In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. Thechanges (before-after) in their levels of LDL cholesterol (in mg/dL) have a mean of 4.5 and a standard deviation of 17.6. Construct a 99% confidence interval estimate of the mean net change inLDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?Click here to view at distribution table.Click here to view page 1 of the standard normal distribution table.Click here to view page 2 of the standard normal distribution table.2What is the confidence interval estimate of the population mean u?mg/dL << mg/dL(Round to two decimal places as needed.)What does the confidence interval suggest about the effectiveness of the treatment?OA. The confidence interval limits do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.OB. The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.OC. The confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.OD. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.3*WSXHommandw #380F3EDC$4aF4RFVde L%5FSTGA6BGF6YH&r7N4F7UJ*8M1-DIIFBK9V-DDFOOL)0^.4F10P:;Icommand optionTime Remaining: 01:16:01{[?14F11+ 11=}]F12Nextdelete11returnshift

Given,sample size(n)=42sample mean(x¯)=4.5standard deviation(s)=17.6degrees of…

Answered: In a test of the effectiveness of…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Find the following probabilities. It may help to sketch the pdf for Standard Normal Distributionand shade in the area that corresponds to the probability of interest. 1) Z ~ Norm(0, 1), find Pr(-2.34 ≤ Z < 1.45)
In a test of the effectiveness of garlic for lowering cholesterol, 36subjects 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) have a mean of 0.3 and a standard deviation of 2.39. Use a 0.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
The shelf life of a battery produced by one major company is known to be Normally distributed, with a mean life of 3.5 years and a standard deviation of 0.75 years. What is the upper quartile of battery shelf life? Use the fact that the middle 50% of the data falls within 0.67 standard deviations from the mean.
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The combined math and verbal scores for students taking a national standardized examination for college admission, is normally distributed with a mean of 800 and a standard deviation of 140. If a college requires a minimum score of 900 for admission, what percentage of student do not satisfy that requirement?
Á study was conducted that examined the relationship between alcohol consumption and risk of heart attack. In all, 417 men, age 40 or older, were tracked over ten years and the number of that experienced a heart attack was recorded. The participants were grouped by those who totally abstained from alcohol and those that drank moderately. For the results given in the table below, use the chi-squared test to find the p value for NHST at the .05 level. (In SPSS use the Yates continuity correction value.) Heart attack Alcohol Crosstabulation Count Alcohol Abstainer Moderate Total Heart attack No 197 192 389 Yes 9. 19 28 Total 206 211 417 Op= .090 Op= 121 !! Op=.067 Op = .048
Suppose that a Normal model describes furl economy (mile per gallon) for automobile and that a Saturn has a standardized score ( z score) of +2.2. This meatn taht Satturn.... gets 2.2 miler sper gallon. get 2.2 times the gas mileage of the avereage car. get 2.2 mpg more than the average car. have a standard deviation of 2.2 mpg achieve fuel eceonomy that is 2.2 standard deviations better thean the average car.
Diabetes Plasma-glucose levels are used to determine the pres- ence of diabetes. Suppose the mean In (plasma-glucose) concentration (mg/dL) in 35- to 44-year-olds is 4.86 with standard deviation = 0.54. A study of 100 sedentary people in this age group is planned to test whether they have a higher or lower level of plasma glucose than the general population. 7.7 If the expected difference is 0.10 In units, then what is the power of such a study if a two-sided test is to be used with a = .05? 7.8 Answer Problem 7.7 if the expected difference is 0.20 In units. 7.9 How many people would need to be studied to have 80% power under the assumptions in Problem 7.7?
The Merck Manual states that, for healthy adults, the mean number of milliliters of oxygen per deciliter of blood is 19.0. A company that sells vitamins claims that its multivitamin complex will increase the oxygen capacity of the blood. A random sample of 28 adults took the vitamin for six months. After blood tests, it was found that the sample mean was 20.7 ml of oxygen per deciliter of blood with a standard deviation of 6.7ml. a. At the 0.05 level, test the claim that the average oxygen capacity has increased. b. How much power do you have to detect a 3ml difference (from the null) in the average amount of oxygen in the blood? c. What sample size would you need to have 90% power to detect this observed difference?
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