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In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects w

ID: 3429228 • Letter: I

Question

In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in thier levels of LDL cholesterol (in mg/dl) have a mean of 4.6 and a standard deviation of 18.9.

A) What is the best point estimate of the population mean net change in LDL Cholesterol after the garlic treatment? (Type Integer or decimal)

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 effetiveness of garlic in reducing LDL Cholesterol. What is the confidence interval estimate of the population mean

_____mg/dL < mean < ____ mg/dL   (round 2 decimal places)

_____ A) do not contain 0, suggesting that the garlic treatment did not affect the LDL Cholesterol levels

_____B) contain 0, suggesting that he garli treatment did affect the LDL Cholesterol levels

_____C) do not contain 0, suggesting that the garlic treatment did affect the LDL Choleterol level

_____D) contain 0, suggesting that the garlic treatment did not affect the LDL Cholesterol levels

Explanation / Answer

In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in thier levels of LDL cholesterol (in mg/dl) have a mean of 4.6 and a standard deviation of 18.9.

sample size >30, z distribution used.

Confidence Interval Estimate for the Mean

Data

Standard Deviation

18.9

Sample Mean

4.6

Sample Size

44

Confidence Level

99%

Intermediate Calculations

Standard Error of the Mean= 18.9/sqrt(44) =

2.8493

Z Value

2.576

Interval Half Width= 2.8493*1.96=

7.3398

Confidence Interval

Interval Lower Limit=( 4.6-7.3398)=

-2.74

Interval Upper Limit=( 4.6+7.3398)=

11.94

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 effetiveness of garlic in reducing LDL Cholesterol. What is the confidence interval estimate of the population mean

-2.74 mg/dL < mean < 11.94 mg/dL   (round 2 decimal places)

_____ A) do not contain 0, suggesting that the garlic treatment did not affect the LDL Cholesterol levels

_____B) contain 0, suggesting that he garli treatment did affect the LDL Cholesterol levels

_____C) do not contain 0, suggesting that the garlic treatment did affect the LDL Choleterol level

_____D) contain 0, suggesting that the garlic treatment did not affect the LDL Cholesterol levels

Confidence Interval Estimate for the Mean

Data

Standard Deviation

18.9

Sample Mean

4.6

Sample Size

44

Confidence Level

99%

Intermediate Calculations

Standard Error of the Mean= 18.9/sqrt(44) =

2.8493

Z Value

2.576

Interval Half Width= 2.8493*1.96=

7.3398

Confidence Interval

Interval Lower Limit=( 4.6-7.3398)=

-2.74

Interval Upper Limit=( 4.6+7.3398)=

11.94

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