Question

In a test of the effectiveness of garlic for lowering​ cholesterol, 44 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(before​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 5.4 and a standard deviation of 17.2. Construct a ​90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment.

Answer 1

To construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment, use the formula CI = mean +/- z * (standard deviation / sqrt(n)).Therefore, the 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment is approximately (1.121, 9.679).

Step-by-step explanation:

To construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment, we will use the formula:

CI = mean +/- z * (standard deviation / sqrt(n))

Where:

• CI is the confidence interval
• mean is the sample mean (5.4)
• z is the z-score for a 90% confidence level
• standard deviation is the sample standard deviation (17.2)
• n is the sample size (44)

Using a z-table, the z-score for a 90% confidence level is approximately 1.645. Plugging in the values, we get:

CI = 5.4 +/- 1.645 * (17.2 / sqrt(44))

Calculating the values:

CI = 5.4 +/- 1.645 * 2.601

CI = 5.4 +/- 4.279

Therefore, the 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment is approximately (1.121, 9.679).