In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (beforeminusafter) in their levels of LDL cholesterol (in mg/dL) have a mean of 2.7 and a standard deviation of 17.8. Construct a 90% 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?

The 90% confidence interval for the mean net change in LDL cholesterol after the garlic treatment is (-1.7, 7.1).

As there are negative values included in the interval, at 90% confidence, we can not conclude there is no enough evidence about the effectiveness of garlic in reducing LDL cholesterol, as the true mean net change can be 0 or negative.

Step-by-step explanation:

We have to calculate a 90% confidence interval for the mean net change in LDL cholesterol after the garlic treatment.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=2.7.

The sample size is N=47.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

The degrees of freedom for this sample size are:

The t-value for a 90% confidence interval and 46 degrees of freedom is t=1.679.

The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 90% confidence interval for the mean net change in LDL cholesterol after the garlic treatment is (-1.7, 7.1).

As there are negative values included in the interval, at 90% confidence, we can not conclude there is no enough evidence about the effectiveness of garlic in reducing LDL cholesterol, as the true mean net change can be 0 or negative.

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