In a test of the effectiveness of garlic for lowering cholesterol, 48 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 LDLcholesterol (in mg/dL) have a mean of 3.6 and a standard deviation of 17.5. 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 theeffectiveness 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 u?mg/dL <µ<mg/dL(Round to two decimal places as needed.)

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Answered: hat is the confidence interval estimate…

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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One group of accounting students took a distance learning class, while another group took the same course in a traditional classroom. At α = .10, is there a significant difference in the mean scores listed below? Exam Scores for Accounting Students Statistic Distance Classroom Mean scores x¯1x¯1 = 9.7 x¯2x¯2 = 10.0 Sample std. dev. s1 = 2.7 s2 = 2.9 Number of students n1 = 23 n2 = 23 Specify the decision rule. (Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.) Find the test statistic tcalc. (Round your answer to 4 decimal places. A negative value should be indicated by a minus sign.) Use Excel to find the p-value. (Round your answer to 3 decimal places.)
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What does a statistically significant t test indicates?
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The standard error of estimate is the standard deviation of the observed x-values about the predicted x values TRUE OR FALSE
Chapter 9, Section 2, Exercise 032 Use the following ANOVA table for regression to answer the questions. Response: Y Source DF Sum Sq Mean Sq F-value Pr(>F) Regression 1 355.21 355.21 14.76 0.000 Residual Error 359 8641.13 24.07 Total 360 8996.34 Give the F-statistic and p-value.Enter the exact answers.The F-statistic is=The p-value is=
Suppose you had obtained a test statistic of 2.45 (this is incorrect). What would the P-value be? Draw a conclusion based on a = 0.01. Select one: O a. 0.01 < P-value < 0.02, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O b. 0.005 < P-value < 0.01, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O c. 0.01 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O d. 0.005 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet…
Determine the cri tical z-score corresponding to a = 0.2 in a left-tail test. crititical z-score = [three decimal accuracy]
The researcher would like to determine if these data provide convincing evidence that the true mean amount of time volunteers who were given training held their breath is greater than volunteers without training. Let μ1 = the true mean amount of time that volunteers who were given training held their breath and μ2 = the true mean amount of time that volunteers without training held their breath. What are the appropriate hypotheses?
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hat is the confidence interval estimate of the population mean u? ]mg/dL