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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.

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.

The formula for confidence interval is $\stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E$.

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

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

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

The degree of freedom is,

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

The critical value ${\mathrm{t}}_{\frac{\mathrm{伪}}{2}}$ 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,

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

For 98% confidence interval for mean is,

$$\begin{gathered} \stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E=0.4-7.23<渭<0.4+7.23 \\ =-6.83<渭<7.63 \end{gathered}$$

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

.

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

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