252y0572 12 01 05 Page layout view ECO252 QBA2 THIRD HOUR EXAM Dec 1 2005 Name KEY Hour of Class Registered MWF 2 MWF3 TR 12 30 TR2 I 40 points Do all the following 2 points each unless noted otherwise Do not answer question yes or no without giving reasons Show your work in questions that are not multiple choice 1 Turn in your computer problems 2 and 3 marked to show the following 5 points 2 point penalty for not doing a In problem 2 what is tested and what are the results b In problem 3 what coefficients are significant What is your evidence c In the last graph in problem 3 where is the regression line 5 2 Dummeldinger As part of a study to investigate the effect of helmet design on football injuries head width measurements were taken for 30 subjects randomly selected from each of 3 groups High school football players college football players and college students who do not play football so that there are a total of 90 observations with the object of comparing the typical head widths of the three groups If the researchers assume that the data in each of these three groups comes from a Normally distributed population they should use the following method a The Kruskal Wallis test b One way ANOVA c The Friedman test d Two Way ANOVA 2 7 3 Sandy Which of the following is not an assumption required for 1 way ANOVA wth 4 columns a 1 2 3 4 b All of the columns are random samples c All of the population have to be Normally distributed d 1 2 3 4 2 9 4 If we are comparing the means of 5 random samples and find the following x1 10 x 2 12 x 3 11 x 4 13 x 5 14 s1 1 0 s 2 1 2 s 3 1 3 s 4 1 5 s 5 1 5 n1 7 n 2 7 n3 7 n 4 7 n 5 7 The appropriate test statistic is a D x1 x 2 x 3 x 4 x 5 s12 s 22 s 32 s 42 s 52 n1 n 2 n3 n 4 n5 b F with 7 and 4 degrees of freedom 0 5 c F with 4 and 30 degrees of freedom 2 d F with 4 and 7 degrees of freedom 0 5 e 2 with 18 degrees of freedom f 2 with 34 degrees of freedom 2 Solution We use ANOVA for multiple comparison of means 11 2 is only used in comparing medians and proportions n n j 35 so total degrees of freedom are 34 There are 5 columns so degrees of freedom between are 4 Thus there are 34 4 30 degrees of freedom MSB within and for an ANOVA F has 4 and 30 DF MSW 252y0572 12 01 05 Page layout view 5 If we are doing a 2 way ANOVA and find the following Two way ANOVA C5 versus C6 C7 Source Rows Columns Interaction Error Total DF 3 2 6 60 71 S 1 955 SS 32 374 7 861 28 999 229 406 298 639 R Sq 23 18 MS 10 7914 3 9304 4 8331 3 8234 F 2 82 1 03 1 26 P 0 046 0 364 0 288 R Sq adj 9 10 The following are significant at the 5 level 3 a Differences between Row means only b Differences between Column means only c Both differences between Column means and Interaction d Interaction only d All are significant at the 5 level e None are significant at the 5 level f Not enough information 14 Explanation Note that only the p value for Rows is below 5 6 If we do a 1 way ANOVA and find the following One way ANOVA C1 C2 C3 C4 Source Factor Error Total Level C1 C2 C3 C4 DF 3 68 71 N 18 18 18 18 SS 32 37 266 27 298 64 Mean 11 916 12 436 12 927 13 736 MS 10 79 3 92 F 2 76 P 0 049 Individual 95 CIs For Mean Based on Pooled StDev StDev 1 095 2 195 1 929 2 434 11 0 12 0 13 0 14 0 Give a 1 Tukey confidence interval or equivalent test for 1 3 and explain whether this shows a significant difference between these two means 3 17 Extra Credit do the same with a Scheffe interval 2 Extra Credit Do the same for an individual confidence interval for the difference and explain why it is more likely to show a significant difference than the other two 2 Solution From the printout n m 68 m 4 s 2 MSW 3 92 n1 18 n 2 18 x 1 11 916 and x 3 12 927 1 1 1 1 1 1 3 92 s 2 0 4356 0 65997 18 18 n1 n 3 n1 n 3 x 1 x 3 11 916 12 979 1 063 First s m n m s a Tukey Confidence Interval 1 3 x 1 x 3 q 2 q 4 68 1 4 59 4 592 q m n m 01 3 2456 So 2 2 2 1 1 n1 n 3 2 1 3 1 063 3 2456 0 65997 1 063 2 142 2 252y0572 12 01 05 Page layout view b Scheff e Confidence Interval 1 4 x 1 x 3 Note m 1 F m 1 n m s 1 1 n n3 1 3 68 3 65 4 10 3 70 4 07 3 68 that F 01 is between F 01 and F 01 So F 01 must be about 4 08 m 1 F m 1 n m 3 F 3 68 3 4 08 3 4986 Our interval is now 1 3 1 063 3 4986 0 65997 1 063 2 309 n m c Individual Confidence Interval 1 3 x 1 x 3 t 2 s 1 1 n1 n3 t n m t 68 005 2 650 Our interval is now 2 1 3 1 063 2 650 0 65997 1 063 1 749 Looking back recall that the four means were significantly different at the 5 level but not the level In this case not even the individual confidence interval shows a significant difference between the means We know that as confidence levels go up confidence intervals have to get wider The individual confidence interval by itself has a confidence level of 99 but since the Tukey and Scheff intervals have a collective confidence level of 99 the individual confidence intervals must have confidence levels above 99 1 7 If we do a 1 way ANOVA and find the following Sandy 12 50 12 51 One way ANOVA Source DF SS Factor 7 30310 Error 101 358 Total 116 108 661 MS F 1 46062 1 60 0 913131 P The degrees of freedom for the F test are 2 a 4 100 b 5 111 c 4 111 d 5 115 e 5 116 f 4 115 …
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