The data in table is from a randomized block design represent bloodpressure measurements from an experiment involving 4 age groups,each with 3 persons. The 3 persons within each age group were randomlyassigned to drugs, A, B and C with one person per drug. This was done to keepthe drug assignments balanced within age groups. Comparing the three drugs inmanner that provides balance for or control for possible age effects. Considerthe data given in table below and compute;a. Mean Square Due to Treatments (MSTR)b. Mean Square Due to Blocks (MSBL)Mean Square Due to Error (MSE)C.d. F-testTreatmentsAge Groups (Blocks) AB17072276843828649092Total318 334CTotal Blocks802228224284 25288 270334 986

The question is about ANOVA.Given :

Answered: The data in table is from a randomized…

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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The leader of two postpartum women’s support groups is interested in the depression levels of the women in her groups. She administers the Center for Epidemiologic Studies Depression Scale (CES-D) screening test to the members of her groups. The CES-D is a 20-question self-test that measures depressive feelings and behaviors during the previous week. The mean depression level from the screening test for the 10 women in the first group is μ₁ = 16; the mean depression level for the 14 women in the second group is μ₂ = 10. Without calculating the weighted mean for the combined group, you know that the weighted mean is:
A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =
An investigator is using a chi-square test to compare the proportion of patients with a successful recovery after receiving one of 4 treatments. There were 25 subjects in each group. What is the critical value for testing no association or treatment and response?
The data is presented in parallel format. Before running the ANOVA in Excel, you will need to put it into serial (column) format. You will need to create a categorical variable for the first factor (days) and for the second factor (time of day). This is a critical step. The data presented shows exam scores for students randomly assigned to the sections meeting on these days at these times of the day: Thursday Friday Morning Afternoon Morning Afternoon 59.1 60.6 50.2 34.7 50.6 60.5 39 44 57.1 59 71.2 43.7 58.6 45.7 57.5 52.9 51.1 72.6 50 53.3 40.1 39.8 58.1 34.6 27.3 59.3 38.9 53.2 27.3 72.9 58.2 52.9 39.5 45.9 57.9 46.8 56.6 40 71 46.7 Conduct a 2-way ANOVA. First, find the group means for each level of the factors. (Report answers accurate to 2 decimal places.) Days: MThursday MFriday Time of day: Mmorning Mafternoon Report the results of the ANOVA for the main & interaction effects. (Report P-values accurate to 4 decimal places and F-ratios accurate to 3 decimal places.) Day: FA = p…
Pn.n.m.
Among 1815 randomly selected high school students surveyed by the Centers for Diseases Control, 300 said they were current smokers. b. Estimate the variability in such sample proportion by finding SE (p).
The data is presented in parallel format. Before running the ANOVA in Excel, you will need to put it into serial (column) format. You will need to create a categorical variable for the first factor (days) and for the second factor (time of day). This is a critical step.The data presented shows exam scores for students randomly assigned to the sections meeting on these days at these times of the day: Thursday Friday Morning Afternoon Morning Afternoon 70.878.15978.246.646.476.559.376.370 65.678.359.359.778.680.892.780.865.392.3 90.890.569.177.36977.558.677.877.558.6 63.372.854.254.472.166.26372.56672.2 Conduct a 2-way ANOVA. First, find the group means for each level of the factors. (Report answers accurate to 2 decimal places.)Days: MThursday=MThursday= MFriday=MFriday= Time of day: Mmorning=Mmorning= Mafternoon=Mafternoon= Report the results of the ANOVA for the main & interaction effects. (Report P-values accurate to 4 decimal places and…
For which of the following would it be most appropriate to use an ANOVA to analyze the data? A researcher is interested in determining the most effective format for advertising. He randomly assigns 60 people to one of three groups: (1) television commercial, (2) radio commercial, or (3) magazine advertisement. Group members review the commercial and then report whether or not they would buy the product. In order to determine the effectiveness of different diet plans, 90 people are randomly assigned to one of three groups: (1) low calorie diet, (2) low-fat diet, or (3) low-carb diet. At the end of 60 days, the number of pounds lost was recorded for each of the participants. Both A and B. Neither A nor B.
The General Social Survey (GSS) collects data on demographics, eduction and work, among many other characteristics of US residents. Suppose we want to estimate the difference between the average number of hours worked by all Americans with a college degree and those without a college degree. Is there sufficient evidence that there is a significant difference between the average number of hours worked by those Americans with a college degree vs. those Americans without a college degree? Use the following output: Welch Two Sample t-testdata: yes and not = 3.1181, df = 1098.5, p-value = 0.001867alternative hypothesis: true difference in means is not equal to 095 percent confidence interval: 1.011652 4.445822sample estimates:mean of x mean of y 42.81574 40.08701 Using the provided output, find the 95% confidence interval for the difference of the average amount of hours worked for those that have a college degree vs. those that do not have a college degree (ie: for the difference of two…
he data on the number of chocolate chips per bag for 42 bags of Chips Ahoy! cookies were obtained by the students in an introductory statistics class at the United States Air Force Academy in response to the Chips Ahoy! 1,000 Chips Challenge sponsored by Nabisco, the makers of Chips Ahoy! Use the data collected by the students to answer the following questions and to conduct the analyses required in each part. They found ¯xx¯ = 1261.6 and s = 117.6 a. Determine a 95% confidence interval for the mean number of chips per bag for all bags of Chips Ahoy! cookies. ≤μ≤≤μ≤ b. Interpret your result in words. c. Can you conclude that the average bag of Chips Ahoy! Cookies contain at least 1000 chocolate chips? yes no
An analysis of variance produces SSb = 40 and MSb = 20. In this analysis, how many treatment conditions are being compared?
Show the impact of sample size on power?

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