27. a.
The scatter diagram suggests a negative linear relationship between the two variables.
b. Let x = stress tolerance and y = average annual salary ($)
866 660
86.6 66
10 10
ii
xy
xy
nn

= = = = = =
2
c. SSE =
2
ˆ
( ) 51.7949
ii
yy − =
SST =
= 129.18
Thus, SSR = SST – SSE = 129.18 – 51.7949 = 77.3851
MSR = SSR/1 = 77.3851
50
55
60
65
70
75
50 60 70 80 90 100 110
Stress Toleracne
Average Annual Salary ($1000s)
e. The relationship between the average annual salary and stress tolerance is counterintuitive because
one would think that jobs that pay more are most likely going to require more time and will likely
involve a more stressful environment. One possibility is that the limited size of the data set is
28. The sum of squares due to error and the total sum of squares are
22
ˆ
SSE ( ) 1.4379 SST ( ) 3.5800
i i i
y y y y= − = = − =
Thus, SSR = SST – SSE = 3.5800 – 1.4379 = 2.1421
s2 = MSE = SSE / (n – 2) = 1.4379 / 9 = .1598
Using t table (9 degrees of freedom), area in tail is less than .005; p-value is less than .01
Using Excel, the p-value corresponding to t = 3.66 is .000.
Because p-value
, we reject H0:
1
= 0
Because we can reject H0:
1
= 0 we conclude that speed of execution and overall satisfaction are
related.
Next we illustrate the use of the F test.
Because p-value
, we reject H0:
1
= 0
1
Source
of Variation
Sum
of Squares
Degrees
of Freedom
Mean
Square
F
p-value
Regression
2.1421
1
2.1421
13.4
.000
Error
1.4379
9
.1598
Total
3.5800
10
29. SSE =
2
ˆ
()
ii
yy − =
233,333.33 SST =
2
()
i
yy−
= 5,648,333.33
Thus, SSR = SST – SSE = 5,648,333.33 –233,333.33 = 5,415,000
MSE = SSE/(n – 2) = 233,333.33/(6 – 2) = 58,333.33
Source of
Variation
Sum
of Squares
Degrees of
Freedom
Mean
Square
F
p-value
Regression
5,415,000.00
1
5,415,000
92.83
.0006
Error
233,333.33
4
58,333.33
Total
5,648,333.33
5
Using F table (1 degree of freedom numerator and 4 denominator), p-value is less than .01
Using Excel, the p-value corresponding to F = 92.83 is .0006.
30. SSE =
2
ˆ
()
ii
yy − =
1043.03 SST =
2
()
i
yy−
= 10,568
Thus, SSR = SST – SSE = 10,568 – 1043.03 = 9524.97
s2 = MSE = SSE/(n–2) = 1043.03/4 = 260.7575
260.7575 16.1480s==
2
()
i
xx−
= 56.655
MSR = SSR/1 = 45,017,877
MSE = SSE/(n – 2) = 7,102,922.54/8 = 887,865.3
F = MSR / MSE = 45,017,877/887,865.3 = 50.7
2
3 ( ) 10
i
x x x= − =
*
22
*
2
ˆ
1 ( ) 1 (4 3)
2.033 1.11
( ) 5 10
yi
xx
ss
n x x
−−
= + = + =
−
*
*ˆ
/2
ˆy
y t s
10.6 3.182 (1.11) = 10.6 3.53
or 7.07 to 14.13
( ) 5 10
i
−
d.
*/2 pred
ˆ
y t s
10.6 3.182 (2.32) = 10.6 7.38
33. a. s = 8.7560
*
22
*
2
ˆ
1 ( ) 1 (8 11)
8.7560 4.3780
( ) 5 180
yi
xx
ss
n x x
−−
= + = + =
−
**
ˆ0.2 2.6 0.2 2.6(4) 10.6yx= + = + =
or 12.85 to 75.15
34. s = 6.5141
2
10 ( ) 190
i
x x x= − =
*
22
*
2
ˆ
1 ( ) 1 (12 10)
6.5141 3.0627
( ) 5 190
yi
xx
ss
n x x
−−
= + = + =
−
**
ˆ7.6 .9 7.6 .9(12) 18.40yx= + = + =
2
3.2 ( ) 0.74
i
x x x= − =
*
22
*
2
ˆ
1 ( ) 1 (3 3.2)
145.89 68.54
( ) 6 0.74
yi
xx
ss
n x x
−−
= + = + =
−
116 2.306(4.8963) = 116 11.2909
37. The point estimate of
*
ˆ
y
is given by
*
ˆ
y
= 27.5 .3
*
x
= 27.5 .3(25) = 20
We now compute the margin of error.
2
35 ( ) 1000
i
x x x= − =
s2 = SSE/(n – 2) = 16/6 = 2.6667 s = 1.633
*
22
*
2
ˆ
1 ( ) 1 (25 35)
1.633 0.7746
( ) 8 1000
yi
xx
ss
n x x
−−
= + = + =
−
22
*
pred 2
1 ( ) 1 (500 575)
1 241.52 1 267.50
( ) 6 93,750
i
xx
ssn x x
−−
= + + = + + =
−
*/2 pred
ˆ
y t s
5046.67 4.604 (267.50) = 5046.67 1231.57
220.2 14.391s==
22
*
1 ( ) 1 (89 105)
xx
−−
Actual p-value = .000
Because p-value
, we reject H0:
1 = 0; monthly maintenance expense is related to usage.
c. r2 = SSR/SST = 1575.76/1924.90 = 0.82. A good fit.
42. a.
y
= 80.0 + 50.0x
Branch office sales are related to the salespersons.
c. t =
50
5.482
= 9.12
Because p-value
, we reject H0:
1 = 0
d. $680,000
43. a.
b. There appears to be a positive relationship between the two variables. Students that graduate from
the schools with higher tuition and fees tend to receive a higher starting salary and bonus.
50
60
70
80
90
100
110
120
130
140
20 25 30 35 40 45 50
Salary & Bonus ($1000s)
Tuition & Fees ($1000s)
The Excel output is shown below:
Regression Statistics
Multiple R
0.8591
R Square
0.7381
Adjusted R Square
0.7236
Standard Error
7.6087
Observations
20
ANOVA
df
SS
MS
F
Significance F
Regression
1
2937.1252
2937.1252
50.7336
1.22948E-06
Residual
18
1042.0748
57.8930
Total
19
3979.2
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
33.7880
9.3398
3.6176
0.0020
14.1658
53.4102
Tuition & Fees
($1000s)
1.9154
0.2689
7.1228
1.22948E-06
1.3505
2.4804
d. The p-value = .000 <
= .05 (t or F); significant relationship
f.
ˆ
y
= 33.788 + 1.9154(43) = 116.15 or approximately $116,000.
Note to Instructor: The average starting salary and bonus reported by U.S. News & World Report for
the University of Virginia was $121,000.