1) A regression model relating x, number of sales persons at a branch office, to y, annual
sales at the office ($1000s), has been developed. The computer output from a regression
analysis of the data follows.
The regression equation is
Y = 80.0 + 50.0X
Predictor
Constant
X
Analysis of Variance
SOURCE
Regression
Error
Total
Coef
80.0
50.0
Stdev
11.333
5.482
t-ratio
7.06
9.12
DF
1
28
29
SS
6828.6
2298.8
9127.4
MS
6828.6
82.1
a.
Write the estimated regression equation
Annual Sales ($1000s) = 80.0 + 50.0(number of sales persons)
b.
How many branch offices were involved in the study?
30
c.
Compute the F statistic and test the significance of the relationship at a .05 level of
significance.
Ho: There is no relationship
Ha: There is a relationship
F = 6828.6/82.1 = 83.17
F critical = 4.196
Since the test statistic is greater than the critical value, we reject the null hypothesis.
We have sufficient evidence to conclude that there is a relationship.
d.
Predict the annual sales at the Memphis branch office. This branch has 12 sales
persons.
Annual Sales ($1000s) = 80.0 + 50.0(12)
= $680,000
2) Find the following:
(a) Compute the sample regression coefficients bo and b1.
b0 = 80.0 and b1 = 50
(b) Compute the estimated variance of the regression.
SSE
2298.8
 1
 0.748
SST
9127.4
(1  r 2 ) SST (1  0.748)9127.4
S e21 

 79.269
nk
30  1
r2  1
(c) Compute the standard error of the regression.
S  S e21  79.27  8.903
(d) Compute the estimated variance of b1.
r 2 (n  k )
0.748(30  1)

 9.278
2
1  0.748
1 r
b12
50 2
2
S b1  2 
 29.043
t
9.278 2
t
(e) Compute the standard error of b1.
S b1  S b21  29.043  5.389
4) Calculate r2 for the following sets of data:
a)   y  y 2  210 and   yˆ  y 2 = 140
r2 = 140/210 = 0.67
b)
 y  y
2
 100 and
 y  ŷ
2
= 10
2
r = 10/100 = 0.10
c)
 yˆ  y 
2
 75 and
 y  ŷ
2
= 25
2
r = 25/75 = 0.33
9)
The following regression equation was obtained using the five independent variables.
The regression equation is
sales = - 19.7 - 0.00063 outlets + 1.74 cars + 0.410 income + 2.04 age
- 0.034 bosses
Predictor
Constant
outlets
cars
income
age
bosses
Coef
-19.672
-0.000629
1.7399
0.40994
2.0357
-0.0344
SE Coef
5.422
0.002638
0.5530
0.04385
0.8779
0.1880
T
-3.63
-0.24
3.15
9.35
2.32
-0.18
P
0.022
0.823
0.035
0.001
0.081
0.864
S = 1.507
R-Sq = 99.4%
R-Sq(adj) = 98.7%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
5
4
9
SS
1593.81
9.08
1602.89
MS
318.76
2.27
F
140.36
P
0.000
(Minitab Software)
a) What percent of the variation is explained by the regression equation?
99.4%
b) What is the standard error of regression?
1.507
c) What is the critical value of the F-statistic?
6.256
d) What sample size is used in the print out?
10
e) What is the variance of the slope coefficient of income?
0.043852 = 0.001923
f) Conduct a global test of hypothesis to determine if any of the regression coefficients are
not zero.
Ho: β1 = β2 = β3 = β4 = β5 = 0
Ha: At least one of the coefficients is different than zero
Since the test statistic (140.36) is greater than the critical value (6.256) we reject the null
hypothesis. We have sufficient evidence to conclude that at least one of the regression
coefficients is different than zero.
g) Conduct a test of hypothesis on each of the independent variables. Would you consider
eliminating outlets and bosses?
Looking at the p-values for each independent variable, outlets, bosses, and age are not
significant at a level of 0.05. So they can be eliminated from the model.
11)Fill in the missing values in following ANOVA table
Source
Factor
Error
Total
df
SS
5
20
25
MS
1027.5
637
1664.5
a. In the above ANOVA table, is the factor significant
at 5% level of significant?
F
205.5
31.85
6.452
Answer
Yes
12) Fill in the missing values in following ANOVA table
Source
Factor
Error
Total
df
SS
2
17
19
MS
6.48
34.5
40.98
F
3.24
2.03
1.597
Answer
No
a. In the above ANOVA table, is the factor significant at 5%
level of significant?
13) Fill in the missing values in following ANOVA table
Source
Factor
Error
Total
df
SS
3
16
19
MS
346.2
88.81
453.0
a. In the above ANOVA table, is the factor significant at 5%
level of significant?
F
115.4
5.55
Answer
Yes
20.79
21) For n = 6 data points, the following quantities have been calculated:
Σxy = 400
Σx = 40
Σy = 76
Σx2 = 346
Σy2 = 1160
(a) Determine the least squares regression line.
 X  40  6.67
X 
n
6
 Y  76  12.67
Y 
n
6
n XY    X  Y 6(400)  (40)(76)
b1 

 1.345
2
6(346)  (40) 2
n  X 2   X 


b0  Y  b1 ( X )  12.67  (1.345)(6.67)  21.63
Yˆ  21.63  1.345 X
(b) Determine the standard error of estimate.
 Y 

2
SST   Y
2
n
 1160 
762
6
 197.33
2

2

X 

2
 (1.345) 2 346  40 
b  X 


n 

6 


2
r 

 0.727
2
2






76


Y
 Y 2  

1160 

6 
n 



2
(1  r ) SST (1  0.727)197.33
S e21 

 10.763
nk
6 1
2
1
S  S e21  10.763  3.281
(c) Construct the 95% confidence interval for the mean of y when x = 7.0.
1
Yˆ  t (n  2,  / 2) S

n
(X p  X )2
( X ) 
2
21.63  1.345(7.0)  2.777(3.281)
12.215  3.735
(8.48,15.95)
( X ) 2
n
1 (7.0  6.67) 2

6
(40) 2
(346) 
6
(d) Construct the 95% confidence interval for the mean of y when x = 9.0.
1
Yˆ  t (n  2,  / 2) S

n
(X p  X )2
( X 2 ) 
21.63  1.345(9.0)  2.777(3.281)
( X ) 2
n
1 (9.0  6.67) 2

6
(40) 2
(346) 
6
9.525  4.418
(5.11,13.94)
(e) Compared the width of the confidence interval obtained in part (c) with that obtained
in part (d). Which is wider and why?
Width of the confidence interval for x = 7.0 is 2*3.735 = 7.47
Width of the confidence interval for x = 9.0 is 2*4.418 = 8.84
Second one is larger because x value is more different than the average. As we go away
from the average x value we are less confident of our prediction.