Question

In a test of the effectiveness of garlic for lowering cholesterol, 48 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 5.8 and a standard deviation of 16.4. Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment.

a) -4.17 to 15.77 mg/dL
b) -2.34 to 13.94 mg/dL
c) 1.22 to 18.22 mg/dL
d) 5.37 to 16.23 mg/dL

Answer 1

The 95% confidence interval for the mean net change in LDL cholesterol after garlic treatment is calculated using the mean, standard deviation, and sample size. The correct interval is approximately 1.163 to 10.437 mg/dL, which is not among the provided options.

Step-by-step explanation:

To answer the student's question about constructing a 95% confidence interval estimate of the mean net change in LDL cholesterol after garlic treatment, we need to use the mean, standard deviation, and the sample size provided. The sample size is 48 subjects, the mean change in LDL cholesterol is 5.8 mg/dL, and the standard deviation is 16.4 mg/dL. Since the sample size is less than 30, we should use the t-distribution for our calculations. However, for larger sample sizes as in this case, the t-distribution is approximately equal to the z-distribution.

To calculate the confidence interval, we first need to determine the critical value for the 95% confidence level. Using the z-table, we find that the critical value for a two-tailed test at the 95% confidence level is approximately 1.96.

Next, we calculate the standard error of the mean by dividing the standard deviation by the square root of the sample size (n):

• SE = 16.4 / √48 = 2.366

Now we can calculate the margin of error:

• Margin of error = Critical value * Standard error = 1.96 * 2.366 = 4.637

Adding and subtracting the margin of error from the mean, we obtain the confidence interval:

• Lower limit = Mean - Margin of error = 5.8 - 4.637 = 1.163
• Upper limit = Mean + Margin of error = 5.8 + 4.637 = 10.437

The 95% confidence interval estimate of the mean net change in LDL cholesterol after garlic treatment is approximately 1.163 to 10.437 mg/dL. Therefore, the correct option is not listed among the choices provided in the question.