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 one year ago
PLEASE OFFER HELP!! WILL GIVE MEDAL! In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 3.2 and a standard deviation of 18.6.
(a) What is the best point estimate of the population mean net change in LDL cholesterol after the garlic treatment?
The best point estimate is __ mg/dL.
(b) Construct a 99% 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?
What is the confidence interval estimate of the population mean µ?
__mg/dL < µ < __ mg/dL (round to two decimal places as needed)
 one year ago
PLEASE OFFER HELP!! WILL GIVE MEDAL! In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 3.2 and a standard deviation of 18.6. (a) What is the best point estimate of the population mean net change in LDL cholesterol after the garlic treatment? The best point estimate is __ mg/dL. (b) Construct a 99% 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? What is the confidence interval estimate of the population mean µ? __mg/dL < µ < __ mg/dL (round to two decimal places as needed)

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StClowers
 one year ago
Best ResponseYou've already chosen the best response.0i know what (a) is but I need to know what (b) is or how to do it. Thanks!!

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0b) The 99% Confidence Interval is: $$ \bar x z_{(1c)/2}\times \cfrac{\sigma}{\sqrt n}\le \bar x \le \bar x+z_{(1c)/2}\times \cfrac{\sigma}{\sqrt n}\\ 3.22.576\cfrac{18.6}{\sqrt {48}}\le 3.2\le 3.2+2.576\cfrac{18.6}{\sqrt {48}}\\ 3.72\le 3.2\le 10.12 $$ Since 0 is included in this interval, these results suggest that a difference of 3.2 could have resulted from mere random variation. \(\large z_{(10.99)/2}\) was determined using this calculator  http://www.wolframalpha.com/input/?i=pvalue+%3D+.01 Note that we divided (1.99) by 2 because we are interested in the twotailed confidence interval. We can validate our results using the confidence interval for mean calculator  http://www.wolframalpha.com/input/?i=confidence+interval+for+mean&f1=0.99&f=ZInterval.c_0.99&f2=48&f=ZInterval.n_48&f3=18.6&f=ZInterval.sigma_18.6&f4=3.2&f=ZInterval.xbar_3.2
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