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# Solucionario Walpole 8 ED

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```Ip is the p × p identity matrix. Here we use the property of tr(AB) = tr(BA)
in linear algebra.
Solutions for Exercises in Chapter 12 181
12.62 (a) y\u2c6 = 9.9375 + 0.6125x1 + 1.3125x2 + 1.4625x3.
(b) The ANOVA table for all these single-degree-of-freedom components can be dis-
played as:
Source of Degrees of Mean Computed
Variation Freedom Square f P -value
x1
x2
x3
Error
1
1
1
4
3.00125
13.78125
17.11125
3.15625
0.95
4.37
5.42
0.3847
0.1049
0.0804
Total 7
Only x3 is near significant.
12.63 (a) For the completed second-order model, we have
PRESS = 9, 657, 641.55,
n\u2211
i=1
|yi \u2212 y\u2c6i,\u2212i| = 5, 211.37.
(b) When the model does not include any term involving x4,
PRESS = 6, 954.49,
n\u2211
i=1
|yi \u2212 y\u2c6i,\u2212i| = 277.292.
Apparently, the model without x4 is much better.
(c) For the model with x4:
PRESS = 312, 963.71,
n\u2211
i=1
|yi \u2212 y\u2c6i,\u2212i| = 762.57.
For the model without x4:
PRESS = 3, 879.89,
n\u2211
i=1
|yi \u2212 y\u2c6i,\u2212i| = 220.12
Once again, the model without x4 performs better in terms of PRESS and
n\u2211
i=1
|yi\u2212
y\u2c6i,\u2212i|.
12.64 (a) The stepwise regression results in the following fitted model:
y\u2c6 = 102.20428\u2212 0.21962x1 \u2212 0.0723\u2212 x2 \u2212 2.68252x3 \u2212 0.37340x5 + 0.30491x6.
(b) Using the Cp criterion, the best model is the same as the one in (a).
12.65 (a) Yes. The orthogonality is maintained with the use of interaction terms.
182 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models
(b) No. There are no degrees of freedom available for computing the standard error.
12.66 The fitted model is y\u2c6 = 26.19333+0.04772x1+0.76011x2\u22120.00001334x11\u22120.00687x22+
0.00011333x12. The t-tests for each coefficient show that x12 and x22 may be eliminated.
So, we ran a test for \u3b212 = \u3b222 = 0 which yields P -value = 0.2222. Therefore, both x12
and x22 may be dropped out from the model.
12.67 (a) The fitted model is y\u2c6 = \u22120.26891 + 0.07755x1 + 0.02532x2 \u2212 0.03575x3. The
f -value of the test for H0 : \u3b21 = \u3b22 = \u3b23 = 0 is 35,28 with P -value < 0.0001.
Hence, we reject H0.
(b) The residual plots are shown below and they all display random residuals.
7.5 8.0 8.5 9.0 9.5 10.0
\u2212
2
\u2212
1
0
1
2
x1
St
ud
en
tiz
ed
R
es
id
ua
l
0 1 2 3 4 5
\u2212
2
\u2212
1
0
1
2
x2
St
ud
en
tiz
ed
R
es
id
ua
l
0 2 4 6 8
\u2212
2
\u2212
1
0
1
2
x3
St
ud
en
tiz
ed
R
es
id
ua
l
(c) The following is the summary of these three models.
Model PRESS Cp
x1, x2, x3 0.091748 24.7365
x1, x2, x3, x
2
1, x
2
2, x
2
3 0.08446 12.3997
x1, x2, x3, x
2
1, x
2
2, x
2
3, x12, x13, x23 0.088065 10
It is difficult to decide which of the three models is the best. Model I contains
all the significant variables while models II and III contain insignificant variables.
However, the Cp value and PRESS for model are not so satisfactory. Therefore,
some other models may be explored.
12.68 Denote by Z1 = 1 when Group=1, and Z1 = 0 otherwise;
Denote by Z2 = 1 when Group=2, and Z2 = 0 otherwise;
Denote by Z3 = 1 when Group=3, and Z3 = 0 otherwise;
Solutions for Exercises in Chapter 12 183
(a) The parameter estimates are:
Variable DF Parameter Estimate P -value
Intercept
BMI
z1
z2
1
1
1
1
46.34694
\u22121.79090
\u221223.84705
\u221217.46248
0.0525
0.0515
0.0018
0.0109
Yes, Group I has a mean change in blood pressure that was significantly lower
than the control group. It is about 23.85 points lower.
(b) The parameter estimates are:
Variable DF Parameter Estimate P -value
Intercept
BMI
z1
z3
1
1
1
1
28.88446
\u22121.79090
\u22126.38457
17.46248
0.1732
0.0515
0.2660
0.0109
Although Group I has a mean change in blood pressure that was 6.38 points lower
than that of Group II, the difference is not very significant due to a high P -value.
12.69 (b) All possible regressions should be run. R2 = 0.9908 and there is only one signifi-
cant variable.
(c) The model including x2, x3 and x5 is the best in terms of Cp, PRESS and has all
variables significant.
