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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%.

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 ${蠂}_L^2=28.177$and ${蠂}_R^2=73.6826$.

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

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

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

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