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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 ​(beforeminus​after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.4 and a standard deviation of 18.2. Construct a 90​% 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?

Respuesta :

Answer:

90% Confidence interval:  (-1.21,8.01)

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 42

Sample mean = 3.4

Sample standard Deviation = 18.2

90% Confidence interval:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.645[/tex]

[tex]3.4 \pm 1.645(\frac{18.2}{\sqrt{42}} ) = 3.4 \pm 4.61 = (-1.21,8.01)[/tex]

Since, the confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.

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