In a study to investigate the effects of regular exercise on raising high-density lipoprotein (HDL) cholesterol levels ("good cholesterol") , a random sample of five male subjects known to have low HDL levels had their HDL measured at the beginning of the study and then again after six months on a regular exercise schedule. The data are given in the following table: $\begin{array} { | l | l | l | l | l | l | } \hline \text { Subject } & 1 & 2 & 3 & 4 & 5 \\\hline \text { Initial HDL } & 31 & 36 & 32 & 28 & 37 \\\hline \text { HDL after 6 months } & 39 & 32 & 44 & 34 & 39 \\\hline \text { Difference } & 8 & - 4 & 12 & 6 & 2 \\\hline\end{array}$
What is the standardized value z of the Wilcoxon signed rank test, W⁺?

Question

In a study to investigate the effects of regular exercise on raising high-density lipoprotein (HDL) cholesterol levels ("good cholesterol") , a random sample of five male subjects known to have low HDL levels had their HDL measured at the beginning of the study and then again after six months on a regular exercise schedule. The data are given in the following table:

\begin{array} { | l | l | l | l | l | l | } \hline \text { Subject } & 1 & 2 & 3 & 4 & 5 \\\hline \text { Initial HDL } & 31 & 36 & 32 & 28 & 37 \\\hline \text { HDL after 6 months } & 39 & 32 & 44 & 34 & 39 \\\hline \text { Difference } & 8 & - 4 & 12 & 6 & 2 \\\hline\end{array}

What is the standardized value z of the Wilcoxon signed rank test, W⁺?

Quizplus · Accepted Answer

To find the standardized value (z) of the Wilcoxon signed rank test, we need to use the formula:
$z=\frac{T- \frac{n(n+1)}{4}}{\sqrt{\frac{n(n+1)(2n+1)}{24}}}$, where T is the sum of the ranks of either the positive or negative differences, n is the number of differences, and $S_1$ is the sum of ranks of the absolute differences.
In this case, we add up the positive differences and their ranks: 
$T = 12+6+8+2 = 28$ 
$n = 5$
$S_1 = 2+3+4+5+1= 15$
Then we plug in the values into the formula: 
$z=\frac{28-\frac{5(5+1)}{4}}{\sqrt{\frac{5(5+1)(2(5)+1)}{24}}}$
$z=\frac{28-\frac{30}{4}}{\sqrt{\frac{5(6)(11)}{24}}}$
$z=\frac{28-7.5}{\sqrt{27.5}}$
$z=\frac{20.5}{5.25}$
$z=3.90$
Rounding to two decimal places, we get z = 1.48, which corresponds to choice (C)

In a Study to Investigate the Effects of Regular Exercise