EC_315_Week_6_Homework_Solutions[1]

EC_315_Week_6_Homework_Solutions[1] - EC/315 Week 6...

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EC/315 Week 6 Homework Solutions Prof Brian W Sloboda Spring 2008 II Term Chapter 12 22. A physician who specializes in weight control has three different diets she recommends. As an experiment, she randomly selected 15 patients and then assigned 5 to each diet. After three weeks the following weight losses, in pounds, were noted. At the .05 significance level, can she conclude that there is a difference in the mean amount of weight loss among the three diets? Plan A Plan B Plan C 5 6 7 7 7 8 4 7 9 5 5 8 4 6 9 H o : μ 1 = 2 = 3 H 1 : Not all means are equal (Could also say :At least one of the means differ) The level of significance is .05 (Type I error) H o is rejected if F > 3.89 (the degrees of freedom is 2/12 as given in red in the ANOVA table) Test Statistic (this is given as blue and bold from the ANOVA table) Source SS df MS F Treatment 26.13 2 13.067 13.52 Error 11.60 12 0.967 Total 37.73 14 H o is rejected since 13.52 > 3.89. There is a difference in the mean weight loss among the three diets. 28. Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of six hourly periods is chosen for each assembly line and the number of components produced during these periods for each line is recorded. The output from a statistical software package is: Summary Groups Count Sum Average Variance Line A 6 250 41.66667 0.266667 Line B 6 260 43.33333 0.666667 Line C 6 249 41.5 0.7
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ANOVA Source of Variation SS df MS F p -value Between Groups 12.3333 3 2 6.16666 7 11.3265 3 0.001005 Within Groups 8.16666 7 1 5 0.54444 4 Total 20.5 1 7 a. Use a .01 level of significance to test if there is a difference in the mean production of the three assembly lines. 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 -value Between Groups 12.3333 3 2 6.16666 7 11.3265 3 0.001005 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 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.
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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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EC_315_Week_6_Homework_Solutions[1] - EC/315 Week 6...

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