Question

A 1992 article in the Journal of the American Medical Association ("A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body temperature, and Other Legacies of Carl Reinhold August Wundrlich") reported body temperature, gender, and heart rate for a number of subjects. The temperatures for 25 female subjects follow: 97.5 97.2 97.4 97.6 97.8 97.9 98.0 98.0 98.1 98.1 98.2 98.3 98.3 98.4 98.4 98.4 98.5 98.6 98.6 98.7 98.8 98.8 98.9 98.9 99.0 Test the hypothesis Upper HO: mu =98.6 versus H1: mu not = 98.6 using alpha=0.05. Round your answers to three decimal places (e.g. 98.765). (a) Calculate the sample mean and standard deviation. sample mean and standard deviation (b) Determine the test statistic to (c) Draw conclusion HO.

Answer 1

a. Sample mean is

=98.256 n

Create the following table.

data	data-mean	(data - mean)2
97.5	-0.756	0.571536
97.2	-1.056	1.115136
97.4	-0.85599999999999	0.73273599999999
97.6	-0.65600000000001	0.43033600000001
97.8	-0.456	0.207936
97.9	-0.35599999999999	0.126736
98	-0.256	0.065536
98	-0.256	0.065536
98.1	-0.15600000000001	0.024336000000002
98.1	-0.15600000000001	0.024336000000002
98.2	-0.055999999999997	0.0031359999999997
98.3	0.043999999999997	0.0019359999999997
98.3	0.043999999999997	0.0019359999999997
98.4	0.14400000000001	0.020736000000002
98.4	0.14400000000001	0.020736000000002
98.4	0.14400000000001	0.020736000000002
98.5	0.244	0.059536
98.6	0.34399999999999	0.118336
98.6	0.34399999999999	0.118336
98.7	0.444	0.197136
98.8	0.544	0.295936
98.8	0.544	0.295936
98.9	0.64400000000001	0.41473600000001
98.9	0.64400000000001	0.41473600000001
99	0.744	0.553536

Find the sum of numbers in the last column to get.

.

(xi – 7)2 = 5.9016

So sample standard deviation is

Στί – Τ)2 V - η -1 = 0.496

b. Test statistics is

T- = t= 98.256 - 98.6 0.496 - = -3.47

c. P value using excel formula is TDIST(3.47,24,2)=0.002<alpha=0.05

Hence we reject the null hypothesis

Reject H0