Tutorial Sheet 5 Solutions
Question 1
Ho: median 1 = median 2 = median 3
Ha: not all the medians are equal
Reject Ho if FR > 5.9915
This crosstab shows the rankings within each block as well as the ranks sums:
|
|
Brand
A |
Brand
B |
Brand
C |
|
Person
#1 |
1 |
2.5 |
2.5 |
|
Person
#2 |
3 |
2 |
1 |
|
Person
#3 |
3 |
2 |
1 |
|
Person
#4 |
2 |
3 |
1 |
|
Person
#5 |
3 |
2 |
1 |
|
Person
#6 |
3 |
2 |
1 |
|
Person
#7 |
3 |
1 |
2 |
|
Person
#8 |
3 |
1.5 |
1.5 |
|
Person
#9 |
3 |
2 |
1 |
|
Person
#10 |
2.5 |
2.5 |
1 |
|
Total |
26.5 |
20.5 |
13 |
FR = [12/(10(3)(4))](26.52 + 20.52 + 132) – 3(10)(4) = 9.15
Reject Ho. Conclude that at least 2 brands are rated differently.
Question 2
We are looking at 3 pairs of Wilcoxon tests. This is my strategy: it appears there is no significant difference between Brands A and B. However, there does appear to be a significant difference between Brands B and C. If there is, I can then conclude a significant difference between Brands A and C. All tests will be conducted as 2-tail tests.
For the first test, I let d = Brand A – Brand B. We have n=9 since there is 1 tie. Reject Ho if T £ 6.
|
|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
|d| |
1 |
1 |
1 |
2 |
2 |
3 |
3 |
4 |
5 |
|
sign |
+ |
+ |
- |
+ |
- |
+ |
+ |
+ |
+ |
|
rank |
2 |
2 |
2 |
4.5 |
4.5 |
6.5 |
6.5 |
8 |
9 |
T+ = 2 + 2 + 4.5 + 6.5 + 6.5 + 8 + 9 = 38.5
T- = 2 + 4.5 = 6.5
Then T = 6.5.
Do not reject Ho. Conclude there is no significant difference in how Brands A and B are rated.
For the second test, I let d = Brand B – Brand C. We have n=9 since there are 2 ties. Reject Ho if T £ 4.
|
|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
|d| |
1 |
1 |
1 |
2 |
3 |
3 |
5 |
8 |
|
sign |
+ |
+ |
+ |
- |
+ |
+ |
+ |
+ |
|
rank |
2 |
2 |
2 |
4 |
5.5 |
5.5 |
7 |
8 |
T+ = 2 + 2 + 2 + 5.5 + 5.5 + 7 + 8 = 32
T- = 4
Then T = 4.
Reject Ho. Conclude there is a significant difference in how Brands B and C are rated.
By the way, for those who are interested in how Brands A and C compare, there were no ties so n=10. We would reject Ho if T £ 8. In this case, T- = 2 (which turns out to be the value of T as well). Clearly, we reject Ho and conclude there is a significant difference in how Brands A and C are rated.
Question 3
Ho: median 1 = median 2 = median 3
Ha: not all the medians are equal
Reject Ho if KW > 5.9915
This crosstab shows the rankings from lowest to highest and the respective rank sums:
|
Firm
A |
Firm
B |
Firm
C |
|
1 |
5 |
16.5 |
|
2.5 |
7 |
16.5 |
|
2.5 |
9 |
23 |
|
5 |
13.5 |
23 |
|
5 |
16.5 |
25 |
|
8 |
16.5 |
26 |
|
11 |
19.5 |
27 |
|
11 |
19.5 |
28 |
|
11 |
21 |
29 |
|
13.5 |
23 |
30 |
|
total = 70.5 |
total = 150.5 |
total = 244 |
KW = [12/(30)(31)](70.52/10 + 150.52/10 + 2442/10) – 3(31) = 19.46
Reject Ho. Conclude the completion rates of at least 2 firms are significantly different.
Question 4
We are looking at 3 pairs of Mann-Whitney tests. It appears that all 3 firms are significantly different based on the rank sums for each firm. I will first compare firms B and A and then Firms C and B. If these are significantly different, I can then conclude that Firms C and A are significantly different. All tests will be conducted as 2-tail tests.
Comparing Firms B and A, let Firm A be group 1:
|
|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
|
Firm
A |
Firm
A |
Firm
A |
Firm
A |
Firm
A |
Firm
B |
Firm
B |
Firm
A |
Firm
B |
Firm
A |
|
Value |
2 |
2.2 |
2.2 |
2.6 |
2.6 |
2.6 |
2.7 |
2.9 |
3.2 |
3.3 |
|
Rank |
1 |
2.5 |
2.5 |
5 |
5 |
5 |
7 |
8 |
9 |
11 |
|
|
11 |
12 |
13 |
14 |
15 |
16 |
17 |
18 |
19 |
20 |
|
|
Firm
A |
Firm
A |
Firm
A |
Firm
B |
Firm
B |
Firm
B |
Firm
B |
Firm
B |
Firm
B |
Firm
B |
|
Value |
3.3 |
3.3 |
3.4 |
3.4 |
3.6 |
3.6 |
3.7 |
3.7 |
3.8 |
3.9 |
|
Rank |
11 |
11 |
13.5 |
13.5 |
15.5 |
15.5 |
17.5 |
17.5 |
19 |
20 |
T1 = 1 + 2.5 + 2.5 + 5 + 5 + 8 + 11 + 11 + 11 + 13.5 = 70.5
T2 = 5 + 7 + 9 + 13.5 + 15.5 + 15.5 + 17.5 + 17.5 + 19 + 20 = 139.5
U1 = 10(10) + 10(11)/2 – 70.5 = 84.5
U2 = 10(10) + 10(11)/2 – 139.5 = 15.5
Then U = 15.5 which we round up to 16.
