Question

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?

Answer 1

`3.1-2.68(17.4)/(√(49))=-3.562`

`3.1+2.68(17.4)/(√(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.

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)(s)/(√(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(17.4)/(√(49))=-3.562`

`3.1+2.68(17.4)/(√(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.