This document discusses the analysis of variance (ANOVA) technique for analyzing data from single-factor engineering experiments. It defines key concepts like treatments, replicates, and the linear statistical model used. The document provides examples of calculating sums of squares, performing an ANOVA, making comparisons between treatment means, and checking model assumptions. It demonstrates how to test hypotheses, calculate confidence intervals, and interpret results for a experiment on the effect of hardwood concentration on paper tensile strength.

1. KING FAHD UNIVERSITY OF PETROLEUM AND MINERALS
DEPARTMENT OF SYSTEMS ENGINEERING
ISE 315: ENGINEERING PROBABILITY AND STATISTICS
(Term 202)
Dr. Al-Turki
CHAPTER 13
DESIGN AND ANALYSIS OF SINGLE-FACTOR
EXPERIMENTS: THE ANALYSIS OF VARIANCE
Applied Statistics and Probability for engineers
Douglas Montgomery and George Runger
Sixth Edition 2014
8/6/2022 1

2. Design of experiment is used for:
1. Improving processes – Improve yield, reduce variability, reduce cost, …
2. Improve engineering product design – material selection, determine
design parameters.
Design of experiment is usually done sequentially:
Screening: experiments include many possible controllable variables to
determine the most important ones.
Refine: experiments to refine the values of the variables resulting from
screening
Optimization: experiments on the selected variables to determine the
optimum values for best performance.
2
13-1: Designing Engineering Experiments

3. Every experiment involves a sequence of activities:
1. Conjecture – the original hypothesis that motivates the experiment.
2. Experiment – the test performed to investigate the conjecture.
3. Analysis – the statistical analysis of the data from the experiment.
4. Conclusion – what has been learned about the original conjecture from
the experiment. Often the experiment will lead to a revised conjecture,
and a new experiment, and so forth.
3

4. 13-2: The Completely Randomized Single-Factor Experiment
Example: Tensile Strength
4
• A manufacturer of paper grocery bags is interested in improving the tensile strength of the
product. Product engineering thinks that tensile strength is a function of the hardwood
concentration in the pulp and that the range of hardwood concentrations of practical interest
is between 5 and 20%.
• A study is conducted to investigate four levels of hardwood concentration: 5%, 10%, 15%,
and 20%. They decide to make up six test specimens at each concentration level, using a
pilot plant. All 24 specimens are tested on a laboratory tensile tester, in random order.
Hardwood Observations
Concentration () 1 2 3 4 5 6 Totals Averages
5 7 8 15 11 9 10 60 10.00
10 12 17 13 18 19 15 94 15.67
15 14 18 19 17 16 18 102 17.00
20 19 25 22 23 18 20 127 21.17
383 15.96
•The levels of the factor are sometimes called treatments.
• Each treatment has six observations or replicates.
• The runs (experiments) are run in random order.
• The outcome of the experiment is the Response
Tensile Strength of Paper (psi)

5. (a) Box plots of hardwood concentration data. (b) Display of the model in Equation 13-1 for the
completely randomized single-factor experiment.
5
Variability within a treatment
Variability
between
treatments

6. 13-2.2 The Analysis of Variance
Definition
6
yij represents the jth observation under treatment i (level)
We may describe the relationship by the
Linear statistical model 𝑌𝑖𝑗 = 𝜇 + 𝜏𝑖 + 𝜖𝑖𝑗
𝑖 = 1,2, … , 𝑎
𝑗 = 1,2, … , 𝑛
 is a parameter common to all treatments (overall mean)
ti is a parameter representing the ith treatment effect. The deviation from 
eij is a random error component assumed to be normally and independently
distributed with mean zero and variance s2 distributed
𝑌𝑖𝑗 = 𝜇𝑖 + 𝜖𝑖𝑗
𝑖 = 1,2, … , 𝑎
𝑗 = 1,2, … , 𝑛
An alternative model of writing the model is
Treatment Observation Totals Averages
1
2
:
a
𝑦11 𝑦12 … 𝑦1𝑛
𝑦21
⋮
𝑦22 …
⋮
𝑦2𝑛
⋮
𝑦𝑎1 𝑦𝑎2 … 𝑦𝑎𝑛
y1.
y2.
:
ya.
𝑦1.
𝑦2.
:
𝑦𝑎.
y.. 𝑦..
Typical data for a single factor experiment

