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Information: 44 subjects were treated with garlic. A mean of 3.8 and standard deviation of 18.7. 99% confidence interval.

Information: 44 subjects were treated with garlic. A mean of 3.8 and standard deviation-example-1
User Adarshr
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1 Answer

4 votes

To calculate the confidence interval, we have the formula:


CI=\bar{x}\pm z(\frac{s}{\sqrt[]{n}})

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.


CI=3.8\pm2.58(\frac{18.7}{\sqrt[]{44}})

Then, solve. Let's start by getting the quotient of 18.7 and square root of 44.


CI=3.8\pm2.58(2.8191)

Next, multiply 2.58 and 2.8191.


CI=3.8\pm7.2733

Lastly, separate the plus and minus sign and do the operation.


\begin{gathered} CI=3.8+7.2733=11.0733\approx11.07 \\ CI=3.8-7.2733=-3.4733\approx-3.47 \end{gathered}

Hence, at 99% confidence interval, the population mean falls between -3.47 mg/dL and 11.07 mg/dL.


-3.47<\mu<11.07

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)

User TJ Tang
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