In Exercises 9–16, assume that each sample is a simplerandom sample obtained from a population with a normal distribution.

Garlic for Reducing Cholesterol In a test of the effectiveness of garlic for lowering cholesterol, 49 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 (based on data from “Effect of Raw Garlic vs Commercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia,” by Gardner et al.,Archives of Internal Medicine,Vol. 167).Construct a 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment. Does the result indicate whether the treatment is effective?

Short Answer

Expert verified

The 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment is 16.9<σ<27.4.

And, the confidence interval does not indicate that the treatment is effective.

Step by step solution

01

Given information

The sample number of subjects who were treated with raw garlic is n=49.

The mean level of LDL cholesterol (in mg/Dl) is 0.4.

The sample standard deviation is s=21min.

The level of confidence is 98%.

02

Compute the critical values and confidence interval

The degrees of freedom is computed as,

df=n-1=49-1=48

The level of confidence is 98%, which implies that the level of significance is 0.02.

Using the Chi-square table, the critical values at 0.02 level of significance and at 48 degrees of freedom are χL2=28.177and χR2=73.6826.

The 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment is computed as,

n-1s2χR2<σ<n-1s2χL249-121273.6826<σ<49-121228.17716.9<σ<27.4

Therefore, the 98% confidence interval estimate of the standard deviation of the changes in LDL cholesterol after the garlic treatment is 16.9<σ<27.4.

And, the confidence interval does not give any information about the effectiveness of the treatment.

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Most popular questions from this chapter

Chickenpox : You plan to conduct a survey to estimate the percentage of adults who have had chickenpox. Find the number of people who must be surveyed if you want to be 90% confident that the sample percentage is within two percentage points of the true percentage for the population of all adults.

a. Assume that nothing is known about the prevalence of chickenpox.

b. Assume that about 95% of adults have had chickenpox.

c. Does the added knowledge in part (b) have much of an effect on the sample size?

In Exercises 9–16, assume that each sample is a simple

random sample obtained from a population with a normal distribution.

Comparing Waiting Lines

a. The values listed below are waiting times (in minutes) of customers at the Jefferson Valley Bank, where customers enter a single waiting line that feeds three teller windows. Construct a95% confidence interval for the population standard deviation .

6.5 6.6 6.7 6.8 7.1 7.3 7.4 7.7 7.7 7.7

b. The values listed below are waiting times (in minutes) of customers at the Bank of Providence, where customers may enter any one of three different lines that have formed at three teller windows. Construct a 95% confidence interval for the population standard deviation .

4.2 5.4 5.8 6.2 6.7 7.7 7.7 8.5 9.3 10.0

c. Interpret the results found in parts (a) and (b). Do the confidence intervals suggest a difference in the variation among waiting times? Which arrangement seems better: the single-line system or the multiple-line system?

Garlic for Reducing Cholesterol In a test of the effectiveness of garlic for lowering cholesterol, 49 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 (based on data from “Effect of Raw Garlic vs Commercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia,” by Gardner et al., Archives of Internal Medicine, Vol. 167). Construct a 98% 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?

Confidence Interval with Known σ. In Exercises 37 and 38, find the confidence interval using the known value of σ.

Birth Weights of Girls Construct the confidence interval for Exercise 9 “Birth Weights of Girls,” assuming that σis known to be 7.1 hg.

Bootstrap Sample Given the sample data from Exercise 2, which of the following are not possible bootstrap samples?

a. 12, 19, 13, 43, 15

b. 12, 19, 15

c. 12, 12, 12, 43, 43

d. 14, 20, 12, 19, 15

e. 12, 13, 13, 12, 43, 15, 19

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