Clinical Trials of Lipitor The sample data below are changes in LDL cholesterol levels in clinical trials of Lipitor (atorvastatin). It was claimed that Lipitor had an effect on LDL cholesterol. (The data are based on results given in a Parke-Davis memo from David G. Orloff, M.D., the medical team leader for clinical trials of Lipitor. Pfizer declined to provide the author with the original data values.) Negative values represent decreases in LDL cholesterol. Use a 0.05 significance level to test the claim that for those treated with 20 mg of Lipitor and those treated with 80 mg of Lipitor, changes in LDL cholesterol have the same median. What do the results suggest?

GroupTreatedwith20 mgofLipitor
-28	-32	-29	-39	-31	-35	-25	-36	-35	-26	-29	-34	-30
GroupTreatedwith80 mgofLipitor
-42	-41	-38	-42	-41	-41	-40	-44	-32	-37	-41	-37	-34	-31

Two samples are given showing the decrease in LDL cholesterol levels when 20 mg of Lipitor and 80 mg of Lipitor are administered.

The researcher wants to test theclaim that for those treated with 20 mg of Lipitor and those treated with 80 mg of Lipitor, the changes in LDL cholesterol have the same median.

The Wilcoxon rank-sum test is performed to test the difference in the medians of the given two samples.

The null hypothesis is as follows:

There is no difference in the median values of the decrease in LDL cholesterol level between those treated with 20 mg of Lipitor and those treated with 80 mg of Lipitor.

The alternative hypothesis is as follows:

There is a difference in the median values of the decrease in LDL cholesterol level between those treated with 20 mg of Lipitor and those treated with 80 mg of Lipitor.

It is a two-tailed test.

• Combine the two samples and label each observation with the sample name/number it comes from.
• The smallest observation is assigned rank 1; the next smallest observation is assigned rank 2, and so on until the largest value.
• If two observations have the same value, the mean of the ranks is assigned to them. For example, if threeobservations are the same and should have been ranked as 2,3, and 4, then the mean of 2,3, and 4 equal to 3 is assigned to the three observations.

The following table shows the ranks:

Compute the sum of the ranks for sample 1 for the group treated with 20 mg Lipitor.

\(\begin{array}{l}25 + 18.5 + 23.5 + .... + 22 = 253.5\\\end{array}\)

Thus, the sum of the ranks for the group treated with 20 mg Lipitoris equal to 253.5 and is denoted by R.

Let\({n_1}\)be the sample size of the group treated with 20 mg Lipitor.

Let\({n_2}\)be the sample size of the group treated with 20 mg Lipitor.

Here,

\(\begin{array}{l}{n_1} = 13\\{n_2} = 14\end{array}\)

Compute the mean value\(\left( {{\mu _R}} \right)\)as follows:

\(\begin{array}{c}{\mu _R} = \frac{{{n_1}\left( {{n_1} + {n_2} + 1} \right)}}{2}\\ = \frac{{13\left( {13 + 14 + 1} \right)}}{2}\\ = 182\end{array}\)

Compute the standard deviation\(\left( {{\sigma _R}} \right)\)as follows:

\(\begin{array}{c}{\sigma _R} = \sqrt {\frac{{{n_1}{n_2}\left( {{n_1} + {n_2} + 1} \right)}}{{12}}} \\ = \sqrt {\frac{{13 \times 14\left( {13 + 14 + 1} \right)}}{{12}}} \\ = 20.61\end{array}\)

The test statistic is:

\(\begin{array}{c}z = \frac{{R - {\mu _R}}}{{{\sigma _R}}}\; \sim N\left( {0,1} \right)\\ = \frac{{253.5 - 182}}{{20.61}}\\ = 3.47\end{array}\)

The absolute value of the z score is equal to3.47.

The critical value of the z from the standard normal table for a two-tailed test with\(\alpha = 0.05\)is equal to 1.96.

As the test statistic value is greater than the critical value, the null hypothesis isrejected.

There is enough evidence to conclude that there is a difference in the median values of the decrease in LDL cholesterol level between those treated with 20 mg of Lipitor and those treated with 80 mg of Lipitor.

Therefore, it can be observed that the dosage amount of Lipitor does have a significant effect on LDL cholesterol.