18.32 (EX) More on pharmacokinetics 1/3: The following exercise is based on summ
ID: 3310028 • Letter: 1
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18.32 (EX) More on pharmacokinetics 1/3: The following exercise is based on summary statistics rather than raw data. This information is typically all that is presented in published reports. Inference procedures can be calculated by hand from the summaries. You must trust that the authors understood the conditions for inference and verified that they apply. This isn't always true.
Avandia (rosiglitazone maleate) is an oral antidiabetic drug produced by the pharmaceutical company GlaxoSmithKline. Before a drug can be prescribed, we must know how the body absorbs and excretes it. Patients were given a single dose of either 1 mg or 2 mg of rosiglitazone, and the maximum plasma concentration of the drug (in ng/ml) was assessed. The study also looked at how fast the body gets rid of the drug by measuring the elimination half-life (in hours). Here are the results.
a) Is there significant evidence of a difference in elimination half-life depending on drug dosage? State hypotheses for the test (1 and 2 are the means for 1 mg and 2 mg respectively).
b) 18.32 (EX) More on pharmacokinetics 2/3: Give the test statistic, the conservative number of degrees of freedom and the P-value for the test.
c) 18.32 (EX) More on pharmacokinetics 3/3: True or False:
There is some evidence that elimination half-life for Avandia depends on drug dosage.
Treatment n mean standard deviation 1 mg 32 3.16 0.72 2 mg 32 3.15 0.39Explanation / Answer
a)
H0: 1 – 2 = 0
H1: 1 – 2 # 0
Assuming population variances are equal, we would have to calculate pooled-variance t-Test
Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)
= (32-1)*0.72^2+(32-1)*0.39^2/31+31
=0.3325
tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)
=(3.16-3.15)-0/0.3325(1/32+1/32)
=0.01/0.14415
=0.069
tCRIT is +/-2 and hence, tSTAT<tSTAT and hence, we cannot reject the null hypothesis. Thus, we conclude that there is no significant evidence of a difference in elimination half-life depending on drug dosage
b)
62 degrees of freedom
The P-Value is .945212. The result is not significant at p < .05.
c)
False
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