12.70 Using the formula of R2adj on page 467, we have
SSE/(n\u2212 k \u2212 1)
SST/(n\u2212 1) = 1\u2212
MSE
MST
.
Since MST is fixed, maximizing R2adj is thus equivalent to minimizing MSE.
12.71 (a) The fitted model is p\u2c6 = 1
1+e2.7620\u22120.0308x
.
(b) The \u3c72-values for testing b0 = 0 and b1 = 0 are 471.4872 and 243.4111, respec-
tively. Their corresponding P -values are < 0.0001. Hence, both coefficients are
significant.
(c) ED50 = \u2212\u22122.76200.0308 = 89.675.
12.72 (a) The fitted model is p\u2c6 = 1
1+e2.9949\u22120.0308x
.
(b) The increase in odds of failure that results by increasing the load by 20 lb/in.2 is
e(20)(0.0308) = 1.8515.
Chapter 13
One-Factor Experiments: General
13.1 Using the formula of SSE, we have
SSE =
k\u2211
i=1
n\u2211
j=1
(yij \u2212 y¯i.)2 =
k\u2211
i=1
n\u2211
j=1
(\u1ebij \u2212 \u1eb¯i.)2 =
k\u2211
i=1
[
n\u2211
j=1
\u1eb2ij \u2212 n\u1eb¯2i.
]
.
Hence
E(SSE) =
k\u2211
i=1
[
n\u2211
j=1
E(\u1eb2ij)\u2212 nE(\u1eb¯2i.)
]
=
k\u2211
i=1
[
n\u3c32 \u2212 n\u3c3
2
n
]
= k(n\u2212 1)\u3c32.
Thus E
[
SSE
k(n\u22121)
]
= k(n\u22121)\u3c3
2
k(n\u22121) = \u3c3
2.
13.2 Since SSA = n
k\u2211
i=1
(y¯i. \u2212 y¯..)2 = n
k\u2211
i=1
y¯2i. \u2212 kny¯2.., yij \u223c n(y;µ + \u3b1i, \u3c32), and hence
y¯i. \u223c n
(
y;µ+ \u3b1i,
\u3c3\u221a
n
)
and y¯.. \u223c n
(
µ+ \u3b1¯, \u3c3\u221a
kn
)
, then
E(y¯2i.) = V ar(y¯i.) + [E(y¯i.)]
2 =
\u3c32
n
+ (µ+ \u3b1i)
2,
and
E[y¯2..] =
\u3c32
kn
+ (µ+ \u3b1¯)2 =
\u3c32
kn
+ µ2,
due to the constraint on the \u3b1\u2019s. Therefore,
E(SSA) = n
k\u2211
i=1
E(y¯2i.)\u2212 knE(y¯2..) = k\u3c32 + n
k\u2211
i=1
(µ+ \u3b1i)
2 \u2212 (\u3c32 + knµ2)
= (k \u2212 1)\u3c32 + n
k\u2211
i=1
\u3b12i .
185
186 Chapter 13 One-Factor Experiments: General
13.3 The hypotheses are
H0 : µ1 = µ2 = · · · = µ6,
H1 : At least two of the means are not equal.
\u3b1 = 0.05.
Critical region: f > 2.77 with v1 = 5 and v2 = 18 degrees of freedom.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Treatment
Error
5.34
62.64
5
18
1.07
3.48
0.31
Total 67.98 23
with P -value=0.9024.
Decision: The treatment means do not differ significantly.
13.4 The hypotheses are
H0 : µ1 = µ2 = · · · = µ5,
H1 : At least two of the means are not equal.
\u3b1 = 0.05.
Critical region: f > 2.87 with v1 = 4 and v2 = 20 degrees of freedom.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Tablets
Error
78.422
59.532
4
20
19.605
2.977
6.59
Total 137.954 24
with P -value=0.0015.
Decision: Reject H0. The mean number of hours of relief differ significantly.
13.5 The hypotheses are
H0 : µ1 = µ2 = µ3,
H1 : At least two of the means are not equal.
\u3b1 = 0.01.
Computation:
Solutions for Exercises in Chapter 13 187
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Shelf Height
Error
399.3
288.8
2
21
199.63
13.75
14.52
Total 688.0 23
with P -value=0.0001.
Decision: Reject H0. The amount of money spent on dog food differs with the shelf
height of the display.
13.6 The hypotheses are
H0 : µA = µB = µC ,
H1 : At least two of the means are not equal.
\u3b1 = 0.01.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Drugs
Error
158.867
393.000
2
27
79.433
14.556
5.46
Total 551.867 29
with P -value=0.0102.
Decision: Since \u3b1 = 0.01, we fail to reject H0. However, this decision is very marginal
since the P -value is very close to the significance level.
13.7 The hypotheses are
H0 : µ1 = µ2 = µ3 = µ4,
H1 : At least two of the means are not equal.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Treatments
Error
119.787
638.248
3
36
39.929
17.729
2.25
Total 758.035 39
with P -value=0.0989.
Decision:```