Question

Question 4 4 pts In a test of the effectiveness of garlic for lowering cholesterol, 53 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) had a mean of 0.4 and a standard deviation of 21.0. Assuming the sample is a simple random sample obtained from a population with a normal distribution, construct a 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment. Keep your numerical answers rounded to 3 decimal places. A. B. C. 98% CI estimate of o = Lower Limit: Upper Limit: Do the results of the study indicate whether the treatment is effective in lowering LDL cholesterol? i. Yes, as the mean changes in LDL cholesterol are positive. ii. Yes, as the estimate for the standard deviation do not include zero. iii. Maybe, but we need more information to say definitively. iv. No, as the mean is too small to be meaningful. v. None of the above. HTML Editora BIVA-A- IEE 3 1 1 X X =

Answer 1

Friday

31/07/2020

Q4

Interpretation = Given that n= 53 SE 21 CI= 98 n-1= 52 Therefore, the 98 % confidence interval for population SD is between= Option 2nd Yes, as the estimate for the standard deviation do not include o At 98 % confidence level = Lower bonda = V(n-1)s-2/x^2U = 152*2142/78.616 = V22932 /78.616 = V291.6963 = 17.08 a = 1-9896 = 1-0.98 = 0.02 Therefore, a/2 = 0.01 & Za /2 = 2.33 Upper bonda = v(n-1)s-2/X^2L = V5221^2/31.246 = V22932/31.246 = 7733.9179 = 27.09 Now lets calculate, Critical value of chi square distribution = a) Lower critical valueſX^2L) = 31.25 b) Upper critical valueſX^2U) - 78.62 {by using chi square critical value table}

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