P-value/2 = 0.0045 from which we derive p-value = 0.009 in which we reject Ho. We conclude there is a significant difference in the completion rates of Firms A and B.
Comparing Firms C and B, let Firm B be group 1:
|
|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
|
Firm
B |
Firm
B |
Firm
B |
Firm
B |
Firm
B |
Firm
B |
Firm
C |
Firm
C |
Firm
B |
Firm
B |
|
Value |
2.6 |
2.7 |
3.2 |
3.4 |
3.6 |
3.6 |
3.6 |
3.6 |
3.7 |
3.7 |
|
Rank |
1 |
2 |
3 |
4 |
6.5 |
6.5 |
6.5 |
6.5 |
9.5 |
9.5 |
|
|
11 |
12 |
13 |
14 |
15 |
16 |
17 |
18 |
19 |
20 |
|
|
Firm
B |
Firm
B |
Firm
C |
Firm
C |
Firm
C |
Firm
C |
Firm
C |
Firm
C |
Firm
C |
Firm
C |
|
Value |
3.8 |
3.9 |
3.9 |
3.9 |
4 |
4.2 |
4.3 |
4.6 |
4.8 |
5 |
|
Rank |
11 |
13 |
13 |
13 |
15 |
16 |
17 |
18 |
19 |
20 |
T1 = 1 + 2 + 3 + 4 + 6.5 + 6.5 + 9.5 + 9.5 + 11 + 13 = 66
T2 = 6.5 + 6.5 + 13 + 13 + 15 + 16 + 17 + 18 + 19 + 20 = 144
U1 = 10(10) + 10(11)/2 – 66 = 89
U2 = 10(10) + 10(11)/2 – 144 = 11
P-value/2 = 0.001 from which we derive p-value = 0.002 in which we reject Ho. We conclude there is a significant difference in the completion rtes of Firms B and C.
If you really want to compare Firms A and C, be my guest but it is no contest: Firm A forms the lowest 10 values, Firm C the top 10. We have T1 = 10(11)/2 = 55 from which we get T2 = 210 – 55 = 155. Then U2 = 10(10) + 10(11)/2 – 155 = 0. The p-value/2 = 0 so then the p-value = 0. Like I said: no contest.
Since Firm C has the highest rank sum, go with Firm C.
Question 5
You should be able to plug these values into your calculator to get the model: vacation = 0.1878 + 0.0434*income
Question 6:
Ditto question 5. From your calculator, you get r = 0.9672 which we square to get 0.9355 = 93.55%.
Question 7:
Ho: B1 = 0
Ha: B1 ¹ 0
Reject Ho if F > 5.99
Here is the ANOVA table:
|
|
df |
SS |
MS |
F |
|
Regression |
1 |
2.8147 |
2.8147 |
86.9988 |
|
Residual |
6 |
0.1941 |
0.0324 |
|
|
Total |
7 |
3.0088 |
|
|
F = 86.9988; Reject Ho and conclude the model is significant.
Question 8
From the ANOVA table the estimate of the common variance of the residuals is the MSE of 0.0324. Thus, the estimate of the common standard deviation of the residuals is the square root of the MSE which is 0.1799.
Question 9
The t value is 2.447 since we are working with 6 degrees of freedom. The MSE = 0.0324 and Sxx = 7(14.6109)2 = 1493.34075.
Lower limit = 0.0434 – 2.447sqrt(0.0324/1493.34075) = 0.0434 – 0.0114 = 0.032
Upper limit = 0.0434 + 0.0114 = 0.0548
0.032 < B1 < 0.0548
If we were to use this confidence interval to test the hypothesis in question 7, we would reject Ho at a 5% level of significance since B1 = 0 does not fall inside the 95% confidence interval.
Question 10
Vacation = 0.1878 + 0.0434(75) = 3.4428 which we multiply by 1000 to get 3442.80 which rounds to 3400.
Question 11
Lower limit = 3.4428 – 2.447sqrt(0.0324)sqrt(1/8 + (75 – 56.2725)2/1493.34075) = 3.4428 – 0.2642 = 3.1786
Upper limit = 3.4428 + 0.2642 = 3.707
We multiply the limits by 1000 to get 3,178.60 and 3,707. These round to 3200 and 3700.
Question 12
Lower limit = 3.4428 – 2.447sqrt(0.0324)sqrt(1 + 1/8 + (75 – 56.2725)2/1493.34075) = 3.4428 – 0.5136 = 2.9292
Upper limit = 3.4428 + 0.5136 = 3.9564
We multiply the limits by 1000 to get 2,929.20 and 3,956.40. These round to 2900 and 4000.