In a study evaluating the impact of garlic on cholesterol reduction, 42 subjects were administered garlic in processed tablet form. Cholesterol levels were recorded both before and after treatment. The difference (before - after) in LDL cholesterol levels (measured in mg/dL) showed a mean of 4.1 and a standard deviation of 19.1. Construct a 99% confidence interval estimate for the mean change in LDL cholesterol post-treatment. Analyze what this confidence interval indicates about garlic's effectiveness in reducing LDL cholesterol levels.Links for reference:- [Click here to view a t 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.](#)**Question:** What is the confidence interval estimate of the population mean μ?\[ \text{mg/dL} < \mu < \text{mg/dL} \](Round to two decimal places as needed.)

Given,sample size(n)=42sample mean(x¯)=4.1standard deviation(s)=19.1degrees of…

veness of ganic 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 effectiveness of garlic in reducing LDL cholesterol? Click here to view at distribution table. Click here to view paga 1 of the standard normal distribution table. Click here to view paga 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? Omg/dl

veness of ganic 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 effectiveness of garlic in reducing LDL cholesterol? Click here to view at distribution table. Click here to view paga 1 of the standard normal distribution table. Click here to view paga 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? Omg/dl

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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na 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 (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.4 and a standard deviation of 19.8. Construct a 95% 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? Click here to view a t 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. What is the confidence interval estimate of the population mean µ? Omg/dL
In a poll, 1100 adults in a region were asked about their online vs. in-store clothes shopping. One finding was that 44% of respondents never clothes-shop online. Find and interpret a 95% confidence interval for the proportion of all adults in the region who never clothes-shop online. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. The 95% confidence interval is from to (Round to three decimal places as needed.)
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A sample mean, sample size, population standard deviation, and confidence level are provided. Use this information to complete parts (a) through (c) below. x = 32, n = 27, o = 3, confidence level = 90% Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The confidence interval is from 31 to 33 (Type integers or decimals rounded to one decimal place as needed.) b. Obtain the margin of error by taking half the length of the confidence interval. What is the length of the confidence interval? 2 (Type an integer or decimal rounded to one decimal place as needed.) c. Obtain the margin of error by using the formula E =z« Vn 2 Identify the critical value. Z, = || |(Type an integer or decimal rounded to two decimal places as needed.) 0. 2
In a random sample of 36 criminals convicted of a certain crime, it was determined that the mean length of sentencing was 54 months, with a standard deviation of 13 months. Construct and interpret a 90% confidence interval for the mean length of sentencing for this crime. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Click here to view the table of critical t-values. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to one decimal place as needed.) O A. There is a 90% probability that the mean length of sentencing for the crime is between and months. O B. We can be 90% confident that the mean length of sentencing for the crime is between and months. O C. 90% of the sentences for the crime are between and months.
A poll asked whether states should be allowed to conduct random drug tests on elected officials. Of 21,327 respondents, 56% said "yes." a. Determine the margin of error for a 99% confidence interval. b. Without doing any calculations, indicate whether the margin of error is larger or smaller for a 90% confidence interval. Explain your answer. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. a. The margin of error for a 99% confidence interval is (Round to three decimal places as needed.) b. Without doing any calculations, indicate whether the margin of error is larger or smaller for a 90% confidence interval. A. The margin of error will be smaller for a 90% confidence interval. B. The margin of error will be about the same for a 90% confidence interval. C. The margin of error will be larger for a 90% confidence interval. D. There is insufficient information to determine the…
Salaries of 49 college graduates who took a statistics course in college have a mean, x, of $63,500. Assuming a standard deviation, o, of $14,619, construct a 90% confidence interval for estimating the population mean u. 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. $
The accompanying data represent the running times of files produced by two motion-picture companies. Assume that the running-time differences are approximately normally distributed. Construct a 90% confidence interval for 90% confidence interval for μ - μ were created, should it be assumed that o² = = If a Company | 106 96 105 Time (minutes) 91 99 II 101 78 123 89 171 89 120 Click here to view page 1 of the table of critical values of the F-distribution, for alpha = 0.05. Click here to view page 2 of the table of critical values of the F-distribution, for alpha = 0.05. The confidence interval is < (Round to three decimal places as needed.)
The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Stance duration (ms) was measured in separate samples of younger and older adults. younger: n=22 mean=801 s=117 older: n=16 mean=780 s=81 Test the claim that the true average stance is different between younger and older adults. Let population 1= younger population 2= older a. Construct a 95% confidence interval estimate, what is the margin of error? Use 2 decimal places.