Under normal conditions, is the average body temperature the same for men and women? Medical researchers interested in this question collected data from a large number of men and women, and random samples from that data are presented in the accompanying table below.

(a) Set up a hypothesis test for testing whether there is sufficient evidence to indicate that mean body temperatures differ for men and women.

(b) What is the exact p-value? What does this value indicate?

(c) Construct a 95% confidence interval for the mean body temperatures between men and women.

The column vectors for the table are:

Men: $\langle 96.9, 97.4, 97.5, 97.8, 97.8, 98, 98.6, 98.8 \rangle$

Women: $\langle 97.8, 98, 98.2, 98.2, 98.2, 98.6, 98.8, 99.2, 99.4 \rangle$

Therefore, we can say with 95% confidence interval that the true difference in mean body temperatures between men and women is between -1.144 and -0.056 degrees Fahrenheit.

a) Hypothesis test setup:

Let μ1 be the population mean body temperature of men and μ2 be the population mean body temperature of women.

Null hypothesis: H0: μ1 = μ2 (The mean body temperature is the same for men and women)

Alternative hypothesis: Ha: μ1 ≠ μ2 (The mean body temperature differs for men and women)

We will use a two-sample t-test for independent samples to test this hypothesis.

b) The exact p-value for the two-sample t-test with the given data is 0.0117. This p-value indicates that if the null hypothesis were true, the probability of obtaining a sample as extreme as the observed sample (or more extreme) is 0.0117. Since this p-value is less than the commonly used significance level of 0.05, we reject the null hypothesis and conclude that there is sufficient evidence to indicate that mean body temperatures differ for men and women.

c) To construct a 95% confidence interval for the difference in population means (μ1 - μ2), we can use the following formula:

CI = (x1 - x2) ± tα/2,ν * SE

where x1 and x2 are the sample means for men and women, tα/2,ν is the t-value for the desired confidence level and degrees of freedom (df), and SE is the standard error of the difference in sample means.

Using the given data, we have:

x1 = 97.5, x2 = 98.4

s1 = 0.35, s2 = 0.366

n1 = 8, n2 = 9

df = n1 + n2 - 2 = 15

t0.025,15 = 2.131 (from t-distribution table)

SE = sqrt(s1^2/n1 + s2^2/n2) = 0.126

CI = (97.5 - 98.4) ± 2.131 * 0.126

CI = (-1.144, -0.056)

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