A pharmaceutical company conducted a study to test the effect of a new high blood medication. The company selected 15 subjects from a population. Each subject was randomly assigned to one ofthree treatment groups equally (i.e. each group consists of five subjects): Group 1 (10mg/day), Group 2 (50mg/day) and Group 3 (100mg/day). After 30 days, doctors measured the blood pressure of eachsubject. Summary of Analysis of Variance table is shown in Table 3.Table 3Sum ofDegree ofFreedomMeanSourceFSquareSquares(i)Treatments(iv)3203.98(ii)(iii)(v)(vi)Error(vii)Total(a)Calculate the value for (i)-(vii) from the above table.(b)Conduct a hypothesis test to test whether all treatment groups have the same mean using a 5% level of significance.

One way ANOVA is a type of hypothesis test, and it helps to know whether the means of three (or)…

Answered: A pharmaceutical company conducted a…

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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This research was conducted with Korean adolescents in the Los Angeles area. Would you be willing to generalize the results of this study to Korean adolescents who live in other regions of the country? Why or why not.
A social psychologist is interested in knowing whether people are more aggressive if they are in the presence of a weapon, such as a gun Treatment Condition- T1, as compared to when they are in a room with an innocuous item such as a tennis racket Control Condition-T2. The study used one sample of 17 college students and had all of the students in the sample participate in both treatment conditions. The mean of the difference scores was 18.0 and the variance (s2) was 612. An alpha level of .01 was set. Calculate the t. A 3.000 B 2.780 C 2.152 D 1.765
Records from previous years for a casualty insurance company show that its clients average a combined total of 1.9 auto accidents per day, with a variance of 0.31. The actuaries of the company claim that the current variance, oʻ, of the number of accidents per day is not equal to 0.31. A random sample of 17 recent days had a mean of 2 accidents per day with a variance of 0.62. If we assume that the number of accidents per day is approximately normally distributed, is there sufficient evidence to conclude, at the 0.05 level of significance, that the actuaries are correct? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H,. p H, :0 H, :0 (b) Determine the type of test statistic to use. (Choose one) ▼ D=0 OSO O20 (c) Find the value of the test statistic. (Round…
The variability in the amounts of impurities present in a batch of chemical used for a particular process depends on the length of time the process is in operation. A manufacturer using two production lines, 1 and 2, has made a slight adjustment to line 2, hoping to reduce the variability as well as the average amount of impurities in the chemical. Samples of 25 and 25 measurements from the two batches yield 1.04, respectively 0.51 variances. Do the data present sufficient evidence to indicate that the process variability is less for line 2? Use 5% significance level. Conduct a hypothesis test of the airline executive’s belief. [α = 0.05]
A study aims to compare the social media usage rates of men and women. Nine men and 9 women are drawn from the general population and the average time (in minutes) they spent on social media each day was compared (men’s mean = 40, sum of squares = 153; women’s mean = 48, sum of squares = 216). What is the (a) estimated population variance for each random sample, and (b) pooled variance? Estimated population variances = 19.13 and 27.00; Pooled variance = 23.07 Estimated population variances = 19.13 and 27.00; Pooled variance = 20.50 Estimated population variances = 17.00 and 24.00; Pooled variance = 20.50 Estimated population variances = 17.00 and 24.00; Pooled variance = 23.07
A researcher is interested whether a new drug is effective in changing patients' cholesterol levels. She divided 33 patients into 3 groups according to the dosage the patient received. The cholesterol levels of all patients are measured and used as the data for the Analysis of Variance (ANOVA). If she reported the test statistic F = 2.1 What is the degree of freedom for the numerator of the F statistic? What is the degree of freedom for the denominator of the F statistic? What is the P-value of the F-test? (Round to 3 decimals) Given the significance level of 0.01, what can she conclude from the test? *The patients in differnt groups have simliar average cholesterial levels. The drug made no difference. *At least one group of patients have an average cholesterol level that is different from the other groups. The drug may have an effect.
According to Zero Population Growth, the average urban U.S. resident consumes 3.3 pounds of food per day. Is this figure accurate for rural U.S. residents? Suppose 64 rural U.S. residents are identified by a random procedure and their average consumption per day is 3.60 pounds of food. Assume a population variance of 1.31 pounds of food per day. Use a 5% level of significance to determine whether the Zero Population Growth figure for urban U.S. residents also is true for rural U.S. residents on the basis of the sample data. Appendix A Statistical Tables (Round your answer to 2 decimal places, e.g. 15.75.) The value of the test statistic is z = and we
What correctly represents the coefficient of determination in terms of the variance that is an output from the analysis of variance table?
chester butchery supplies vienna sausages in the entire kzn province. a meat packer is investigating the marked mass shown on vienna sausages. a pilot study showed a mean of 11.8kg per pack and a variance of 0.49kg. how many packs should be sampled in order to be 92% confident that the sample mean will differ by at most 0.2kg.
Please use ANOVA

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