A company manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. To measure how much employees at these plants know about quality management, a random sample of 6 employees was selected from each plant and the employees selected were given a quality awareness examination. The examination scores for these 18 employees are shown in the followingtable. The sample means, sample variances, and sample standard deviations for each group are also provided. Managers want to use these data to test the hypothesis that the mean examination score is the same for all three plants.Sourceof VariationTreatmentsErrorSamplemeanTotalPlant 1 Plant 2 Plant 3Atlanta DallasSeattle85768275о но: М1 =H2=Изна: M1 #M2 + M37290Sample 46.8varianceSamplestandard 6.84deviation807176737368897554.07.35SumDegreesof Squares of Freedom6065о но: М1 =H2= изH₂: Not all the population means are equal.62Set up the ANOVA table for these data. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)6876716735.25.93MeanSquareTest for any significant difference in the mean examination score for the three plants. Use α = 0.05.State the null and alternative hypotheses.о но: H1 #M2 # ИзHa: H1 = M2 = M3O Ho: At least two of the population means are equal.H₂: At least two of the population means are different.FO Ho: Not all the population means are equal.на: М1 = M2 = M3Find the value of the test statistic. (Round your answer to two decimal places.)Find the p-value. (Round your answer to four decimal places.)p-value =p-valueState your conclusion.Reject Ho. There is not sufficient evidence to conclude that the means for the three plants are not equal.Do not reject Ho. There is sufficient evidence to conclude that the means for the three plants are not equal.O Do not reject Ho. There is not sufficient evidence to conclude that the means for the three plants are not equal.O Reject Ho. There is sufficient evidence to conclude that the means for the three plants are not equal.

Given information:k= 3 groups or treatments Each group has 6 observation

Answered: Sample mean Error Sample variance Plant…

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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Treatment II III 2 N = 24 G = 72 EX2 = 292 3 M = 2 M = 3 T = 16 T= 24 T= 32 SS = 16 SS = 24 SS = 20 Use an ANOVA with a = .05 to determine whether there is a significant mean difference between the two treatments. (Round to two decimal places where needed.) Source df MS Feritical Between treatments Within treatments Total F Distribution Numerator Degrees of Freedom = 6 Denominator Degrees of Freedom = 16 0.0 1.0 3.0 4.0 5.0 6.0 7.0 9.0 11.0 120 Conclusion: Fail to reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are significant differences among the three treatments. o Fail to reject the null hypothesis; there are significant differences among the three treatments.
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