Reserve Supplemental Exercises Chapter 13 Problem 5
Consider the following computer output.
Source
DF
SS
MS
F
P
Factor
?
?
?
?
?
Error
15
167.5
?
Total
19
326.2
S = 3.342
R − Sq = ?
R − Sq(adj)=34.96%
(a) How many levels of the factor were used in this experiment?
(b) How many replicates were used?
(c) Fill in the missing information.
(d) What conclusions would you draw if
0.05
=
? What if
0.01
=
?
SOLUTION
0.01
=
Reserve Supplemental Exercises Chapter 13 Problem 6
An article in Agricultural Engineering (December 1964, pp. 672-673) described an experiment
in which the daily weight gain of swine is evaluated at different levels of housing temperature.
The mean weight of each group of swine at the start of the experiment is considered to be a
nuisance factor. The data from this experiment are as follows:
Mean weight (lbs)
Housing Air Temperatures (°F)
50
60
70
80
90
100
100
1.37
1.58
2.00
1.97
1.40
0.39
150
1.47
1.75
2.16
1.82
1.14
−0.19
200
1.19
1.91
2.22
1.67
0.88
−0.77
(a) Does housing air temperature affect mean weight gain? Test the null hypothesis, using
0.05
=
.
(b) Use Fisher’s LSD method to determine if these pairs of temperature levels are significantly
different. Use
0.05
=
.
Mean air temperature levels
Difference
60 and 90
80 and 100
50 and 60
70 and 90
80 and 50
(c) Analyze the residuals from this experiment. Is the variance constant?
SOLUTION
(a) Analysis of variance of Weight Gain
Source
DF
SS
MS
F
P
Mean Weight
2
0.2227
0.1113
1.48
0.273
(b) Fisher’s pairwise comparisons
( )
0.025,12
/2, 1
22 0.0751 0.4875
3
E
an
MS
LSD t t
n
−
= = =
Air Temperature
5
10.1852
2.0370
27.13
0.000
Error
10
0.7509
0.0751
Total
17
11.1588
(c) There appear to be some problems with the assumption of constant variance.
Reserve Supplemental Exercises Chapter 13 Problem 7
An article in Marine Biology [“Allozymes and Morphometric Characters of Three Species of
Mytilus in the Northern and Southern Hemispheres” (1991, Vol. 111, pp. 323-333)] discussed
the ratio of the anterior adductor muscle scar length to shell length for shells from five different
geographic locations. The following table is part of a much larger data set from their research.
Location
Ratio
Tillamook, Oregon
0.057
0.081
0.083
0.097
0.081
0.086
Newport, Oregon
0.087
0.066
0.067
0.081
0.074
0.065
Petersburg, Alaska
0.097
0.135
0.081
0.101
0.096
0.106
Magadan, Quebec
0.103
0.091
0.078
0.068
0.067
0.070
Tvarminne, Finland
0.070
0.102
0.095
0.097
0.103
0.105
(a) Are there any differences in the mean ratios due to different locations at
0.05
=
? Calculate
the P-value.
(b) Construct a normal probability plot of the residuals. Does the normality assumption seem
reasonable?
SOLUTION
(a)
Source
DF
SS
MS
F
P
Factor
4
0.003562
0.000891
4.66
0.006
(b)
Reserve Supplemental Exercises Chapter 13 Problem 8
An article in Bioresource Technology [“Preliminary Tests on Nisin and Pediocin Production
Using Waste Protein Sources: Factorial and Kinetic Studies” (2006, Vol. 97(4), pp. 605-613]
described an experiment in which pediocin was produced from waste protein. Nisin and pediocin
are bacteriocins (compounds produced by bacteria that inhibit related strains) used for food
preservation. Three levels of protein (g/L) from trout viscera extracts were compared.
Protein (g/L)
Pediocin production (ratio to baseline)
1
2
3
4
5
6
7
8
9
10
1.67
2.90
3.42
3.18
3.25
2.50
2.64
4.52
2.95
4.35
4.65
4.54
4.29
4.42
4.47
3.98
3.33
3.77
4.5
4.62
4.87
Error
25
0.004782
0.000191
Total
29
0.008345
(a) Calculate the P-value for the protein factor
(b) Does the level of protein have an effect on mean pediocin production? Use
0.05
=
.
(c) Would you draw a different conclusion if
0.01
=
had been used?
(d) Construct a normal probability plot of the residuals. Does the normality assumption seem
reasonable?
(e) Find a 95% confidence interval on mean pediocin production when the level of protein is
2.50 g/L.
SOLUTION
(a)
Source
DF
SS
MS
F
P
(b) The P-value for the protein factor = 0.023 < 0.05 so that the level of protein has a significant
(d)
Factor
2
3.455
1.727
4.90
0.023
Error
15
5.286
0.352
Total
17
8.741
The normality plot has some deviations from a line.
(e)
0.05/2,15 2.1315t=
, and the pooled standard deviation is 0.5936.
Reserve Supplemental Exercises Chapter 13 Problem 9
An electronics engineer is interested in the effect on tube conductivity of five different types of
coating for cathode ray tubes in a telecommunication system display device. The five coating
types are selected at random from a large number of types. The following conductivity data are
obtained.
Coating type
Conductivity
1
143
141
150
146
2
152
149
137
143
3
134
133
132
127
4
129
127
132
129
5
147
148
144
142
(a) What is an appropriate statistical model for this experiment?
(b) Estimate the parameters of this model.
SOLUTION
(a) The experiment includes random effects.
(b) Summary
Groups
Count
Sum
Average
Variance
1
4
580
145
15.33
2
4
581
44.25
3
4
526
9.67
4
4
517
4.25
5
4
581
7.58
Reserve Supplemental Exercises Chapter 13 Problem 10
(a) Find the expected value for
E
MS
in the fixed effects model analysis of variance.
(b) Would it change if the random effects model had been specified?
SOLUTION
(a)
( )
( )
1
12
1
j
i
ij i
an
E
yy
MS an
=
=
−
=−
and
ij i ij
ya
= + +
. Then
.ij i ij i
yy
− = −
and
( )
12
.
1
j
ij i
n
n
=
−
−
is
Reserve Supplemental Exercises Chapter 13 Problem 11
Consider testing the equality of the means of two normal populations for which the variances are
unknown but are assumed to be equal. The appropriate test procedure is the two sample t-test.
Show that the two-sample t-test equivalent to the single-factor analysis of variance F-test.
SOLUTION
The two-sample t-test rejects equality of means if the statistic
12
11
p
yy
t
snn
−
=
+
is too large.
Reserve Supplemental Exercises Chapter 13 Problem 12
Consider the ANOVA with a = 2 treatments. Is the MSE in this analysis equal to the pooled
variance estimate used in the two sample t-test?
SOLUTION
Reserve Supplemental Exercises Chapter 13 Problem 13
We have linear combination
1
a
iI
i
cY
=
. What is its variance?
SOLUTION
Reserve Supplemental Exercises Chapter 13 Problem 14
In a fixed-effects model, suppose that there are n observations for each of four treatments. Let
222
1 2 3
,,QQQ
be single-degree-of-freedom sums of squares for orthogonal contrasts. A contrast is a
linear combination of the treatment means with coefficients that sum to zero. The coefficient
vectors of orthogonal contrasts are orthogonal vectors.
What is the treatments sum of squares?
SOLUTION
Reserve Supplemental Exercises Chapter 13 Problem 15
Consider the single-factor completely randomized design with a treatments and n replicates.
If the difference between any two treatment means is as large as D, what is the minimum value
that the OC curve parameter Φ2 can take?
SOLUTION
2
21
2
()
a
i
i
n
a
=−
=
, we need to show that
212
()
2
i
i
a
D
=
−
Reserve Supplemental Exercises Chapter 13 Problem 16
Consider the single-factor completely randomized design with the total number of observation N.
Construct the
( )
100 1
−
percent confidence interval for
2
.
SOLUTION
( )
22
2
..
1 1 1
2
1
( ) ( )
,
11
a n n
a
ij i ij i
i
i j j
i
Ei
y y y y
s
MS where s
a n a n
= = =
=
−−
= = =
−−
Reserve Supplemental Exercises Chapter 13 Problem 17
Consider the random-effects model for the single-factor completely randomized design.
( )
100 1
−
% confidence interval on the ratio of variance components
2
2
is
SOLUTION
Construct
( )
100 1
−
% confidence interval on the ratio of variance components
2
2
.
Reserve Supplemental Exercises Chapter 13 Problem 18
Consider the random-effects model for the single-factor completely randomized design.
(a)
( )
100 1
−
% confidence interval on the ratio
2
22
+
is _________.
(b) Find a
( )
100 1
−
% confidence interval for
2
22
+
. _________.
SOLUTION
(a) Find
( )
100 1
−
% confidence interval on the ratio
2
22
+
.
Reserve Supplemental Exercises Chapter 13 Problem 19
Consider the fixed-effects model of the completely randomized single-factor design. The model
parameters are restricted by the constraint
1
0
a
i
i
=
=
. (Actually, other restrictions could be used,
but this one is simple and results in intuitively pleasing estimates for the model parameters.) For
the case of unequal sample size n1, n2, . . . , na, the restriction is
1
0
i
ii
a
n
=
=
.
Considering the mean square for treatments we can show, that _________.
Does this suggest the null hypothesis in this model is
0 1 1 2 2
: … 0
ii
H n n n
= = = =
?
SOLUTION
( )
+
2
22
+
Considering the mean square for treatments we can show, that
( )
12
2
i
ii
a
T
n
E MS a
=
=+−
.
Reserve Supplemental Exercises Chapter 13 Problem 20
Sample Size Determination. In the single-factor completely randomized design, the accuracy of a
( )
100 1 %
−
confidence interval on the difference in any two treatment means is
( )
/2, 1 2/
E
an
t MS n
−
.
(a) If A is the desired accuracy of the interval, the sample size required is
(b) Suppose that in comparing
5a=
means you have a preliminary estimate of
2
of 3. If you
want the 95% confidence interval on the difference in means to have an accuracy of 2, how
many replicates should you use?
SOLUTION
(a)
If A is the accuracy of the interval, then
(b)
Because n determines one of the degrees of freedom of the tabulated F value on the right-side of
the equation in part (a), some approximation is needed. Because the value for a 95% confidence
interval based on a normal distribution is 1.96, we approximate
( )
/2, 1an
t
−
by 2 and