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 0.010.01 significance level for both parts.			Men	Women
		muμ	mu 1μ1	mu 2μ2
		n	1111	5959
		x overbarx	97.6997.69degrees°F	97.2597.25degrees°F
		s	0.920.92degrees°F	0.710.71degrees°F

FIND P VALUE

T VALUE

CONFIDENCE INTERVAL FOR TESTING MEN HAVE A HIGER BODY TEMP THAN WOMEN

Answer 1

a) The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                             = (97.69 - 97.25)/sqrt((0.92)^2/11 + (0.71)^2/59)

                             = 1.505

DF = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

      = ((0.92)^2/11 + (0.71)^2/59)^2/(((0.92)^2/11)^2/10 + ((0.71)^2/59)^2/58)

     = 12

P-value = P(T > 1.505)

             = 1 - P(T < 1.505)

             = 1 - 0.9209

              = 0.0791

Since the P-value is greater than the significance level (0.0791 > 0.01), so we cannot conclude that men have a higher body temperature than women.

b) Confidence level = 1 - 0.01 = 0.99

So at 99% confidence interval the critical value is t^* = 3.055

The 99% confidence interval for difference in population means () is

() +/- t^* * sqrt(s1^2/n1 + s2^2/n2)

= (97.69 - 97.25) +/- 3.055 * sqrt((0.92)^2/11 + (0.71)^2/59)

= 0.44 +/- 0.893

= -0.453, 1.333

Since the confidence interval contains both positive and negative values, so we cannot conclude that the men have a higher body temperature than women.