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?

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".

[tex]\bar X=3.1[/tex] represent the sample mean

[tex]\mu[/tex] 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:

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

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

[tex]df=n-1=49-1=48[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], 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 [tex]t_{\alpha/2}=2.68[/tex]

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

[tex]3.1-2.68\frac{17.4}{\sqrt{49}}=-3.562[/tex]

[tex]3.1+2.68\frac{17.4}{\sqrt{49}}=9.762[/tex]

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.