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 4.1 and a standard deviation of 18.7. construct a 95% 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?
click here to view a t distribution table.
click here to view page 1 of the standard normal distribution table.
click here to view page 2 of the standard normal distribution table.
what is the confidence interval estimate of the population mean μ?
μ<μ<μ mg/dl
(round to two decimal places as needed.)

Explanation:

Step1: Determine the critical value

For a 95% confidence interval, the significance level \(\alpha = 1 - 0.95=0.05\), and \(\alpha/2 = 0.025\). Using the standard normal distribution table (or \(z\)-table), the critical value \(z_{\alpha/2}=1.96\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a population mean when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 18.7\), \(n = 48\)

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Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu <\bar{x} + E\). Assume the sample mean \(\bar{x}\) is 0 (since the problem does not give the sample - mean value, but if we assume the net - change mean is 0 for the sake of showing the formula application. In a real - world scenario, if the sample mean was given, say \(\bar{x}=x\), we would use that value).

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Answer:

\(-5.30<\mu<5.30\)