{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EC_315_Week_6_Homework_Solutions[1]

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

This preview shows pages 1–3. Sign up to view the full content.

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 = μ 2 = μ 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online