Question

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 mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?

Short answer

The 98% confidence intervalof the mean net change in their LDL cholesterol after the garlic treatmentis$-6.82\text{mg/dl}<\mu <7.62\text{mg/dl}$.

The confidence interval limits contain 0, which suggests that the garlic treatment did not affect the LDL cholesterol levels.

Step-by-step solution

Given information

There are 49 subjects who were treated in a test of the effectiveness of garlic for lowering their cholesterol. The mean of changes in their LDL cholesterol is 0.4 mg/dl and the standard deviation is 21.0 mg/dl.

Check the requirements

The sample size (n) of the test is 49 which is greater than 30. Assume that the samples are selected randomly.

As the population standard deviation is unknown, the t-distribution would be used.

Describe the formula for confidence interval.

The formula for confidence interval is $\overline{x}-E<\mu <\overline{x}+E$.

Here, E is margin of error given by, $E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$

Where, $t_{\frac{\alpha }{2}}$ is the critical value.

Here, $\overline{x}$represents the sample mean of test of effectiveness of garlic for lowering cholesterol and $\mu$ represents the population mean of test of effectiveness of garlic for lowering cholesterol.

Calculate margin of error

The degree of freedom is,

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

The critical value ${\mathrm{t}}_{\frac{\mathrm{\alpha }}{2}}$ is obtained from t-table using 0.02 level of significance and 48 degrees of freedom as 2.41.

Margin of error is given by,

$$\begin{gathered} E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}} \\ =2.41\times \frac{21}{\sqrt{49}} \\ =7.23 \end{gathered}$$

Construct the confidence interval

For 98% confidence interval for mean is,

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}+E=0.4-7.23<\mu <0.4+7.23 \\ =-6.83<\mu <7.63 \end{gathered}$$

Therefore, 98% confidence interval for mean net change in LDL is $-6.83mg/dl<\mu <7.62mg/dl$

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Interpret the result

The 98% confidence interval of the mean net change in their LDL cholesterol after the garlic treatment is$-6.82\text{mg/dl}<\mu <7.62\text{mg/dl}$.

The confidence interval limits contain 0, which suggests that the garlic treatment did not affect the LDL cholesterol levels.