EC3315_HW6_Jamie Gossett

EC3315_HW6_Jamie Gossett - than the critical value (6.36)....

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Jamie Gossett EC315 Homework – Week 6 22. H0:u1=u2=u3; H1: not all treatment means are equal Plan A X=EX/n=5+7+4+5+4/5=5 Plan B X=EX/n=6+7+7+5+6/5=6.2 Plan C X=EX/n=7+8+9+8+9/5=8.2 H0 is rejected. There is a difference in treatment means. Accept H1 28. a. Use a .01 level of significance to test if there is a difference in the mean production of the three assembly lines. : = = H0 μA μB μC : . H1 The mean production of the three assembly lines is not equal The level of significance is .01 (Type I error) H o is rejected if F > 6.36 (the degrees of freedom is 2/15 as given in red in the ANOVA table) Test Statistic (this is given as blue and bold from the ANOVA table) ANOVA Source of Variation SS df MS F p - v al u e Between Groups 12.3333 3 2 6.16666 7 11.3265 3 0. 0
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0 1 0 0 5 Within Groups 8.16666 7 1 5 0.54444 4 Total 20.5 1 7 The null hypothesis of equal means is rejected because the F statistic (11.33) is greater
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Unformatted text preview: than the critical value (6.36). Also you could interpret using the p value. The p-value (0.0010) is also less than the significance level (0.01). The mean production rates are different. b. Develop a 99 percent confidence interval for the difference in the means between Line B and Line C. t for a 99 percent confidence interval and 15 degrees of freedom is 2.947- + = .-. . . + X1 X2 tMSE1n1 1n2 43 33333 41 5 2 9470 54444416 16 = . . 1 83333 1 25544 The confidence intervals is 0.57789 to 3.08877 and since both values are positive, this means that there is a difference between the two means. Remember if there is a 0 contained in the confidence interval, then it means there is no conclusion can be drawn between the two means....
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This note was uploaded on 04/23/2011 for the course EC 315 taught by Professor Barcus during the Spring '10 term at Park.

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EC3315_HW6_Jamie Gossett - than the critical value (6.36)....

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