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 in their levels of LDL cholesterol (in mg/dL) have a mean of 3.2 and a standard deviation of 18.6.
(a) What is the best point estimate of the population mean net change in LDL cholesterol after the garlic treatment?

The best point estimate is __ mg/dL.

(b) Construct a 99% 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?

What is the confidence interval estimate of the population mean µ?
__mg/dL < µ < __ mg/dL (round to two decimal places as needed)

Answer 1

b) The 99% Confidence Interval is:

x¯−z(1−c)/2×σ/√n≤x¯≤x¯+z(1−c)/2×σ/√n

3.2−2.57618.6/√48≤3.2≤3.2+2.57618.6/√48

−3.72≤3.2≤10.12
Since 0 is included in this interval, these results suggest that a difference of 3.2 could have resulted from mere random variation. z(1−0.99)/2 was determined using this calculator.

Note that we divided (1-.99) by 2 because we are interested in the two-tailed confidence interval. We can validate our results using the confidence interval for mean calculator

Answer 2

Answer:

Explanation:

Hello!

a)

To check if garlic lowers LDL cholesterol, 48 subjects were treated with garlic. Their cholesterol levels were measured before and after the treatment recording the changes in the LDL cholesterol levels obtaining a mean of 3.2 mg/dL and a standard deviation of 18.6 mg/dL.

So the study variable is X " Change in LDL cholesterol level of a subject after being treated with garlic.

a) The best point estimate of the population mean is the sample mean X[bar].

For this example, the point estimate of the population average LDL cholesterol level change is X[bar]= 3.2mg/dL.

b) To construct a 99% CI for the population mean you have to make sure that the study variable has at least a normal distribution. There is no information about the distribution of the population, but considering that the sample is big enough, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal X[bar] ≈ N(μ; σ²/n).

With this you can use an aproximation of the standard normal to construct the interval. The formula to use for the interval is:

[X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$]

$Z_{1-\alpha /2} = Z_{0.995} = 2.576$

[3.2 ± 2.576 * $\frac{18.6}{\sqrt{48} }$]

[3.2 ± 2.576 * 2.684]

[-3.714; 10.114]

With a confidence level of 99% youd expect the interval [-3.714; 10.114]mg/dL to contain the true value of the average LDL cholesterol level changed of subjects that were treated with garlic.

Let's say that if garlic lowers the cholesterol level, the average cholesterol level change will be less than cero, symbolically μ < 0. Now since the interval contains the cero, at a significance level of 1% (complementary to the 99% confidence level for the interval), you could conclude that garlic does not lower the LDL cholesterol levels.

I hope this helps!