Question

In Exercises 9–16, assume that each sample is a simplerandom sample obtained from a population with a normal distribution.

Garlic for Reducing Cholesterol In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol(in mg/dL) had a mean of 0.4 and a standard deviation of 21.0 (based on data from “Effect of Raw Garlic vs Commercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia,” by Gardner et al.,Archives of Internal Medicine,Vol. 167).Construct a 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment. Does the result indicate whether the treatment is effective?

Short answer

The 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment is $16.9<\sigma <27.4$.

And, the confidence interval does not indicate that the treatment is effective.

Step-by-step solution

Given information

The sample number of subjects who were treated with raw garlic is $n=49$.

The mean level of LDL cholesterol (in mg/Dl) is 0.4.

The sample standard deviation is $s=21$min.

The level of confidence is 98%.

Compute the critical values and confidence interval

The degrees of freedom is computed as,

$$\begin{gathered} df=n-1 \\ =49-1 \\ =48 \end{gathered}$$

The level of confidence is 98%, which implies that the level of significance is 0.02.

Using the Chi-square table, the critical values at 0.02 level of significance and at 48 degrees of freedom are ${\chi }_L^2=28.177$and ${\chi }_R^2=73.6826$.

The 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment is computed as,

$$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{\chi }_R^2}}<\sigma <\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_L^2}} \\ \sqrt{\frac{\left(49-1\right){21}^2}{73.6826}}<\sigma <\sqrt{\frac{\left(49-1\right){21}^2}{28.177}} \\ 16.9<\sigma <27.4 \end{gathered}$$

Therefore, the 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment is $16.9<\sigma <27.4$.

And, the confidence interval does not give any information about the effectiveness of the treatment.