A study was done on body temperatures of men and women. The results are shown in

Question

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 normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. A. B. Reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women. C. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women. D. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women. b. Construct a confidence interval suitable for testing the claim that men have a higher mean body temperature than women.

Explanation / Answer

Solution:

Here, we have to find the confidence interval for difference between two population means.

We are given

X1bar = 97.55

X2bar = 97.39

S1 = 0.86

S2 = 0.73

N1 = 11

N2 = 59

Degrees of freedom = N1 + N2 – 2 = 11+59 – 2 = 68

Alpha = = 0.05

Confidence level = 1 – = 1 – 0.05 = 0.95 or 95%

Critical t value = t = 1.9955

Confidence interval = (X1bar – X2bar) -/+ t*sqrt[(S1^2/N1)+(S2^2/N2)]

Confidence interval = (97.55 - 97.39) -/+ 1.9955*sqrt((0.86^2/11)+(0.73^2/59))

Lower limit = (97.55 - 97.39) - 1.9955*sqrt((0.86^2/11)+(0.73^2/59)) = -0.39109

Upper limit = (97.55 - 97.39) + 1.9955*sqrt((0.86^2/11)+(0.73^2/59)) = 0.711093

Confidence interval = (-0.39109, 0.711093)

-0.391 < µ1 - µ2 < 0.711

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