THIRD HOUR EXAM Hour of Class Registered

252y0571 11/28/05 (Page layout view!)

ECO252 QBA2Name KEY

Dec 1 2005MWF 2, MWF3, TR 12:30, TR2

I. (40 points) Do all the following (2points 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[7]

3. (Sandy) Which of the following is not an assumption required for 1-way ANOVA wth 4 columns..

a)*.

b) All of the columns are random samples

c) All of the population have to be Normally distributed.

d)[9]

4. If we are comparing the means of 4 random samples and find the following:

The appropriate test statistic is:

a) with 12 degrees of freedom

b) with 19 degrees of freedom

c) with 5 and 4 degrees of freedom (0.5)

d) * with 3 and 16 degrees of freedom (2)

e) with 4 and 5 degrees of freedom. (0.5)

f) [11]

Solution: We use ANOVA for multiple comparison of means. is only used in comparing medians and proportions. so 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, has 3 and 16 DF.

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 P

Rows 3 8.2963 2.76542 2.95 0.040

Columns 2 3.7183 1.85916 1.98 0.147

Interaction 6 25.8108 4.30180 4.58 0.001

Error 60 56.3071 0.93845

Total 71 94.1325

S = 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 only

b) Differences between Column means only

c) Differences between both Row and Column means

d) *Interaction only

e) All are significant at the 1% level

f) None are significant at the 1% level

g) 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 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 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 and .

First. .

a) Tukey Confidence Interval. So

b) Scheff Confidence Interval. Note that is between and . So must be about 4.08. . Our interval is now

c) Individual Confidence Interval Our interval is now

Looking 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 P

Factor ? 6.76792 0.615264 0.636

Error ? 162.448 0.966951

Total 179169.216

The degrees of freedom for the F test are(2)

a) 10, 168

b) 11, 158

c) 10, 158

d) *11, 168.

e) 9, 178

f) 10, 178

Explanation: 6.76792/11 = 0.615264. 162.448/168 = 0.96695. 11+168 = 179.

8. If we do a 1-way ANOVA and assume that your answer in 7 is correct, pick an appropriate value for with a 10% significance level from the table and explain your results. (2) [21]

Solution: From the F table - is between 1.63 and 1.65; is between 1.60 and 1.62; is between 1.63 and 1.65; is between 1.60 and 1.62; is between 1.66 and 1.68; is between 1.63 and 1.65. 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)

, , , The predicted value of when is:a) 10

b) 11

c) 12

d) *13

e) Answer can’t be obtained with information given. (4)[25]

So the equation is and if ,

10. Assume the following data:

4 2

3 0

7 5

2 1

16 8

Find the following. Show your work! , , (4)[29]

Row

1 4 2 16 4 8

2 3 0 9 0 0

3 7 5 49 25 35

4 2 1 4 1 2

16 8 78 30 45Starred quantities must be positive.

252y0571 11/28/05 (Page layout view!)

* So or

This must be between zero and one.

11. The percentage of the total (squared) variation of the variable around its mean accounted for by the variable is measured by the

a) Coefficient of Correlation

b) Coefficient of Explanation

c) *Coefficient of Determination

d) Standard error of the estimate (2)[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 = 12.5 + 5.10 X

Predictor Coef SE Coef T P

Constant 12.464 2.465 5.06 0.001

X 5.0990 0.6282 8.12 0.000

Analysis of Variance

Source DF SS MS F P

Regression 1 998.38 998.38 65.89 0.000

Residual Error 8 121.22 15.15

Total 9 1119.60

Obs

1 0.00 11.00 12.46

2 5.00 34.00 37.96

3 3.00 34.00 27.76

4 6.00 48.00 43.06

5 3.00 23.00 27.76

6 6.00 40.00 43.06

7 3.00 26.00 27.76

8 1.00 19.00 17.56

9 5.00 39.00 37.96

10 2.00 24.00 22.66

298, 34and154

12. From the computer output above, find the following:

a) (2)

b) (2)

c) A 95% confidence interval for (2)

d) A 95% confidence interval for when (3)[40]

General Comment: Note that the table giving the regression equation and the ANOVA are something that you should understand. The values and appear there. To their right are and . 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 below any significance level that you might use, the constant is highly significant. For the coefficient of X the p-value is below any significance level that you might use, so the coefficient is highly significant too.

a) Solution: The ANOVA table says , and .

b) Solution:This can be copied from the ANOVA table.

c) A 95% interval for . The regression output has

Predictor Coef SE Coef T P

Constant 12.464 2.465 5.06 0.001

X 5.0990 0.6282 8.12 0.000

d) A 95% confidence interval for when Solution: The Confidence Interval is , where . The table above says that if , . .