:

df = [s1^2/n1 + s2^2/n2]^2 / [s1^2/n1]^2/n1-1 + [s2^2/n2]^2/n2-1

df = [0.84^2/11 + 0.73^2/59]^2 / [0.84^2/11]^2/11-1 + [0.73^2/59]^2/59-1

df = 12.966

df = 13

P-value = 0.205

Since, P-value (0.205) > α (0.01) significance level.

We do not reject the null hypothesis Ho.

Conclusion:

Fail to reject the null hypothesis Ho. There is not sufficient evidence to support the claim that the men have higher body temperature than women.

the confidence interval:

At α = 0.01

t critical = t(0.005,13) = 3.0123

CI = x̄1 - x̄2 ± t critical * SE

CI = 97.65 - 97.42 ± 3.0123 * 0.2705

CI = 0.23 ± 0.8148

CI = (-0.5848, 1.0448)

Therefore, confidence interval is

-0.585 < μ1 - μ2 < 1.045

Since confidence interval contains 0. It does not support the claim.

No, because the confidence interval contains zero