TABLE 11-0TABLE 13-6TABLE 13-8252x0333 11/18/03 ECO252 QBA2 Name (Page layout view!) THIRD HOUR EXAM Hour of Class Registered (Circle) Nov 25 2003I. (30+ points) Do all the following (2points each unless noted otherwise).TABLE 11-0Shiffler and Adams present the partially complete ANOVA table below that resulted from the analysis of a problem with 3 rows and 3columns. ANOVASource of Variation SS df MS FF Columns 18 Rows 40InteractionWithin (Error) 208Total 296 62 1. Complete the table. Assume a 5% significance level. You may not be able to get exactly the degrees of freedom you are looking for, but you should be able to come close. (4)2. Is there significant interaction? Explain your answer. TABLE 13-6The following Minitab table (with many parts deleted) was obtained when "Score received on an exam (measured in percentage points)" (Y) is regressed on "percentage attendance" (X) for 22 students in a Statistics for Business and Economics course. Regression Analysis: Orders versus WeightThe regression equation isScore = …… + ….. AttendancePredictor Coef SE Coef T PConstant 39.3927 37.2435 1.0576 0.3028Attendance 0.34058 0.52852 0.6444 0.5266S = 20.2598 R-Sq = 2.034% R-Sq(adj) = -2.864%Analysis of VarianceSource DF SS MS F PRegression 1 0.523Residual Error 20 Total 21 3. Referring to Table 13-6, which of the following statements is true?a) -2.86% of the total variability in score received can be explained by percentage attendance. b) -2.86% of the total variability in percentage attendance can be explained by score received.c) 2% of the total variability in score received can be explained by percentage attendance. d) 2% of the total variability in percentage attendance can be explained by score received. 252x0333 11/18/034. Referring to Table 13-6, which of the following statements is true?a) If attendance increases by 0.341%, the estimated average score received will increase by 1 percentage point.b) If attendance increases by 1%, the estimated average score received will increase by 39.39 percentage points.c) If attendance increases by 1%, the estimated average score received will increase by 0.341 percentage points. d) If the score received increases by 39.39%, the estimated average attendance will go up by 1%.5. (Text CD problem 12.51)The manager of a commercial mortgage department has collected data over 104 weeks concerning the number of mortgages approved. The data is the x and O columns below (x is the number of mortgages approved and O is the number of weeks that happened, for example there were 32 weeks in which 2 mortgages were approved) and the problem asks if it follows a Poisson distribution.Row x O E 1 0 13 12.7355 2 1 25 26.7445 3 2 32 28.0817 4 3 17 19.6572 5 4 9 10.3200 6 5 6 4.3344 7 6 1 1.5170 8 7 1 0.4551 9 8 0 0.1195 10 9 0 0.0279 11 10 0 0.0059 12 11 0 0.0011 13 12 0 0.0002 104 104.000Since we have no guide as to what the parameter of the distribution is, the x and O columns were multiplied together to tell us that there were 219 mortgages approved over 104 weeks to give us an average of 2.1 mortgages per week. The E above is the computer – generated Poisson distribution multiplied by 104 .In a Kolmogorov – Smirnov procedure we make the O and E into cumulative distributions and compare them as is done below.Row oF eF D 1 0.12500 0.12246 0.0025435 2 0.36538 0.37962 0.0142304 3 0.67308 0.64963 0.0234453 4 0.83654 0.83864 0.0021047 5 0.92308 0.93787 0.0147973 6 0.98077 0.97955 0.0012180 7 0.99038 0.99414 0.0037536 8 1.00000 0.99851 0.0014857 9 1.00000 0.99966 0.0003369 10 1.00000 0.99993 0.0000689 11 1.00000 0.99999 0.0000126 12 1.00000 1.00000 0.0000019 13 1.00000 1.00000 0.0000000Assume this is correct and explain how you would finish this analysis and why you would or would not reject the null hypothesis. (4)2252x0333 11/18/036. Referring to the previous problem, a more direct method of comparing the observed and expecteddata is below. Answer the following questions.a) What method is being used? (1)b) How many degrees of freedom do we have? (1)c) Why are the columns shorter here than in Problem 5? (1)d) Do we reject our null hypothesis? Why? (3) Row O E EO2 1 13 12.7355 13.2700 2 25 26.7445 23.3693 3 32 28.0817 36.4650 4 17 19.6572 14.7020 5 9 10.3200 7.8488 6 6 4.3344 8.3056 7 2 2.1267 1.8808 104 104.0000 105.8415 7. In problems 5 and 6, one of the methods was used improperly. Which one? Why? 8. Random samples of salaries (in thousands) for lawyers in 3 cities are presented by Dummeldinger. They are repeated in the three left columns. Atlanta DC LA rank-At rank-DC rank-LA 1 45.5 41.5 52.0 12.0 7.0 17.5 2 47.9 40.1 72.0 13.0 5.0 21.0 3 43.1 39.0 41.0 11.0 3.5 6.0 4 42.0 56.5 54.0 8.5 20.0 19.0 5 49.0 37.0 33.0 14.5 2.0 1.0 6 52.0 49.0 42.0 17.5 14.5 8.5 7 39.0 43.0 50.0 3.5 10.0 16.0 80.0 62.0 89.0 You are asked to analyze them, which you do using a Kruskal – Wallis procedure. You are aware that the tables you have are only appropriate for columns with 5 or fewer items in them, so you drop the last two items in each column and after ranking the items from 1 to 15 get a Kruskal – Wallis H of 1.82. If you use the tables, What did you test and what is the conclusion? (3) 9. You remember how to work with column sizes that are too large for the table. You rank the data as appears in the three right columns above. Compute the Kruskal – Wallis H and use it to test your null hypothesis at the 5% significance level.(3) 10. The Kruskal – Wallis test above was done on the assumption that the underlying data did not follow the Normal distribution. Let’s assume that you found out that the underlying distributions were Normal and had a common variance. The method to use would be. a) Friedman Testb) Chi – squared test.c) One way ANOVAd) Two – way ANOVA3252x0333 11/18/03TABLE 13-8The regression
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