Solutions to Practice Problems
Problem 10.1:
(a) Cholesterol levels
(b) There is only one qualitative factor, socioeconomic
class.
(c) Since there is only one factor, each level in this factor is
a treatment. There are four treatmentspoverty, low
income,
Solutions to Practice Problems
Problem 12.7:
(a) The residual plot looks fine.
(b) All the residuals for humidity = 0.2 are below the 0 line
and all the residuals for humidity = 0.3 are above the 0
line. The model equal variance assumption of the error
te
Solutions to Practice Problems
Problem 12.9:
Quantitative: Average height of trees in transect, Total number of trees in
transect, Height of hedgerow in transect, Width of hedgerow, Width of
hedgerow, Width of transect verge, Depth of transect ditch, Widt
Solutions to Practice Problems
Problem 12.11:
(a) E ( y ) = 23.81 + 31.30 x1 + 3.84 x 2 0.275 x1 x 2
(b)
Hypothesis:
H 0 : 1 = 2 = 3 = 0
H a : At least one i 0 for i = 1,2,3
Test Statistic:
Fc = 1695.286
Rejection Region: F > F3,11,0.05 = 3.59
Thus, we re
Solution for Practice Problems:
Problem 14.3
1.
Let F ( x ) be the distribution for population A and G ( y ) the
distribution for population B. Then the hypotheses should be
stated as the following
H0 : F ( x ) = G ( y )
Ha : F ( x ) > G ( y )
2.
Sample A
Solutions to Practice Problems
Problem 12.3:
(a)
Hypothesis: H0: 1 = 2 = 3 = 0
Ha: At least one i 0, i = 1, 2, 3
Test Statistic:
Fc =
R2
(1 R )
2
k
=
[ n ( k + 1) ]
0.85 / 3
= 30.22
(1 0.85) /[ 20 (3 + 1)]
Rejection Region: Fc > F0.05, 3, 16 = 3.24
Conclu
Solution to Practice Problem
Problem 11.10:
(a)
1894
SS
1 = xy =
= 0.766
SS xx 2474
= y x = 76 (0.766)(46) = 40.764
0
1
y = 0 + 1 x = 40.764 + 0.766 x
(b) Sure, it looks great.
(c)
SSE = SS yy 1SS xy = 2056 (0.766)(1894) = 605.196
s2 = MSE = SSE (n 2) =
Solutions to Practice Problems
Problem 10.3:
(a)
General Linear Models Procedure
Dependent Variable: RESP
Source
Model
Error
Corrected Total
DF
2
10
12
Sum of
Squares
12.420308
18.332000
30.752308
Mean
Square
6.210154
1.833200
F Value
3.39
Pr > F
.07528
(
Solutions to Practice Problems
Problem 10.6:
(a)
Source
Treat
Block
Error
Total
df
3
5
15
23
ANOVA
SS
28.2
69.0
34.1
131.3
MS
9.4
13.8
2.27
F
4.14
6.08
p-value
0.067
<0.01
(b)
H0: 1 =2 = 3 = 4
Ha: At least two means differ.
Ftreat = 4.14
p-value = 0.067 >
Solutions to Practice Problems
Problem 10.8:
(a)
General Linear Models Procedure
Dependent Variable: RESP
Source
GLASS
TEMP
GLASS*TEMP
Error
Corrected Total
DF
2
2
4
18
26
Sum of
Squares
151966.7
1955410.1
289005.0
5926.0
2402307.9
Mean
Square
75983.4
977
Solutions to Practice Problems
Problem 11.1:
(a) There is a positive linear relationship between x and y.
SS
(b) 1 = xy = 0.918
SS xx
0 = y 1 x = 0.020
Problem 11.2:
(a)
93
SS
(b) 1 = xy = = 10.33
SS xx
9
0 = y 1 x = 64.43 (10.33)(1.5) = 48.93
1
Solutions to Practice Problems
Problem 11.3:
e xj = 140 c h= 43.429
26
2
SSxx = x
2
SSxy = xy
n
x
2
7
e je yj = 129 c h h 39.857
26 c
24
=
n
7
e yj = 124 c h= 41.714
24
2
SS yy = y
2
n
2
7
39.857
SS
1 = xy =
= 0.918
SS xx 43.429
SSE = SS SS = 41.714 (0.
Solutions to Practice Problems
Problem 11.5:
(a) 1 = SSxy SSxx = 0.1055
0 = y 1 x = 67.877
SSE = SS SS = 73.77
yy
1
xy
s2 = SSE (n 2) = 12.29
s = s2 = 3.506
s = s SS xx = 0.022
1
(b)
It might not have a straight-line relationship. However, the city size
a
Solution to Practice Problem
Problem 11.7:
r = SS xy
SS xx SS yy = 2670.04
r 2 = (SS yy SSE) SS yy =
(25300)(355.50) = 0.89
355.50 73.77
= 0.792
355.50
City Size and Expenditure are highly positive correlated.
Solutions to Practice Problems
Problem 11.8:
(a) y = 0 + 1 xp = 17.388852 + (0.657752)(50) = 50.276
t0.005, 8 = 3.355
2
1 ( xp x )
1 (50 51.1) 2
sy = s
+
= 9.9398
+
= 3.147
n
SS xx
10
5638.9
99% C. I. = y t0.005, 8 sy = 50.276 (3.355)(3.147) = [39.72, 60.
Solution for Practice Problems:
Problem 14.5
1. T+ 43
2. T 152
3. T+ 371
Problem 14.6
H0 : The distributions for the schools, A and B are identical
Ha : They are different
Twin pair
1
2
3
4
5
6
A
65
72
86
50
60
81
B
69
72
74
52
47
72
D
-4
0
12
-2
13
9
|D|