MinitabSourceFactorAdj ss0.0479790.004451DFAdj MS0.0159930.000032F-Value503.06P-Value30.000ErrorTotal1401430.0524302. Cola Weights For the four samples described in Exercise 1, the sample of regular Coke hasa mean weight of 0.81682 lb, the sample of Diet Coke has a mean weight of 0.78479 lb, thesample of regular Pepsi has a mean weight of 0.82410 lb. and the sample of Diet Pepsi has amean weight of 0.78386 lb. If we use analysis of variance and reach a conclusion to reject equal-ity of the four sample means, can we then conclude that any of the specific samples have meansthat are significantly different from the others?3. Cola Weights For the analysis of variance test described in Exercise 1, is that test left-tailed,right-tailed, or two-tailed?4. Cola Weights Identify the value of the test statistic in the display included with Exercise 1.In general, do larger test statistics result in larger P-values, smaller P-values, or P-values thatare unrelated to the value of the test statistic?5. Cola Weights The displayed results from Exercise I are from one-way analysis of variance.What is it about this test that characterizes it as one-way analysis of variance instead of two-wayanalysis of variance?6. One-Way ANOVA In gencral, what is onc-way analysis of variance used for?7. One vs. Two What is the fundamental difference between one-way analysis of variance andtwo-way analysis of variance?

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Answered: 2. Cola Weights For the four samples…

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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1. A purification process for a chemical involves passing it, in solution, through a resin on which impurities are adsorbed. A chemical engineer wishing to test the efficiency of 3 different resins took a chemical solution and broke it into 15 batches. She tested each resin 5 times and then measured the concentration of impurities after passing through the resins. The results of the Analysis of Variance on the data are shown below. Analysis of Variance Table Response: concentration Df Sum Sq Mean Sq F value Pr(>F) resin 2 1.45E-05 7.27E-06 0.048737 0.9526 Residuals 12 0.001789 0.000149 a.) Interpret the result of the Analysis of Variance (ANOVA). What is the statistical decision and corresponding conclusion? b.) What is the implication of the results obtained from the analysis of the data?
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Periodically, customers of a financial services company are asked to evaluate the company's financial consultants and services. Higher ratings on the client satisfaction survey indicate better service, with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use ? = 0.05 and test to see whether the consultant with more experience has the higher population mean service rating. Assume the population variances are unequal. Consultant A Consultant B n1 = 16 n2 = 10 x1 = 6.85 x2 = 6.25 s1 = 0.69 s2 = 0.75 (a) State the null and alternative hypotheses. -H0:?1 − ?2 = 0 Ha:?1 − ?2 = 0 -H0:?1 − ?2 > 0 Ha:?1 − ?2 = 0 -H0:?1 − ?2 ≠ 0 Ha:?1 − ?2 = 0 -H0:?1 − ?2 = 0 Ha:?1 − ?2 > 0 -H0:?1 − ?2 = 0 Ha:?1 − ?2 ≠ 0 (b) Compute the value of the test statistic. (Round your answer to three decimal places.)…
Actuaries use various parameters when evaluating the cost of a life insurance policy. The variance of the life spans of a population is one of the parameters used for the evaluation. Each year, the actuaries at a particular insurance company randomly sample 30 people who died during the year (with the samples chosen independently from year to year) to see whether the variance of life spans has changed. The life span data from this year and from last year are summarized below. Current Last Year Year *1 = 75.8 x2 = 76.2 s3 = 47.61 s3= 92.16 (The first row gives the sample means and the second row gives the sample variances.) Assume that life spans are approximately normally distributed for each of the populations of people who died this year and people who died last year. Can we conclude, at the 0.05 significance level, that the variance of the life span for the current year, of, differs from the variance of the life span for last year, o,? Perform a two-tailed test. Then complete the…
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You are testing the claim that having lights on at night increases weight gain (abstract). A sample of 10 mice lived in an environment with bright light on all of the time and 8 mice who lived in an environment with a normal light/dark cycle is given below. Test the claim using a 2% level of significance. Assume the population variances are unequal and that the weight changes are normally distributed. Give answers to 3 decimal places. Data available at StatKey, choose Mice Wgt Gain-2e data set Light (1) 1.71 4.67 4.99 5.33 5.43 6.94 7.15 9.17 10.26 11.67 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: ₁ Test Statistic = Dark (₂) 2.27 2.53 2.83 4 4.21 4.6 5.95 6.52 Ha: ₁ Based on the hypotheses, find the following: p-value = = V Select an answer ✓ Select an answer (Hint: difference in means from H₂) The correct decision is to fail to reject the null hypothesis The correct summary…
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Periodically, customers of a financial services company are asked to evaluate the company's financial consultants and services. Higher ratings on the client satisfaction survey indicate better service, with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use a = 0.05 and test to see whether the consultant with more experience has the higher population mean service rating. Assume the population variances are unequal. Consultant A Consultant B n1 = 16 n2 10 X1 = 6,81 X2 = 6.25 %3D S1 = 0.68 S2 = 0.75 (a) State the null and alternative hypotheses. O Hoi H1 - H2 = 0 Hai H1- Hz>0 O Ho: H1 - 42 = 0 Hại H1 - H2# 0 O Ho: H1 - H2 > 0 %3D Ha: H1 - H2 = 0 O Ho: H1 - Hz # 0 Hai Hy - Hz = 0 Hoi H1 - H2 = Hai H1- H2 = 0 (b) Compute the value of the test statistic. (Round your answer to three decimal places.) 19.191 (c) What is the p-value?…

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