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. Use a 0.05 significance level for both parts.

(Women on right, won't let me label header.)

	Men
μ	μ1		μ2
n	11		59
x overbar	97.63°F		97.37°F
s	0.95°F		0.74°F

Test the claim that men have a higher mean body temperature than women.

Answer 1

n1 = 11

n2=59

s1 = 0.95

s2 = 0.74

Null and alternative hypothesis is

H0 :u1 = u2

Vs

H1 :u1> u2

Level of significance = 0.05

Before doing this test we have to check population variances are equal or not.

Null and alternative hypothesis is

Vs

Test statistic is

F = Larger variance / Smaller variance = 0.9025 / 0.5476 = 1.648

Degrees of freedoms

Degrees of freedom for numerator = n1 - 1 = 14 - 1 = 13

Degrees of freedom for denominator = n2 - 1 = 14 - 1 = 13

Critical value = 1.998( using f-table )

F test statistic < critical value    we fail to reject null hypothesis.

Conclusion: Population variances are equal.

So we have to use pooled variance.

Formula

, d.f = n1 + n2 – 2 = 11 + 59 - 2 = 68

p-value = 0.155 ( using t table )

p-value , Failed to Reject H0

conclusion : There is a insufficient evidence to conclude that  the men have a higher mean body temperature than women.