Question

9.3.9 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: 98.6 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 Ho := 98.6 versus H :u # 98.6, using a = 0.05. Round your answers to three decimal places (e.g. 98.765). (a) Calculate the sample mean and standard deviation. (b) Determine the test statistic to (c) Draw conclusion. Ho Statistical Tables and Charts

Answer 1

Given, Sample size (n) = 25 Population mean(u) = 98.6 Set up the null and the alternative hypothesis. H - u=98.6 Hu=98.6 The significance level, a = 0.05. Critical value: The critical value at the given level of significance with 24 degrees of freedom is. 2.064. Decision rule: Reject H, if the t > 2.064

First find the sample mean. Mean: 24 98.6+97.2 +97.4+...+99 25 = 98.3 Then, the standard deviation is, s-, 2 (x==32 = [(98.6–98.3)* +(97.2–98.3)_+(99–983)*] = 0.4743

Calculate test statistic Under Ho, the test statistic for one sample t test is, S V7 98.3-98.6 0.4743 =-3.163 Conclusion: Here, the test statistic value (11) = 3.163) is greater than the critical value (2.064). So, reject null hypothesis at the 5% level of significance.