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# WCU ECO 252 - ECO 252 Third Hour Exam

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252y0572 12/01/05 (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 (2) [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(2) [9] 4. If we are comparing the means of 5 random samples and find the following:70.110111nsx 72.112222nsx 73.111333nsx 75.113444nsx 75.114555nsxThe appropriate test statistic is:a) 52542432322212154321nsnsnsnsnsxxxxxD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 freedomf) 2with 34 degrees of freedom (2) [11]Solution: We use ANOVA for multiple comparison of means. 2 is only used in comparing medians and proportions. 35jnnso 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 within, and for an ANOVA, MSWMSBF  has 4 and 30 DF.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 DF SS MS F PRows 3 32.374 10.7914 2.82 0.046Columns 2 7.861 3.9304 1.03 0.364Interaction 6 28.999 4.8331 1.26 0.288Error 60 229.406 3.8234Total 71 298.639S = 1.955 R-Sq = 23.18% R-Sq(adj) = 9.10% The following are significant at the 5% level. (3)a) *Differences between Row means onlyb) Differences between Column means onlyc) Both differences between Column means and Interactiond) Interaction onlyd) All are significant at the 5% levele) None are significant at the 5% levelf) 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 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 31 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,4m,92.32MSWs,181n ,182n916.111.x and 927.123.x.First3111nns31211nns18118192.365997.04356.0 . 063.1979.12916.113.1. xx.a) Tukey Confidence Interval    31,3131112nnsqxxmnm    2456.3259.4259.4221268,401.,qqmnm. So  142.2063.165997.02456.3063.1312252y0572 12/01/05 (Page layout view!)b) Scheffe Confidence Interval     31,13141111nnsFmxxmnm. 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   4986.308.43368,3F. Our interval is now  309.2063.165997.04986.3063.131c) Individual Confidence Interval   313131112nnstxxmn  .650.268005.2ttmn Our interval is now 749.1063.165997.0650.2063.131Looking back, recall that the four means were significantly different at the 5% level but not the 1% 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 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 ? 7.30310 1.46062 1.60Error ? 101.358 0.913131 Total 116 108.661 The degrees of freedom for the F test are (2) [19]a) 4, 100b) 5, 111c) 4, 111d) 5, 115.e) 5, 116f) 4, 115

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