Chapter 15, Problem 28SE

The article “Effects of a Rice-Rich Versus Potato-Rich Diet on Glucose, Lipoprotein, and Cholesterol Metabolism in Noninsulin-Dependent Diabetics” (Anter. J. of Clinical Nutr., 1984: 598-606) gives the accompanying data on cholcsterol-synthesis rate for eight diabetic subjects. Subjects were fed a standardized diet with potato or rice as the major carbohydrate source. Participants received both diets for specified periods of time, with cholesterol-synthesis rate (mmol/day) measured at the end of each dietary period. The analysis presented in this article used a distribution-free test. Use such a test with significance level .05 to determine whether the true mean cholesterol-synthesis rate differs significantly for the two sources of carbohydrates.

	Cholesterol-Synthesis Rate
Subject	1	2	3	4	5	6	7	8
Potato	1.88	2.60	1.38	4.41	1.87	2.89	3.96	2.31
Rice	1.70	3.84	1.13	4.97	.86	1.93	3.36	2.15

To determine

Test whether the true mean cholesterol-synthesis rate differs for the two sources of carbohydrates.

Answer to Problem 28SE

There is no sufficient evidence to support that there is a difference between the cholesterol synthesis rates for the two carbohydrate source at 5% level of significance.

Explanation of Solution

Given info:

The cholesterol-synthesis rate are measured at the end of the two diets (potato and rice) are measured for participants. Each participant had received both diets for different period. The level of significance is 0.05.

Calculation:

Conditions for using Wilcoxon signed ranked test for paired observation:

• The difference of the paired data should be continuous
• The distribution of the difference should be symmetric.

Here, the difference of the data is continuous. For checking the symmetry boxplot can be drawn.

Software procedure:

Step by step procedure to draw the boxplot using the MINITAB software:

• Choose Graph > Boxplot.
• Choose Simple, and then click OK.
• In Graph variables, enter the corresponding column of difference.
• Click OK.

Output using the MINITAB software is given below:

From the boxplot, it can be observes that the distribution of difference of two diets is approximately symmetric.

Hence, Wilcoxon signed ranked test can be applied.

Parameter of interest:

Let ${\mu }_D$ be the difference between cholesterol-synthesis rates for the two carbohydrate source.

Hypotheses:

Null hypothesis:

$H_0:{\mu }_D=0$

That is, there is no difference between cholesterol synthesis rates for the two carbohydrate source.

Alternative hypothesis:

$H_{\text{a}}:{\mu }_D\ne 0$

That is, there is a difference between the cholesterol synthesis rates for the two carbohydrate source.

Test statistic:

Software procedure:

Step by step procedure to obtain the test-statistic value using the MINITAB software:

• Choose Stat >Nonparametrics> 1-Sample Wilcoxon.
• In Variables, enter the column of difference
• Choose the Test median, enter the value as 0 and select not equal in Alternative.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the test statistic value is 24

P-value:

For two tailed test, the P-value is $2P_0\left(S_{+}\ge \max \left\{s_{+},\frac{n\left(n+1\right)}{2}-s_{+}\right\}\right)$ where, $s_{+}$= 24, n = 8, the sum of the ranks associated with positive $\left(x_i-{\mu }_0\right)$ ‘s

$\begin{array}{c}2P_0\left(S_{+}\ge \max \left\{s_{+},\frac{n\left(n+1\right)}{2}-s_{+}\right\}\right)=2P_0\left(S_{+}\ge \max \left\{24,\frac{8\left(8+1\right)}{2}-24\right\}\right)\\ =2P_0\left(S_{+}\ge \max \left\{24,36-24\right\}\right)\\ =2P_0\left(S_{+}\ge \max \left\{24,12\right\}\right)\\ =2P_0\left(S_{+}\ge 24\right)\end{array}$

Procedure for finding P-value from table A.13 of Appendix:

• Locate the sample size 8.
• The nearest critical value $\left(c_1\right)$ of 24 is 28.
• The corresponding value is $P_0\left(S_{+}\ge 24\right)=0.098,$

Hence, the P-value is more than $2\times 0.098=0.196$.

Decision:

If $P-\text{value}\le \alpha$, reject the null hypothesis $H_0$

If $P-\text{value>}\alpha$ fail to reject the null hypothesis $H_0$

Conclusion:

For $\alpha =0.05$

Here, the P-value is greater than the level of significance.

That is,$\left\{P-\text{value}>0.196\right\}>\alpha \left(0.05\right)$.

By rejection rule, fail to reject the null hypothesis.

Interpretation:

Thus, it can be concluded that there is no sufficient evidence to support that there is a difference between the cholesterol synthesis rates for the two carbohydrate source at 5% level of significance.