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in a test of the effectiveness of garlic for lowering cholesterol, 42 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 5.7 and a standard deviation of 15.6. construct a 95% confidence interval estimate of the mean net change in LSL cholesterol after the garlic treatment. what does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? 

To calculate the 95% confidence interval estimate for the mean net change in LDL cholesterol levels after the garlic treatment, we can use the formula for the confidence interval of the mean:

\[
CI = \bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right)
\]

Where:
- \( \bar{x} \) = sample mean (5.7 mg/dL)
- \( s \) = sample standard deviation (15.6 mg/dL)
- \( n \) = sample size (42 subjects)
- \( t_{\alpha/2} \) = critical value from the t-distribution for a 95% confidence level with \( n-1 \) degrees of freedom

We'll now calculate the confidence interval based on these values.

Let's compute it.The 95% confidence interval for the mean net change in LDL cholesterol levels after garlic treatment is approximately \( (0.84, 10.56) \) mg/dL.

### Interpretation:
Since the confidence interval does not include zero, it suggests that there is a statistically significant positive change (increase) in LDL cholesterol levels, not a decrease. Therefore, based on this sample, garlic does **not** seem effective in lowering LDL cholesterol.

Let me know if you need more details or have any questions!

Here are 5 related questions:
1. How does the critical value from the t-distribution affect the confidence interval?
2. What are the assumptions behind using a t-distribution for constructing confidence intervals?
3. How would increasing the sample size impact the confidence interval?
4. Could a different confidence level (e.g., 90% or 99%) change the interpretation?
5. How does the standard deviation influence the width of the confidence interval?

**Tip:** Confidence intervals provide a range of values within which the true population mean is likely to lie; they don't guarantee specific outcomes, but offer insights into variability and estimation accuracy.

In a test of the effectiveness of garlic for lowering cholesterol, 42 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 5.7 and a standard deviation of 15.6. Construct a 95% confidence interval estimate of the mean net change in LSL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?

Learn how to construct a 95% confidence interval for the net change in LDL cholesterol after garlic treatment. This page explains the use of the t-distribution and assumptions for estimating the true mean with a sample of 42 subjects. The example demonstrates garlic's effect on LDL cholesterol levels, suggesting significant insights.

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