7. 8/6/2022 7
Completely randomized Design (CRD)
Observations are taken in random order in order to have a uniform environment
(experiment unit) condition as much as possible.
Selection of treatments (levels)
Fixed effect models: (Section 13.2)
• Specific values of treatments are chosen.
• Conclusions are limited to these specific values (can not be extended to other values)
• The test is for the hypothesis about the mean treatment effect
Random effect models: (Section 13.3)
• Specific values of treatments are randomly sampled from a large set of possible values.
• Conclusions are extended to all possible values of the population
• The test is for the hypothesis about the variability of the treatment effect, ti

8. Analysis of variance ANOVA
8
ti represents the deviation of the effect of treatment i from the
overall mean  such that
i represents the effect of factor i, i = +ti
𝑖=1
𝑎
𝜏𝑖 = 0
𝑦𝑖. =
𝑗=1
𝑛
𝑦𝑖𝑗 𝑦𝑖. =
𝑦𝑖.
𝑛
𝑖 = 1,2, … , 𝑎 𝑦.. =
𝑖=1
𝑎
𝑗=1
𝑛
𝑦𝑖𝑗 𝑦.. =
𝑦..
𝑎𝑛
We are interested in testing the equality of the a treatment means 1, 2 ,…, a
This is equivalent to testing
H0: t1 = t2 = … = ta=0 H1: ti ≠0 for at least one i
Define:
If the null hypothesis is true, each observation consists of the overall  plus
a realization of the random error eij.
Which means that all observations have the same random error. i.e. changing
the level has no effect on the observation.

9. 13-2.2 The Analysis of Variance
Definition
9
The sums of squares computing formulas for the ANOVA with equal sample
sizes in each treatment are:
𝑆𝑆𝑇 = 𝑖=1
𝑎
𝑗=1
𝑛
𝑦𝑖𝑗 − 𝑦..
2
𝑆𝑆𝑡𝑟𝑒𝑎𝑡 = 𝑛 𝑖=1
𝑎
𝑦𝑖. − 𝑦..
2
𝑆𝑆𝐸 = 𝑖=1
𝑎
𝑗=1
𝑛
𝑦𝑖𝑗 − 𝑦𝑖.
2
The error sum of squares is SSE = SST – SSTreatments
 
 


a
i
n
j
ij
T
N
y
y
SS
1
2
1
2 ..




a
i
i
N
y
n
y
SS
1
2
2
Treatments
..
.
Source of
Variation Sum of Squares
Degrees of
Freedom Mean Square F0
Treatments SSTreatments a  1 MSTreatments
Error SSE a(n  1) MSE
Total SST an  1
The Analysis of Variance for a Single-Factor Experiment, Fixed-Effects Model

10. EXAMPLE:
Tensile Strength ANOVA Consider the paper tensile strength experiment described earlier.
This experiment is a CRD. We can use the analysis of variance to test the hypothesis that
different hardwood concentrations do not affect the mean tensile strength of the paper.
The hypotheses are
H0: t1 = t2 = t3 = t4 = 0
H1: ti  0 for at least one i
10
We will use a = 0.01.
The sums of squares for the analysis of variance are computed as follows:
96
.
512
24
)
383
(
)
20
(
)
8
(
)
7
(
..
2
2
2
2
4
1
2
6
1
2







  
 

i j
ij
T
N
y
y
SS
17
.
130
79
.
382
96
.
512
79
.
382
24
)
383
(
6
)
127
(
)
102
(
)
94
(
)
60
(
..
.
Treatments
2
2
2
2
2
4
1
2
2
Treatments












 

