252y0571 11/28/05 (Page layout view!) ECO252 QBA2 Name KEY THIRD HOUR EXAM Hour of Class Registered Dec 1 2005 MWF 2, MWF3, TR 12:30, TR2I. (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 thatthere 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 ANOVAc) The Friedman testd) Two-Way ANOVA [7]3. (Sandy) Which of the following is not an assumption required for 1-way ANOVA wth 4 columns..a) *4321.b) All of the columns are random samplesc) All of the population have to be Normally distributed.d) 4321[9] 4. If we are comparing the means of 4 random samples and find the following:50.210111nsx 52.212222nsx 53.211333nsx 55.213444nsxThe appropriate test statistic is:a) 2 with 12 degrees of freedomb) 2with 19 degrees of freedomc) F with 5 and 4 degrees of freedom (0.5)d) *F with 3 and 16 degrees of freedom (2)e) F with 4 and 5 degrees of freedom. (0.5)f) 4243232221214321nsnsnsnsxxxxD[11]Solution: We use ANOVA for multiple comparison of means. 2 is only used in comparing medians and proportions. 20jnnso total degrees of freedom are 19. There are 4 columns, so degrees of freedom between are 3. Thus there are 19 – 3 = 16 degrees of freedom within, and for an ANOVA, MSWMSBF has 3 and 16 DF.252y0571 11/28/05 (Page layout view!) 5. If we are doing a 2-way ANOVA and find the following:Two-way ANOVA: C8 versus C9, C10 Source DF SS MS F PRows 3 8.2963 2.76542 2.95 0.040Columns 2 3.7183 1.85916 1.98 0.147Interaction 6 25.8108 4.30180 4.58 0.001Error 60 56.3071 0.93845Total 71 94.1325S = 0.9687 R-Sq = 40.18% R-Sq(adj) = 29.22%The following are significant at the 1% level. (3)a) Differences between Row means onlyb) Differences between Column means onlyc) Differences between both Row and Column meansd) *Interaction onlye) All are significant at the 1% levelf) None are significant at the 1% levelg) Not enough information. [14]Note that Interaction is the only F with a p-value below .01. 6. If we do a 1-way ANOVA and find the following.One-way ANOVA: C1, C2, C3, C4 Source DF SS MS F PFactor 3 32.37 10.79 2.76 0.049Error 68 266.27 3.92Total 71 298.64 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev +---------+---------+---------+---------C1 18 11.916 1.095 (--------*--------)C2 18 12.436 2.195 (--------*---------)C3 18 12.927 1.929 (--------*---------)C4 18 13.736 2.434 (--------*---------) +---------+---------+---------+--------- 11.0 12.0 13.0 14.0Give a 1% Tukey confidence interval (or equivalent test) for 41 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 ,68 mn,4m,92.32MSWs,181n ,182n916.111.x and 736.134.x.First4111nns21211nns18118192.365997.04356.0 . 82.1736.13916.114.1. xx.a) Tukey Confidence Interval 41,4141112nnsqxxmnm 2456.3259.4259.42121268,401.,qqmnm. So 14.282.165997.02456.382.1412252y0571 11/28/05 (Page layout view!)b) Scheffe Confidence Interval 41,14141111nnsFmxxmnm. Note that 68,301.F is between 10.465,301.F and 07.470,301.F. So 68,301.Fmust be about 4.08. mnmFm,11 4986.308.43368,3F. Our interval is now 31.282.165997.04986.382.141c) Individual Confidence Interval 414141112nnstxxmn .650.268005.2ttmn Our interval is now 75.182.165997.0650.282.141Looking back, recall that the four means were significantly different at the 5% level but not the 1% level. In this case only 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 Scheffè intervals have a collective confidence level of 99%, the individual confidence intervals must have confidence levels above 99%. 7. If we do a 1-way ANOVA and find the following: (Sandy 12.50, 12.51)One-way ANOVA: Source DF SS MS F PFactor ? 6.76792 0.615264 0.636Error ? 162.448 0.966951 Total 179 169.216 The degrees of freedom for the F test are (2)a) 10, 168b) 11, 158c) 10, 158d) *11, 168.e) 9, 178f) 10, 178 Explanation: 6.76792/11 = 0.615264. 162.448/168 = 0.96695. 11+168 = 179.
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