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? □mg/dl < μ < □mg/dl (round to two decimal places as needed.)

Explanation:

Step1: Find the critical value

Since the sample size \(n = 48\), the degrees of freedom \(df=n - 1=48-1 = 47\). For a \(95\%\) confidence interval (\(\alpha=1 - 0.95=0.05\), \(\alpha/2=0.025\)). Using a \(t\) - distribution table (or a calculator), the critical value \(t_{\alpha/2}\approx 2.012\) (for \(df = 47\), we can also use the value for \(df = 50\) as an approximation if the exact \(df\) is not in the table, \(t_{0.025,50}=2.009\)).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 18.7\), \(n = 48\).

$$
LATEXBLOCK0
$$

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is given by \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=4.1\)

$$
LATEXBLOCK1
$$

Answer:

\(-1.33\space mg/dL<\mu<9.53\space mg/dL\)

Since the confidence interval contains \(0\), it suggests that there is not sufficient evidence at the \(95\%\) confidence level to conclude that garlic is effective in reducing LDL cholesterol.