SS
SS
SS
N
y
n
y
SS
T
E
i
i

11. 11
The ANOVA is summarized below. Since f0.01,3,20 = 4.94, we reject H0 and conclude that
hardwood concentration in the pulp significantly affects the mean strength of the paper.
We can also find a P-value for this test statistic as follows:
Since is considerably smaller than a = 0.01, we have strong evidence to
conclude that H0 is not true.
6
~
20
,
3 10
59
.
3
)
60
.
19
( 



 F
P
P
6
~ 10
59
.
3 


P
ANOVA for the Tensile Strength Data
Source of
Variation Sum of Squares
Degrees of
Freedom
Mean
Square f0
P-value
Hardwood
concentration
382.79 3 127.60 19.60 3.59 E-6
Error 130.17 20 6.51
Total 512.96 23
Confidence Interval on a Treatment Mean
A 100(1  a) percent confidence interval on the mean of the ith treatment t is
   
n
MS
t
y
n
MS
t
y E
n
a
i
i
E
n
a
i 1
/2,
1
/2, .
. 
a

a 




For 20% hardwood, the resulting confidence interval on the mean is 19.00 psi  4  23.34 psi

12. Confidence Interval on a Difference in Treatment Means
For the hardwood concentration example, 1.74  3  2  4.40
12
A 100(1  a) percent confidence interval on the difference in two treatment means
μi  μj is
   
n
MS
t
y
y
n
MS
t
y
y E
n
a
j
i
j
i
E
n
a
j
i
2
.
.
2
.
. 1
/2,
1
/2, 
a

a 







An Unbalanced Experiment
The sums of squares computing formulas for the ANOVA with unequal sample sizes ni in
each treatment are
and
SSE = SST  SSTreatments
 
 


a
i
n
j
ij
T
N
y
y
SS
i
1
2
1
2 ..




a
i i
i
N
y
n
y
SS
1
2
2
Treatments
..
.

13. 13-2.3 Multiple Comparisons Following the ANOVA
LSD compares all pairs of means with the null hypothesis H0: i=j for all i≠j
If the sample sizes are different in each treatment:
13
 
n
MS
t E
n
a
2
LSD 1
/2, 
a










 
a
j
i
E
a
N
n
n
MS
t
1
1
LSD /2,
Multiple Comparisons methods are used when some treatments are different. One
method is Fisher’s least significant difference (LSD)
The hypothesis is rejected if 𝑦𝑖. − 𝑦𝑗. >LSD

14. Example
We will apply the Fisher LSD method to the hardwood concentration experiment.
There are a = 4 means, n = 6,
MSE = 6.51, and t0.025,20 = 2.086. The treatment means are
The value of LSD is .
Therefore, any pair of treatment averages that differs by more than 3.07 implies that the
corresponding pair of treatment means are different.
14
psi
17
.
21
.
psi
00
.
17
.
psi
67
.
15
.
psi
00
.
10
.
4
3
2
1




y
y
y
y
  07
.
3
/6
51
.
6
2
086
.
2
/
2
LSD 20
,
025
.
0 

 n
MS
t E
The comparisons among the observed treatment averages are as follows:
4 vs. 1 = 21.17  10.00 = 11.17 > 3.07
4 vs. 2 = 21.17  15.67 = 5.50 > 3.07
4 vs. 3 = 21.17  17.00 = 4.17 > 3.07
3 vs. 1 = 17.00  10.00 = 7.00 > 3.07
3 vs. 2 = 17.00  15.67 = 1.33 < 3.07
2 vs. 1 = 15.67  10.00 = 5.67 > 3.07
Conclusions: From this analysis, we see that there are significant differences between all
pairs of means except 2 and 3.

