In a test of the effectiveness of garlic for lowering​ cholesterol, 4949 subjects were treated with garlic in a processed tablet form. Chole

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?

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Jasmine 21 hours 2021-10-13T04:15:20+00:00 1 Answer 0

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    2021-10-13T04:17:15+00:00

    Answer:

    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)    

    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.

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