MGMT2263
Worksheet #5
Solutions
Question 1
Ho: median 1 = median 2 = median 3
Ha: not all are equal
The degrees of freedom is 2. Reject Ho if KW > 5.9915
Here are the ranks for each class with their respective totals:
T1 = 1 + 5 + 10 + 10 + 15 + 16.5 + 20 = 77.5
T2 = 2 + 3 + 7.5 + 13.5 + 18 + 19 + 21 = 84
T3 = 4 + 6 + 7.5 + 10 + 12 + 13.5 + 16.5 = 69.5
As a check, T1 + T2 + T3 = 231 = 21(22)/2.
KW = (12/(21)(22))(77.52/7 + 842/7 + 69.52/7) – 3(22) = 0.3915
Do not reject Ho.
Conclude there is no significant difference in the grades of the students from the 3 classes.
Question 2
Ho: median 1 = median 2 = median 3 = median 4 = median 5 = median 6
Ha: not all are equal
The degrees of freedom is 5. Reject Ho if FR > 11.0705
This crosstab shows the ranks within each block as well as the rank sums:
|
|
A |
B |
C |
D |
E |
F |
|
Person #1 |
6 |
3 |
1 |
3 |
3 |
5 |
|
Person #2 |
4 |
4 |
4 |
1.5 |
6 |
1.5 |
|
Person #3 |
3 |
5.5 |
3 |
1 |
5.5 |
3 |
|
Person #4 |
3 |
4.5 |
4.5 |
2 |
6 |
1 |
|
Person #5 |
5.5 |
3.5 |
3.5 |
2 |
5.5 |
1 |
|
Person #6 |
6 |
5 |
1 |
3 |
4 |
2 |
|
Person #7 |
3 |
4 |
5.5 |
5.5 |
1 |
2 |
|
Person #8 |
4.5 |
1.5 |
1.5 |
4.5 |
3 |
6 |
|
Person #9 |
3.5 |
3.5 |
3.5 |
3.5 |
3.5 |
3.5 |
|
Person
#10 |
3 |
3 |
1 |
5 |
6 |
3 |
|
Total |
41.5 |
37.5 |
28.5 |
31 |
43.5 |
28 |
As a check the rank sums total to 210. The rank sum for each block is 21 multiplied by 10 blocks to get 210.
FR = (12/[(10)(6)(7)](41.52 + 37.52 + 28.52 + 312 + 43.52 + 282) – 3(10)(7) = 6.5143
Do not reject Ho.
Conclude there is no significant difference in how the exhibits are rated.
Question 3
Sxy = 26,370 – 10(762)(2.85) = 4,653
Sxx = 9(144,206.6667) = 1,297,860
b1 = 4,653/1,297,860 = 0.0036 after rounding to 4 decimals.
b0 = 2.85 – 0.0036(762) = 0.1181 after rounding to 4 decimals.
The model is: time = 0.1181 + (0.0036)(distance)
Question 4
We already have Sxy and Sxx from question 1.
Syy = 9(2.058333) = 18.525
r2 = 4,6532/[(1,297,860)(18.525)] = 0.9005 = 90.05%
Question 5
Ho: B1 = 0
Ha: B1 ¹ 0
The degrees of freedom are 1 and 8. Reject Ho if F > 5.32
Here is the ANOVA table:
|
|
df |
SS |
MS |
F |
|
Regression |
1 |
16.6816 |
16.6816 |
72.3959 |
|
Residual |
8 |
1.8434 |
0.2304 |
|
|
Total |
9 |
18.5250 |
|
|
F = 72.3959
Reject Ho. Conclude the model is significant.
Question 6
Sb1 = sqrt(0.2304/1,297,860) = 0.000421334
The t value is 2.306 since we have 8 degrees of freedom.
Lower limit = 0.0036 – 2.306(0.000421334) = 0.0036 – 0.00097 = 0.0026 after rounding to 4 decimals
Upper limit = 0.0036 + 0.00097 = 0.0046
0.0026 < B1 < 0.0046. Since the hypothesized coefficient of zero is not in the 95% confidence interval for B1, we would reject Ho at a 5% level of significance.