15. 13-2.4 Residual Analysis and Model Checking
15
Residuals for the Tensile Strength Experiment 𝑒𝑖𝑗 = 𝑦𝑖𝑗 − 𝑦𝑖𝑗
Hardwood Concentration (%) Residuals
5 3.00 -2.00 5.00 1.00 -1.00 0.00
10 -3.67 1.33 -2.67 2.33 3.33 -0.67
15 -3.00 1.00 2.00 0.00 -1.00 1.00
20 -2.17 3.83 0.83 1.83 -3.17 -1.17
The residual is the difference between an observation yij and its estimated value, 𝑦𝑖𝑗
For the Completely randomized design 𝑦𝑖𝑗 = 𝑦𝑖𝑗
To check equal variances at each level
To check Normality of residuals
To check independence between
residual and the fitted value

16. 13-2.5 Determining Sample size
OC curves may be used for determining the sample size (number of replications)
needed to achieve adequate sensitivity.
1-b=P(Rejecting H0|H0 is false) = P(F0>fa,a-1,a(n-1) |H0 is false)
The alternative is expressed in terms of F, where ∅2
=
𝑛 𝑖=1
𝑎
𝜏𝑖
2
𝑎𝜎2 =
𝑛
𝑎
𝑖=1
𝑎
𝜏𝑖
2
𝜎2
It can be shown that, under the alternative hypothesis, F0 has a non-central F
distribution. a=0.01
OC-curves are used for the power and the sample size.
Example:
Assume that we need to reject with probability at least 0.9 if the ratio we want
to detect is 𝑖=1
5
𝜏𝑖
2
𝜎2 = 5. Let a5,
Solution:
v1=a-1=4, v2=5(n-1)
F2=
𝑛
5
5 = 𝑛
Try different values of n
So we need at least 6 replications to obtain at least 0.9 power.
n F2 F a(n-1) b Power(1-b)
4 4 2 15 0.38 0.62
5 5 2.24 20 0.18 0.82
6 6 2.45 25 0.06 0.94>0.9
OC
curve

17. 8/6/2022 17
b
F
V1=3
V1=4
Four treatments
Five treatments
F
b
F for a0.05
F for a0.01
F for a0.05
F for a0.01

18. 8/6/2022 18

19. 13-3: The Random-Effects Model
13-3.2 ANOVA and Variance Components
The linear statistical model is
The variance of the response is
We assume that the errors eij are independent N(0, s2)
Also we assume that the treatment effect ti are independent N(0, 𝜎𝜏
2
)
where 𝜎𝜏
2
is the variance between treatments
19





e

t



n
j
a
i
Y ij
i
ij
,
,
2
,
1
,
,
2
,
1


  2
2
s

s
 t
ij
Y
V
When levels are chosen randomly
For a random-effects model, the appropriate hypotheses to test are:
0
:
0
:
2
1
2
0

s

s
t
t
H
H
Under the null hypothesis there is variability between treatments

20. 20
The ANOVA decomposition of total variability is still valid:
SST = SSTreatments + SSE
The expected mean square for treatments is
and the expected mean square for error is
2
2
Treatments
Treatments
1
)
(
t
s

s











n
a
SS
E
MS
E
 
2
1
)
(
s









n
a
SS
E
MS
E E
E
Under the null hypothesis, both MStreatment and MSE estimate s2
And their ratio 𝐹0 =
𝑀𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
𝑀𝑆𝐸
is an F random variable with a-1 and a(n-1)
degrees of freedom.
The null hypothesis would be rejected at the a-level of significance if F0>fa,a-1,a(n-1)
the estimators of the variance components are E
MS

s2
ˆ
n
MS
MS E


s
t
Treatments
ˆ

21. Example: A textile manufacturing company weaves a fabric on a large number of
looms. The company is interested in loom-to-loom variability in tensile strength.
To investigate this variability, a manufacturing engineer selects four looms at random
and makes four strength determinations on fabric samples chosen.
21
Observations
Loom 1 2 3 4 Total Average
1 98 97 99 96 390 97.5
2 91 90 93 92 366 91.5
3 96 95 97 95 383 95.8
4 95 96 99 98 388 97.0
1527 95.45
Analysis of Variance for the Strength Data
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
f0 P-value
Looms 89.19 3 29.73 15.68 1.88 E-4
Error 22.75 12 1.90
Total 111.94 15
we conclude that the looms in the plant differ significantly in their ability to produce
fabric of uniform strength.
The variance components are estimated by and
Therefore, the variance of strength in the manufacturing process is estimated by
𝑉 𝑌𝑖𝑗 = 𝜎𝜏
2
+ 𝜎2
= 6.96 + 1.90 = 8.86
Conclusions: Most of the variability in strength in the output product is attributable to
differences between looms.
90
.
1
ˆ 
s
96
.
6
4
90
.
1
73
.
29
ˆ 


