In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with garic in a processed tablet form. Cholesorol levels were measured before and after the treament. the changes in their levels of LDL cholestorol (in mg/dL have a mean of 3.5 and a standard deviation of 17.7

What is your question?

It says "Construct a 99% confidence interval estimate of the mean net change in

LDL cholestorol after the garlic treatment. What does the confidense interval
suggest about the effectiveness of garlic in reducing LDL cholestorol?

What is the confidence interval estimate of the population mean μ?

___mg/dL < μ < ___mg/dL

I just don't get how to solve this

99% = mean ± 2.575 SEm

SEm = SD/√n

To determine the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with garlic in a processed tablet form. The levels of LDL cholesterol were measured before and after the treatment, and we are provided with the mean and standard deviation of the changes in their LDL cholesterol levels.

Mean: The mean of 3.5 indicates that, on average, the subjects' LDL cholesterol levels decreased by 3.5 mg/dL after the treatment with garlic tablets.

Standard Deviation: The standard deviation of 17.7 signifies the variability or spread of the data points around the mean. A higher standard deviation implies a greater dispersion of the data points, indicating a wider range of changes in LDL cholesterol levels among the subjects.

It's important to note that the given statistics provide information about the average change in LDL cholesterol levels as well as the variability, but they don't reveal the statistical significance or clinical significance of the findings. For a comprehensive evaluation, additional statistical tests or analysis, along with other relevant information, would be needed.