Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects w

ID: 3125968 • Letter: I

Question

In a test of the effectiveness of garlic for lowering cholesterol, 47 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 3.9 and a standard deviation of 18.1. Complete parts (a) and (b) below.

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. (Type an integer or a decimal.) b. Construct a 90% 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 ? ___ mg/dL

___mg/dL < <___mg/dl

What does the confidence interval suggest about the effectiveness of the treatment?

A.The confidence interval limits do not contain 0, suggesting that the garlic treatment

did affect the LDL cholesterol levels.

B. The confidence interval limits do not contain 0, suggesting that the garlic treatment

did not affect the LDL cholesterol levels.

C.The confidence interval limitscontain 0, suggesting that the garlic treatment

did not affect the LDL cholesterol levels.

D. The confidence interval limits contain 0, suggesting that the garlic treatment

did affect the LDL cholesterol levels.

Explanation / Answer

a.
The best point estimate is ___ mg/dL = Mean(x)=3.9

b.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=3.9
Standard deviation( sd )=18.1
Sample Size(n)=47
Confidence Interval = [ 3.9 ± Z a/2 ( 18.1/ Sqrt ( 47) ) ]
= [ 3.9 - 1.64 * (2.64) , 3.9 + 1.64 * (2.64) ]
= [ -0.43,8.23 ]
the confidence interval estimate of the population mean, -0.43 mg/dL < < 8.23 mg/dl

c.
D. The confidence interval limits contain 0, suggesting that the garlic treatment
did affect the LDL cholesterol levels.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote