Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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f
Temperatures
Error
1268.5333
112.8333
4
25
317.1333
4.5133
70.27
Total 1381.3667 29
with P -value< 0.0001.
The results from Tukey\u2019s procedure can be obtained as follows:
y¯0 y¯25 y¯100 y¯75 y¯50
55.167 60.167 64.167 70.500 72.833
194 Chapter 13 One-Factor Experiments: General
The batteries activated at temperature 50 and 75 have significantly longer activated
life.
13.24 The Duncan\u2019s procedure shows the following results:
y¯E y¯A y¯C
0.3300 0.9422 1.0063
Hence, the sorption rate using the Esters is significantly lower than the sorption rate
using the Aromatics or the Chloroalkanes.
13.25 Based on the definition, we have the following.
SSB = k
b\u2211
j=1
(y¯.j \u2212 y¯..)2 = k
b\u2211
j=1
(
T.j
k
\u2212 T..
bk
)2
=
b\u2211
j=1
T 2.j
k
\u2212 2T
2
..
bk
+
T 2..
bk
=
b\u2211
j=1
T 2.j
k
\u2212 T
2
..
bk
.
13.26 From the model
yij = µ+ \u3b1i + \u3b2j + \u1ebij,
and the constraints
k\u2211
i=1
\u3b1i = 0 and
b\u2211
j=1
\u3b2j = 0,
we obtain
y¯.j = µ+ \u3b2j + \u1eb¯.j and y¯.. = µ+ \u1eb¯...
Hence
SSB = k
b\u2211
j=1
(y¯.j \u2212 y¯..)2 = k
b\u2211
j=1
(\u3b2j + \u1eb¯.j \u2212 \u1eb¯..)2.
Since E(\u1eb¯.j) = 0 and E(\u1eb¯..) = 0, we obtain
E(SSB) = k
b\u2211
j=1
\u3b22j + k
b\u2211
j=1
E(\u1eb¯2.j)\u2212 bkE(\u1eb¯2..).
We know that E(\u1eb¯2.j) =
\u3c32
k
and E(\u1eb¯2..) =
\u3c32
bk
. Then
E(SSB) = k
b\u2211
j=1
\u3b22j + b\u3c3
2 \u2212 \u3c32 = (b\u2212 1)\u3c32 + k
b\u2211
j=1
\u3b22j .
13.27 (a) The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = \u3b14 = 0, fertilizer effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.05.
Critical region: f > 4.76.
Computation:
Solutions for Exercises in Chapter 13 195
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Fertilizers
Blocks
Error
218.1933
197.6317
71.4017
3
2
6
72.7311
98.8158
11.9003
6.11
Total 487.2267 11
P -value= 0.0296. Decision: Reject H0. The means are not all equal.
(b) The results of testing the contrasts are shown as:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
(f1, f3) vs (f2. f4)
f1 vs f3
Error
206.6700
11.4817
71.4017
1
1
6
206.6700
11.4817
11.9003
17.37
0.96
The corresponding P -values for the above contrast tests are 0.0059 and 0.3639,
respectively. Hence, for the first contrast, the test is significant and the for the
second contrast, the test is insignificant.
13.28 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0, no differences in the varieties
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.05.
Critical region: f > 5.14.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Treatments
Blocks
Error
24.500
171.333
42.167
2
3
6
12.250
57.111
7.028
1.74
Total 238.000 11
P -value=0.2535. Decision: Do not reject H0; could not show that the varieties of
potatoes differ in field.
13.29 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0, brand effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.05.
Critical region: f > 3.84.
Computation:
196 Chapter 13 One-Factor Experiments: General
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Treatments
Blocks
Error
27.797
16.536
18.556
2
4
8
13.899
4.134
2.320
5.99
Total 62.889 14
P -value=0.0257. Decision: Reject H0; mean percent of foreign additives is not the
same for all three brand of jam. The means are:
Jam A: 2.36, Jam B: 3.48, Jam C: 5.64.
Based on the means, Jam A appears to have the smallest amount of foreign additives.
13.30 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = \u3b14 = 0, courses are equal difficulty
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Subjects
Students
Error
42.150
1618.700
1112.100
3
4
12
14.050
404.675
92.675
0.15
Total 2772.950 19
P -value=0.9267. Decision: Fail to rejectH0; there is no significant evidence to conclude
that courses are of different difficulty.
13.31 The hypotheses are
H0 : \u3b11 = \u3b12 = · · · = \u3b16 = 0, station effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.01.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Stations
Dates
Error
230.127
3.259
44.018
5
5
25
46.025
0.652
1.761
26.14
Total 277.405 35
Solutions for Exercises in Chapter 13 197
P -value< 0.0001. Decision: Reject H0; the mean concentration is different at the
different stations.
13.32 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0, station effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.05.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Stations
Months
Error
10.115
537.030
744.416
2
11
22
5.057
48.821
33.837
0.15
Total 1291.561 35
P -value= 0.8620. Decision: Do not reject H0; the treatment means do not differ
significantly.
13.33 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0, diet effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.01.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Diets
Subjects
Error
4297.000
6033.333
1811.667
2
5
10
2148.500
1206.667
181.167
11.86
Total 12142.000 17
P -value= 0.0023. Decision: Reject H0; differences among the diets are significant.
13.34 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0, analyst effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.05.
Computation:
198 Chapter 13 One-Factor Experiments: General
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Analysts
Individuals
Error
0.001400
0.021225
0.001400
2
3
6
0.000700
0.007075
0.000233
3.00
Total 0.024025 11
P -value= 0.1250. Decision: Do not reject H0; cannot show that the analysts differ
significantly.
13.35 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = \u3b14 = \u3b15 = 0, treatment effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.01.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Treatments
Locations
Error
79630.133
634334.667
689106.667
4
5
20
19907.533
126866.933
34455.333
0.58
Total 1403071.467 29
P -value= 0.6821. Decision: Do not reject H0; the treatment means do not differ
significantly.
13.36 The hypotheses are
H0 : \u3b11 = \u3b12 = \u3b13 = 0, treatment effects are zero
H1 : At least one of the \u3b1i\u2019s is not equal to zero.
\u3b1 = 0.01.
Computation:
Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Treatments
Subjects
Error
203.2792
188.2271
212.8042
2
9
18
101.6396
20.9141
11.8225
8.60
Total 604.3104 29
P -value= 0.0024. Decision: RejectH0; the mean weight losses are different for different
treatments and the therapists had the greatest effect on the weight loss.
Solutions for Exercises in Chapter 13 199
13.37 The total sums of squares can be written as\u2211
i
\u2211
j
\u2211
k
(yijk \u2212 y¯...)2 =
\u2211
i
\u2211
j
\u2211
k
[(y¯i.. \u2212 y¯...) + (y¯.j. \u2212 y¯...) + (y¯..k \u2212 y¯...)
+ (yijk \u2212 y¯i.. \u2212 y¯.j. \u2212 y¯..k + 2y¯...)]2
=r
\u2211
i
(y¯i.. \u2212 y¯...)2 + r
\u2211
j
(y¯.j. \u2212 y¯...)2 + r
\u2211
k
(y¯..k \u2212 y¯...)2
+
\u2211
i
\u2211
j
\u2211
k
(yijk \u2212 y¯i.. \u2212 y¯.j. \u2212 y¯..k + 2y¯...)2
+ 6 cross-product terms,
and all cross-product terms are equal to zeroes.
13.38 For the model yijk = µ+ \u3b1i + \u3b2j + \u3c4k + \u1ebijk, we have
y¯..k = µ+ \u3c4k + \u1eb¯..k, and y¯... = µ+ \u1eb¯....
Hence SSTr = r
\u2211
k
(y¯..k \u2212 y¯...)2 = r
\u2211
k
(\u3c4k + \u1eb¯..k \u2212 \u1eb¯...)2, and
E(SSTr) = r
\u2211
k
\u3c4 2k + r
\u2211
k
E(\u1eb¯2..k)\u2212 r2
\u2211
k
E(\u1eb¯2...)
= r
\u2211
k
\u3c4 2k + r
\u2211
k
\u3c32
r
\u2212 r2\u3c3
2
r2
= r
\u2211
k
\u3c4 2k + r\u3c3
2 \u2212 \u3c32
=