A 1992 article in the Journal of the American Medical Association (\"A Critical

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: 97.4 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 H_0: mu = 98.6 versus H_1: mu notequalto 98.6, using alpha = 0.05. (a) Calculate the sample mean and standard deviation. bar X = 5 = (b) Determine the test statistic t_0- (c) Draw conclusion. H_0.

Explanation / Answer

Solution:-

Numbers -

97.4,97.2,97.4,97.6,97.8,97.9,98.0,98.0,98.1,98.1,98.2,98.3,98.3,98.4,98.4,98.4,98.5,98.6,98.6,98.7,98.8,98.8,98.9,98.9,99.0

(a) Count = 25 (How many numbers)

Sum = 2456.3 (All the numbers added up)

Mean = 98.252 (Arithmetic mean = Sum / Count)

Now calculate the Variance:-

Sum of Differences2 = 6.0624 (Add up the Squared Differences)

Variance = 0.242496 (Sum of Differences2 / Count)

Lastly, take the square root of the Variance, which will give our standard deviation:-

Standard Deviation = 0.4924388287 (The square root of the Variance)

or S = 0.492

(b) Test statistic t0:-

t = [ x - ] / [ s / sqrt( n ) ]
t = (98.252 - 98.6) / [ 0.492 / sqrt(25) ]
t = ( -0.348 ) / ( 0.492 / 5) = ( -0.348 ) / (0.0984) = -3.5365854

or to = -3.537

(c) Conclusion

The P-Value is 0.001685.

The result is significant at p < 0.05.

Reject Null Hypothesis (Ho).

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