s
t

22. 13-4: Randomized Complete Block Design
13-4.1 Design and Statistical Analysis
• The randomized block design is a procedure for comparing two treatment
means when all experiments can not be made under homogeneous
conditions.
• It is a method used for reducing the background noise in the experiment by
blocking out the nuisance factor effect.
• It is an extension of the paired t-test to situations where the factor of interest
has more than two levels (Chapter 10.4).
A randomized complete block design RCBD.
22
Treatment
Method
Block (Girder)
1 2 3 4
1 y11 y12 y13 y14
2 y21 y22 y23 y24
3 y31 y32 y33 y34
ti: treatment i

23. 13-4.1 Design and Statistical Analysis
General procedure for a randomized complete block design:
23
A Randomized Complete Block Design with a Treatments and b Blocks
Blocks
Treatments 1 2  b Totals Averages
1 y11 y12  y1b y1.
2 y21 y22  y2b y2.
     
a ya1 ya2  yab ya.
Totals y.1 y.2  y.b y..
Averages 
.
1
y
.
2
y
.
a
y
1
.
y 2
.
y b
y. ..
y

24. 13-4.1 Design and Statistical Analysis
The appropriate linear statistical model:
We assume
• treatments and blocks are initially fixed effects
• deviation from the mean
• blocks do not interact
24





e

b

t



b
j
a
i
Y ij
j
i
ij
,
,
2
,
1
,
,
2
,
1


0
and
0 1
1

b

t 
 

b
j j
a
i i
We are interested in testing: H0: t1 = t2 = … = tn = 0
H1: ti ≠0 for at least one i

25. 25
The sum of squares identity for the randomized complete block design is
 
 


 
 











a
i
b
j
i
j
ij
a
i
b
j
j
a
i
i
b
j
ij
y
y
y
y
y
y
a
y
y
b
y
y
1 1
2
..
1 1
2
1
2
1
2
)
.
.
(
..)
.
(
..)
.
(
..)
(
or symbolically SST = SSTreatments + SSBlocks + SSE
The computing formulas for the sums of squares in the analysis of
variance for a randomized complete block design are
 
 


a
i
b
j
ij
T
ab
y
y
SS
1
2
1
2 ..
ab
y
y
b
SS
a
i
i
..
.
1 2
1
2
Treatments 
 

ab
y
y
a
SS
b
j
j
..
.
1 2
1
2
Blocks 
 


26. The mean squares are:
26
  
1
1
1
1
Blocks
Blocks
Treatments
Treatments







b
a
SS
MS
b
SS
MS
a
SS
MS
E
E
The expected values of these mean squares are:
2
1
2
2
Blocks
1
2
2
Treatments
)
(
1
)
(
1
)
(
s


b

s


t

s





E
b
j
j
a
i
i
MS
E
b
a
MS
E
a
b
MS
E

27. 13-4.1 Design and Statistical Analysis
27
ANOVA for a Randomized Complete Block Design
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square F0
Treatments SSTreatments a - 1
Blocks SSBlocks b - 1
Error
SSE
(by subtraction)
(a -1)(b-1)
Total SST ab – 1
1
Treatments

a
SS
E
MS
MSTreatments
1
Blocks

b
SS
  
1
1 
 b
a
SSE
If the same number of observations were collected without blocking, the error
degrees of Freedom would be a(b-1). So blocking caused an increase in MSE.
This will increase the critical value from the F distribution. Higher values of SStreat
is needed to detect Differences in treatment effects.

