To calculate the confidence interval, we have the formula:
[tex]CI=\bar{x}\pm z(\frac{s}{\sqrt[]{n}})[/tex]where bar x = mean, z = z-value, s = standard deviation, and n = sample size.
For a 99% confidence interval, the z-value is 2.58. Let's plug in the given information in the question to the formula above.
[tex]CI=3.8\pm2.58(\frac{18.7}{\sqrt[]{44}})[/tex]Then, solve. Let's start by getting the quotient of 18.7 and square root of 44.
[tex]CI=3.8\pm2.58(2.8191)[/tex]Next, multiply 2.58 and 2.8191.
[tex]CI=3.8\pm7.2733[/tex]Lastly, separate the plus and minus sign and do the operation.
[tex]\begin{gathered} CI=3.8+7.2733=11.0733\approx11.07 \\ CI=3.8-7.2733=-3.4733\approx-3.47 \end{gathered}[/tex]Hence, at 99% confidence interval, the population mean falls between -3.47 mg/dL and 11.07 mg/dL.
[tex]-3.47<\mu<11.07[/tex]Since our confidence interval includes zero, this means that at some point if experiment is to be rerun again, there's a chance that the garlic treatment did not affect the LDL cholesterol levels. (Option C)