MGMT2263 final exam practice solutions

 

Question 1

a)      Ho: m1 = m2 = m3 = m4

Ha: not all means are equal

Reject Ho if F > 2.95

Here is the ANOVA table:

Source of Variation

SS

df

MS

F

Between Groups

8179.625

3

2726.5417

10.6383

Within Groups

7176.25

28

256.2946

 

Total

15355.875

31

 

 

Reject Ho. Conclude the average annual expenditures are difference for at least 2 of the categories.

b)     P-value = P(F > 10.6383). The 1% critical value is 4.57. Since P(F > 4.57) = 1%, this means P(F > 10.6383) < 1%. By inference, this means the p-value < 5%. Therefore, reject Ho.

c)      Q = 3.85; D = 3.85sqrt(256.2946/8) = 21.79. Label the groups as follows:

Under 10

10 to 49

50 to 99

100+

group 1

group 2

group 3

group 4

Their means are as follows:

group 1

group 2

group 3

group 4

80.875

94.25

113.5

121.625

The following groups are significantly different:

group 1 from groups 3 and 4

group 2 from group 4

d)     Lower limit = (121.625 – 80.875) – 2.048sqrt(256.2946/8 + 256.2946/8) = 40.75 – 16.39 = 24.36

Upper limit = 40.75 + 16.39 = 57.14

24.36 < m4 - m1 < 57.14

Note that m4 - m1 = 0 does not fall in the confidence interval. From this we would conclude that m1 and  m4 are significantly different. This is consistent with ANOVA since we concluded that at least two means are significantly different.

 

Question 2

a)      Ho: median 1 = median 2 = median 3 = median 4

Ha: not all are equal

Reject Ho if KW > 7.8147

This is the rankings for each group (total at the bottom):

Class 1

Class 2

Class 3

Class 4

16

18

6

1

17

22

25.5

32

7

23

27

2

28

24

10.5

3

14

5

29.5

31

9

19.5

8

25.5

12

19.5

10.5

4

15

21

13

29.5

118

152

130

128

We see that 118 + 152 + 130 + 128 = 528 = 32(33)/2 so the rank sums check out.

KW = [12/(32(33))](1182/8 + 1522/8 + 1302/8 + 1282/8) – 3(33) = 0.875

Do not reject Ho. Conclude there is no significant difference among the pedagogies.

b)     The p-value = P(c2 > 0.875). From the chi-square table, we see the 10% critical value is 6.2514. Thus, the p-value is greater than 10% and under the general rule of thumb we would not reject Ho.

 

Question 3

a)      Sxx = 9(6.397916327)2 = 368.4; Syy = 9(2.442880813)2 = 53.709; Sxy = 1,398.5 – 10(16.4)(7.69) = 137.34

b)     b1 = 137.34/368.4 = 0.3728; b0 = 7.69 – (0.3728)(16.4) = 1.5761

income = 1.5761 + 0.3728(hours)

c)      r = 137.34/(sqrt(368.4)sqrt(53.709)) = 0.9764. This indicates a strong direct relationship between the number of hours cold-calling and income. (Based on what the others teach, a correlation between 75% and 100% is strong, between 50% and 75% is medium and below 50% is weak.)

d)     r2 = 95.33%

e)      Ho: B1 = 0

Ha: B1 Ή 0

Reject Ho if F > 5.32

Here is the ANOVA table:

 

df

SS

MS

F

Regression

1

51.2005

51.2005

163.2885

Error

8

2.5085

0.3136

 

Total

9

53.7090

 

 

Reject Ho. Conclude the model is significant.

f)       Ho: B1 £ 0

Ha: B1 > 0

The degrees of freedom for the critical value is 8.

Reject Ho if t > 1.86.

s(b1) = sqrt(0.3136/368.4) = 0.0292

t = 0.3728/0.0292 = 12.7671 (Note that should be the square root of the F statistic in part e)

Reject Ho and conclude there is a significant positive relationship between hours and income.

g)      Lower limit = 0.3728 – 2.306(0.0292) = 0.3728 – 0.0673 = 0.3055

Upper limit = 0.3728 + 0.0673 = 0.4401

0.3055 < B1 < 0.4401

We would reject Ho since the hypothesis slope of zero does not fall in the confidence interval.

h)      income = 1.5761 + 0.3728(25) = 10.896 which we multiply by 1000 to get $10,896.

i)       Lower limit = 10.896 – 2.306sqrt(0.3136)sqrt(1/10 + (25 – 16.4)2/368.4) = 10.896 – 0.7082 = 10.1878 which we multiply by 1000 to get $10,188

Upper limit = 10.896 + 0.7082 = 11.6042 which we multiply by 1000 to get $11,604.

$10,188 < my < $11,604

 

Question 4

a)      Ho: median 1 = median 2 = median 3

Ha: not all are equal

Reject Ho if FR > 5.9915

Here are the rankings within each block with the rank sums at the bottom:

Judge 1

Judge 2

Judge 3

1

2.5

2.5

3

1.5

1.5

3

1.5

1.5

1.5

1.5

3

2

2

2

1

2.5

2.5

11.5

11.5

13

We see that 11.5 + 11.5 + 13 = 36 = 6(6) so the rank sums are correct.

