Mathematics, 06.03.2020 21:38 Sjanaya853

In a test of the effectiveness of garlic for lowering cholesterol, 4949 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 3.13.1 and a standard deviation of 17.417.4. Construct a 9999% 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?

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Answer from: kelonmazon2492

$3.1-2.68\frac{17.4}{\sqrt{49}}=-3.562$

$3.1+2.68\frac{17.4}{\sqrt{49}}=9.762$

So on this case the 99% confidence interval would be given by (-3.562;9.762)

And for this case we can conclude that we don't have a significant result since the confidence interval contains the value of 0 so we don't have enough evidence to conclude that we have a reducing effect.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=3.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=17.4 represent the sample standard deviation

n=49 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

df=n-1=49-1=48

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,48)".And we see that $t_{\alpha/2}=2.68$

Now we have everything in order to replace into formula (1):

$3.1-2.68\frac{17.4}{\sqrt{49}}=-3.562$

$3.1+2.68\frac{17.4}{\sqrt{49}}=9.762$

So on this case the 99% confidence interval would be given by (-3.562;9.762)

Answer from: Quest

answer: a.

step-by-step explanation: the answer is a. because, you need to converge the fractions to find a solution.

Answer from: Quest

2.56 is the width of the garden

step-by-step explanation:

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