THIRD HOUR EXAM Hour of Class Registered

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, 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 x2  12 x3  11 x4  13 x5  14

s1  1.0 s2  1.2 s3  1.3 s4  1.5 s5  1.5

n1  7 n2  7 n3  7 n4  7 n5  7 The appropriate test statistic is: x  x  x  x  x D  1 2 3 4 5 a) s 2 s 2 s 2 s 2 s 2 1  2  3  4  5 n1 n2 n3 n4 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) [11] Solution: We use ANOVA for multiple comparison of means.  2 is only used in comparing n  n  35 medians and proportions.  j so total degrees of freedom are 34. There are 5 252y0572 12/01/05 (Page layout view!)

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 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 P Rows 3 32.374 10.7914 2.82 0.046 Columns 2 7.861 3.9304 1.03 0.364 Interaction 6 28.999 4.8331 1.26 0.288 Error 60 229.406 3.8234 Total 71 298.639

S = 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 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 DF SS MS F P Factor 3 32.37 10.79 2.76 0.049 Error 68 266.27 3.92 Total 71 298.64

Individual 95% CIs For Mean Based on Pooled StDev Level 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 14.0

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)

2 Solution: From the printout n  m  68, m  4, s  MSW  3.92, n1  18, n2  18,

x.1  11.916 and x.3  12.927 .  1 1   1 1   1 1  s   2   First    s     3.92    0.4356  0.65997 .  n1 n3   n1 n3  18 18 

x.1  x.3  11.916 12.979  1.063 .

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m,nm s 1 1 a) Tukey Confidence Interval 1  3  x1  x3  q   2 n1 n3 1 q4,68 4.59 4.592 m,nm .01 . So q      3.2456 2 2 2 2

1  3  1.063 3.24560.65997  1.063  2.142  1 1      x  x  m 1 F m1,nm  s   b) Scheffe Confidence Interval 1 4  1 3       . Note  n1 n3  3,68 3,65 3,70 3,68 that F.01 is between F.01  4.10 and F.01  4.07 . So F.01 must be about 4.08. m1,nm 3,68 m 1F  3F  34.08  3.4986 . Our interval is now

1  3  1.063 3.49860.65997  1.063  2.309

nm 1 1 c) Individual Confidence Interval 1  3  x1  x3  t s  2 n1 n3 nm 68 t  t.005  2.650. Our interval is now     1.063  2.6500.65997  1.063 1.749 2 1 3

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 P Factor ? 7.30310 1.46062 1.60 Error ? 101.358 0.913131 Total 116 108.661 The degrees of freedom for the F test are (2) [19] a) 4, 100 b) 5, 111 c) 4, 111 d) 5, 115. e) 5, 116 f) 4, 115 Explanation: 7.30310/5 = 1.46062. 101.358/111 = 0.913135. 5 + 111 = 116.

8. If we do a 1-way ANOVA and assume that your answer in 7 is correct, pick an appropriate value for F with a 10% significance level from the table and explain your results. (2) [21] 4,100 5,111 4,111 Solution: From the F table - F.10 is 2.00; F.10 is between 1.89 and 1.91; F.10 is 5,115 5,116 between 1.99 and 2.00; F.10 is between 1.89 and 1.91; F.10 is between 1.89 and 1.91; 4,115 F.10 is between 1.99 and 2.00. In any case, the computed value of F is below the table value, so we cannot reject the null hypothesis of equal factor means.

9. If we do a simple regression and find the following: (Sandy 13.1, 13.2) If we do a simple regression and find the following: (Sandy 13.1, 13.2) 2  xy  1200 , x  5 , y  10, n  10 ,  x  500. The predicted value of y when x  4 is: a) 5.2 b) *7.2 c) 8.6 d) 9.6 e) Answer cant’t be obtained with information given. (4) [25]

x ? y ? Explanation: x     5 y     10 n 10 n 10 SSx   x 2  nx 2  500 1052  500  250  250 Sxy   xy  nx y  1200 10510  1200  500  700 Sxy  xy  nxy 700 b1     2.80 SSx  x 2  nx 2 250

b0  y  b1x  10  2.805  4 So the equation is Yˆ  4  2.8X and if x  4 , Yˆ  4  2.84  7.2

