1. In a pediatric clinic a study is carried out to see how effective aspirin is
ID: 3357246 • Letter: 1
Question
1. In a pediatric clinic a study is carried out to see how effective aspirin is in reducing temperature. Twelve 5-year-old children suffering from influenza had their temperatures taken immediately before and 1 hour after administration of aspirin. The results are given below. Suppose we assume normality and want to test the hypothesis that aspirin is reducing the temperature. Body temperature (°F) before and after taking aspirin Subject Before After Difference Subect Before After Difference 1 102.4 99.6 2.8 2 103.2 100.1 3.1 3 101.9 100.2 1.7 4 103.0 101.1 1.9 5 101.2 99.8 1.4 6 100.7 100.2 0.5 7 102.5 101.0 1.5 8 103.1 100.1 3 9 102.8 100.7 2.1 10 102.3 101.1 1.2 11 101.9 101.3 0.6 12 101.4 100.2 1.2 Total 21 a. What are the null and alternative hypotheses in this situation? b. Test whether there is a difference between mean body temperature before and after taking aspirin. Provide the test statistic, degrees of freedom for the test statistic, the p-value and a full conclusion. c. Calculate a 95% confidence interval for the true mean difference in body temperature before and after taking aspirin.
Explanation / Answer
Here we want to test the effectiveness of aspirin in reducing temperature. Suppose, mu1 and mu2 be the mean body temperatures respectively for before and after taking the aspirin.
a) Then the null hypothesis is H0: mu1=mu2 vs the alternative hypothesis H1: mu1>mu2.
b) Suppose, mean(X1) and mean(X2) be the sample mean of body temperatures of before and after taking aspirin and 'd' = mu1 - mu2. Then, to test the hypothesis, the test statistic is
t = (mean(X1) - mean(X2) - d) / (stdev(d)/sqrt(n))
Under null hypothesis, the test statistic follows a t distribution with (n-1) = 11 df and the observed t = sqrt(12)*(102.2 - 100.45) / 0.8692 = 6.975. The tabulated value of t0.01;11 = 2.718. The p-value for the test is 0.00001. Since the observed value is more than the tabulated value (or the p-value < 0.01), the null hypothesis is rejected at 1% level significance and conclude that there is significant effects of aspirin in reducing temperatures.
c) The 95% CI for the true mean difference in body temperature before and after taking aspirin is given below:
[ {mean(X1) - mean(X2)} - t0.025;11 * stdev(d)/sqrt(n), {mean(X1) - mean(X2)} + t0.025;11 * stdev(d)/sqrt(n) ]
= [ 1.75 - 2.2*0.8692/3.464, 1.75 + 2.2*0.8692/3.464 ] = [ 1.198, 2.302 ].
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