Question 7
From the ANOVA table, the estimate of the common variance of the residuals is 0.2304.
Question 8
Ho: B1 £ 0
Ha: B1 > 0
The degrees of freedom = 8. Reject Ho if t > 1.86
t = 0.003585/0.000421334 = 8.51 (which also equals the square root of 72.3959 from the ANOVA table)
Reject Ho
Conclude that there is a significant direct relationship between the variables.
Question 9
time = 0.1181 + (0.0036)(500) = 1.91 after rounding to 2 decimals. It may be wise to use the values for b0 and b1 from your calculator to derive this answer.
Question 10
Lower limit = 1.91 – 2.306*sqrt(0.2304)*sqrt(1/10 + (500-762)2/1,297,860) = 1.91 – 0.4328 = 1.48 after rounding to 2 decimals.
Upper limit = 1.91 + 0.4328 = 2.34
1.48 < my < 2.34 for a distance of 500 miles
Question 11
Lower limit = 1.91 - 2.306*sqrt(0.2304)*sqrt(1 + 1/10 + (500-762)2/1,297,860) = 1.91 – 1.1885 = 0.72 after rounding to 2 decimals
Upper limit = 1.91 + 1.1885 = 3.1
0.72 < y < 3.1 for a distance of 500 miles
Question 12
Sxy = 3,115.036 – 10(1.76)(175.92) = 18.844
Sxx = 9(0.0278222) = 0.2504
b1 = 18.844/0.2504 = 75.2556 after rounding to 4 decimals.
b0 = 175.92 – 75.2556(1.76) = 43.4702 after rounding to 4 decimals.
The model is: price = 43.4702 + 75.2556(size)
Question 13
We already have Sxy and Sxx from question 8.
Syy = 9(178.004) = 1,602.036
r2 = 18.8842/[(0.2504)(1,602.036)] = 0.8852 = 88.52%
Question 14
Ho: B1 = 0
Ha: B1 ¹ 0
The degrees of freedom are 1 and 8. Reject Ho if F > 5.32
Here is the ANOVA table:
|
|
df |
SS |
MS |
F |
|
Regression |
1 |
1418.1164 |
1418.1164 |
61.6842 |
|
Residual |
8 |
183.9196 |
22.9900 |
|
|
Total |
9 |
1602.0360 |
|
|
F = 61.6842
Reject Ho. Conclude the model is significant.
Question 15
Sb1 = sqrt(22.99/0.2504) = 9.5819
The t value is 2.306 since we have 8 degrees of freedom.
Lower limit = 75.2556 – 2.306(9.5819) = 75.2556 – 22.0959 = 53.1597 after rounding to 4 decimals
Upper limit = 75.2556 + 22.0959 = 97.3515
53.1597 < B1 < 97.3515. Since the hypothesized coefficient of zero is not in the 95% confidence interval for B1, we would reject Ho at a 5% level of significance.
Question 16
price = 43.4702 + 75.2556(1.5) = 156.354. We multiply this by 1000 to get 156,354 which rounds to 156,400.
Question 17
Lower limit = 156.354 – 2.306*sqrt(22.99)*sqrt(1/10 + (1.5-1.76)2/0.2504) = 156.354 – 6.725 = 149.629
Upper limit = 156.354 + 6.725 = 163.079
We multiply these limits by 1000 to get 149,629 and 163,079. We round these to 149,600 and 163,100 respectively.
149,600 < my < 163,100 for an area of 1500 square feet
Question 18
Lower limit = 156.354 – 2.306*sqrt(22.99)*sqrt(1 + 1/10 + (1.5-1.76)2/0.2504) = 156.354 – 12.942 = 143.412
Upper limit = 156.354 + 12.942 = 169.296
We multiply these limits by 1000 to get 143,412 and 169,296. We round these to 143,400 and 169,300 respectively.
143,400 < y < 169,300 for an area of 1500 square feet