10. Assume the following data: x y 7 -2 9 -6 5 -1 8 -9 29 -18 2 Find the following. Show your work!  x ,  xy , R 2 (4) [29] Row x y x 2 y 2 xy 1 7 -2 49 4 -14 2 9 -6 81 36 -54 3 5 -1 25 1 -5 4 8 -9 64 81 -72 29 -18 219 122 -145 Starred quantities must be positive. x 29 y 18 x     7.25 y     4.5 n 4 n 4

SST  SSy   y 2  ny 2  122  44.52  122  81  41 * SSx   x 2  nx 2  219  47.252  219  210.25  8.75 * Sxy   xy  nx y  145  47.254.5  145 130.5  14.5

Sxy  xy  nxy 14.5 b1     1.657 SSx  x 2  nx 2 8.75 SSR 24.0265 SSR  b Sxy  1.657(14.5)  24.0265 * So R 2    .5860 or 1 SST 41 Sxy2 14.52 210.25 R 2     .5861 This must be between zero and one. SSxSSy 8.7541 358.75

11. The coefficient of determination is defined as a) Total (squared) variation in y divided by the explained variation. b) *Explained variation in y divided by the total (squared) variation in y c) Unexplained variation in y divided by the total (squared) variation in y . d) Sum of the explained and unexplained variation in y divided by the total variation in y . [31]

————— 11/28/2005 8:40:25 PM ———————————————————— Welcome to Minitab, press F1 for help.

MTB > Regress c1 1 c2; SUBC> Constant; SUBC> Brief 3.

Regression Analysis: Y versus X The regression equation is Y = 4.53 + 10.2 X

Predictor Coef SE Coef T P Constant 4.531 7.217 0.63 0.548 X 10.198 1.256 8.12 0.000

Analysis of Variance

Source DF SS MS F P Regression 1 3993.5 3993.5 65.89 0.000 Residual Error 8 484.9 60.6 Total 9 4478.4 Obs X Y Yˆ 1 2.00 22.00 24.93 2 7.00 68.00 75.92 3 5.00 68.00 55.52 4 8.00 96.00 86.11 5 5.00 46.00 55.52 6 8.00 80.00 86.11 7 5.00 52.00 55.52 8 3.00 38.00 35.12 9 7.00 78.00 75.92 10 4.00 48.00 45.32 2  y  596 ,  x  54 and  x  330

12. From the computer output above, find the following: a) R 2 (2) b) se (2)

c) A 90% confidence interval for 1 (2) d) A 90% prediction interval for Y when X  5. (3) [40]

General Comment: Note that the table giving the regression equation and the ANOVA are something that you should understand. The values b0  4.531 and b1  10.198 appear there. To their right are sb0  7.217

and sb1  1.256 . These are used in the t ratios that appear next. But you should be able to ignore them and note that because the p-value for the constant is above any significance level that you might use, the constant is not significant. For the coefficient of X the p-value is below any significance level that you might use, so the coefficient is highly significant.

a) R 2 Solution: The ANOVA table says SSR  3993.5 , SST  4478.4 and SSE  SST  SSR  484.9 . SSR 3993.5 R 2    .8917 SST 4478.4 2 ˆ 2 SSE Y Y  SS y  b SS x SST  SSR 484.9 b) se Solution: s 2        60.6125 e n  2 n  2 n  2 n  2 8

se  60.6125  7.7854

c) A 90% interval for 1 . The regression output has

Predictor Coef SE Coef T P Constant 4.531 7.217 0.63 0.548 X 10.198 1.256 8.12 0.000

8 df  n  2  8 t.05  1.860   b 1.860s  10.198 1.860 1.256  10.198  2.336 1 1 b1   d) A 90% prediction interval for Y when X  5. Solution: The Confidence Interval is 2  X  X  ˆ 2 2  1  0   Y  Y  ts , where sY  se  1 . The table above says that if x  5 , Yˆ  55.52 . 0 0 Y  n SS   x 

 x 54 2 2 2 x    5.4 SSx   x  nx  330 105.4  330  291.60  38.4 . n 10 2  1 X  X    1 5  5.42  s 2  s 2   0 1  60.6125  1  60.61251.1 0.00467  66.9263 Y e  n SS  10 38.4   x   

s  66.9263  8.18085 Y  55.52 1.8608.18085  55.52 15.22 Yˆ 0

I won’t comment on the computer assignments except to say that there was a table in the second assignment like that in question 5. There were 3 tests here and you should have said what they were and stated the meaning of the specific p-values. For the regression see the general comment on question 12.