FR = [12/(6(3)(4))](11.52 + 11.52 + 133) – 3(6)(4) = 0.25

Do not reject Ho. Conclude there is no significant difference among the judges in how they rate the couples.

b)     Ho: median 1 = median 2 = … = median 6

Ha: not all are equal

Reject Ho if FR > 11.0705

Here are the rankings within each block with the rank sums at the bottom:

Pair 1

Pair 2

Pair 3

Pair 4

Pair 5

Pair 6

2

6

3.5

1

5

3.5

3

6

2

1

4.5

4.5

3

6

2

1

4.5

4.5

8

18

7.5

3

14

12.5

We see that 8 + 18 + 7.5 + 3 + 14 + 12.5 = 63 = 3(21) so the rank sums are correct.

FR = [12/(3(6)(7))](82 + 182 + 7.52 + 32 + 142 + 12.52) – 3(3)(7) = 13.71

Reject Ho and conclude that at least 2 pairs are rated differently.

c)      The p-value = P(c2 > 13.71). We see the test statistic is between the 2.5% critical value of 12.8325 and 1% critical value of 15.0863. Thus, the p-value is between 1% and 2.5%.

 

Question 5

a)      Ho: m1 = m2 = m3 = m4

Ha: not all means are equal

Reject Ho if F > 3.29

Here is the ANOVA table:

Source of Variation

SS

df

MS

F

Factor

20.635

3

6.8783

1.1296

Block

8756.255

5

1751.2510

287.6090

Error

91.335

15

6.0890

 

Total

8868.225

23

 

 

F = 1.1296.

Do not reject Ho. Conclude there is no significant difference among the office in their average monthly expenses.

b)     Ho: m1 = m2 = m3 = m4 = m5

Ha: not all means are equal

Reject Ho if F > 3.26

             Here is the ANOVA table:

Source of Variation

SS

df

MS

F

Factor

0.1335

3

0.0445

0.6007

Block

4.4630

4

1.1158

15.0607

Error

0.8890

12

0.0741

 

Total

5.4855

19

 

 

             F = 15.0607

             Reject Ho. Conclude there is a significant difference among the categories in average expenses.

c)      Q = 4.51; D = 4.51sqrt(0.0741/4) = 0.6138

Here are the averages for each group ranked from lowest to highest:

Entertainment

0.5

Office supplies

0.875

Transportation

1.1

Utilities

1.275

Miscellaneous

1.925

The following groups are significantly different:

Entertainment from utilities and miscellaneous

Office supplies from miscellaneous

Transportation from miscellaneous

Utilities from miscellaneous

 

Question 6

a)      Ho: gross income does not depend on number of employees

Ha: gross income does depend on number of employees

Reject Ho if test stat > 16.92

Here is the crosstab of expected values:

182.977

285.662

166.591

97.77

88.118

137.57

80.227

47.084

37.444

58.458

34.091

20.007

26.461

41.31

24.091

14.139

As we can see, none of the categories need collapsing since the lowest expected value of 14.139 is well above 5.

The first term of the test statistic is (286 – 182.977)2/182.977 = 58.006 which, by itself, is sufficiently large to reject the null hypothesis and conclude that gross income depends on the number of employees.

b)     If you do all the grunge work, you will find the test statistic to be 764.844 (if we round to 3 decimals). Then Cramer’s V = sqrt(764.844/(1342(3))) = 43.59% which is the degree to which gross income depends on the number of employees.

 

Question 7

a)      The model is price = -11.088 + 21.5768(sq. ft.) + 33.3956(bedroom) – 30.7437(garage) + 38.951(basement). For the information given, price = -11.088 + 21.5768(15) + 33.3956(3) – 30.7437(1) + 38.951(0) = 382.0071 which we multiply by 1000 to get $382,000 after rounding to the nearest thousand.

b)     Suppose we call this variable x4. The reported p-value is for this test:

Ho: B4 = 0

Ha: B4 Ή 0

Since the p-value is 0.1925, we would naturally conclude that the variable does not significantly contribute to the model.

Suppose instead we want to test this hypothesis:

Ho: B4 £ 0
Ha: B4 > 0

How do we get the p-value? Recall that the p-value for a two-tail test is determined by multiplying the p-value for the appropriate one-tail test by 2. Since the p-value for the two-tail test is 0.1925, we get the p-value for the one-tail test by dividing 0.1925 by 2 to get 0.09625. Since this p-value is between 1% and 10%, the results would be inconclusive under the general rule of thumb.

c)      Since the error degrees of freedom = 7, the t critical value is 2.365. The standard error for this variable is 4.593.

Lower limit = 21.5768 – 2.365(4.593) = 21.5768 – 10.8624 = 10.7144

Upper limit = 21.5768 + 10.8624 = 32.4392

(Note that these limits are slightly off the limits generated by Excel. This will be taken into account when marking the exam.)

d)     The table is:

Variables

Adj. R2

ANOVA p-value

t-test: are all variables significant? (yes or no)

presence of multicollinearity (yes/no)

#1 All

variables

0.8338

0.0016

no

no

#2 square footage,

bedrooms

0.8295

0.0001

yes

no

 #3 square footage

 

0.6912

0.0005

yes

no

best

#1

#2

#2, #3

tie

Choose model #2 since:

§        its adjusted r2 is slightly less than that of model #1

§        it has the lowest ANOVA p-value

§        all its variables are significant

§        it has no presence of multicollinearity