28. Example 13-5 Fabric Strength An experiment was performed to determine the effect
of four different chemicals on the strength of a fabric. Five fabric samples were
selected, and a RCBD was run by testing each chemical type once in random order on
each fabric sample.
The data are shown below. We will test for differences in means using an ANOVA with
a = 0.01.
28
Fabric Strength Data—Randomized Complete Block Design
Fabric
Sample
Treatment
Totals
Treatment
Averages
Chemical Type 1 2 3 4 5 yi. 𝑦.𝑗
1 1.3 1.6 0.5 1.2 1.1 5.7 1.14
2 2.2 2.4 0.4 2.0 1.8 8.8 1.76
3 1.8 1.7 0.6 1.5 1.3 6.9 1.38
4 3.9 4.4 2.0 4.1 3.4 17.8 3.56
Block totals y.j 9.2 10.1 3.5 8.8 7.6 39.2
(y..)
Block averages 𝑦𝑖. 2.30 2.53 0.88 2.20 1.90 𝑦.. = 1.96

29. 29
The sums of squares for the analysis of variance are computed as follows:
04
.
18
20
)
2
.
39
(
5
)
8
.
17
(
)
9
.
6
(
)
8
.
8
(
)
7
.
5
(
..
.
69
.
25
20
)
2
.
39
(
)
4
.
3
(
)
6
.
1
(
)
3
.
1
(
..
2
2
2
2
2
4
1
2
2
Treatments
2
2
2
2
4
1
2
5
1
2

















 

 
i
i
i j
ij
T
ab
y
b
y
SS
ab
y
y
SS

96
.
0
04
.
18
69
.
6
69
.
25
69
.
6
20
)
2
.
39
(
4
)
6
.
7
(
)
8
.
8
(
)
5
.
3
(
)
1
.
10
(
)
2
.
9
(
..
.
Treatments
Blocks
2
2
2
2
2
2
5
1
2
2
Blocks















 

SS
SS
SS
SS
ab
y
a
y
SS
T
E
j
j

30. 30
Analysis of Variance for the Randomized Complete Block Experiment
Source of Variation Sum of
Squares
Degrees of
Freedom
Mean Square f0 P-value
Chemical types
(treatments)
18.04 3 6.01 75.13 0.0000
Fabric samples
(blocks)
6.69 4 1.67 20.875 0.0000
Error 0.96 12 0.08
Total 25.69 19
Since f0 = 75.13  f0.01,3,12 = 5.95 (the P-value is 4.79  10-8), we conclude that there is
a significant difference in the chemical types so far as their effect on strength is
concerned.
SSE= 7.644 SST= 25.688 SSt= 18.044 F0= 12.59
f0.01,3,16 = 5.29
f0.01,3,12 = 5.95
If blocking was not used the results would be

31. 13-4.2 Multiple Comparisons
Fisher’s Least Significant Difference for Example 13-5
Results of Fisher’s LSD method.
31
𝐿𝑆𝐷 = 𝑡0.025,12
2𝑀𝑆𝐸
𝑏
= 2.179
2(0.08)
5
= 0.39
4 𝑣𝑠. 1 = 𝑦4. − 𝑦1.=3.56-1.14=2.42>0.39
4 𝑣𝑠. 3 = 𝑦4. − 𝑦3.=3.56-1.38=2.18>0.39
4 𝑣𝑠. 2 = 𝑦4. − 𝑦2.=3.56-1.76=1.80>0.39
2 𝑣𝑠. 1 = 𝑦2. − 𝑦1.=1.76-1.14=0.62>0.39
2 𝑣𝑠. 3 = 𝑦2. − 𝑦1.=1.76-1.38=0.38<0.39
3 𝑣𝑠. 1 = 𝑦3. − 𝑦1. = 1.38-1.14=0.24<0.39

32. 13-4.3 Residual Analysis and Model Checking
Normal probability plot of residuals from
The randomized complete block design.
32
Residuals by treatment from the
randomized complete block design.
Residuals by block from the randomized
complete block design.
Residuals versus ŷij from the
randomized complete block design.
𝑒𝑖𝑗 = 𝑦𝑖𝑗 − 𝑦𝑖𝑗 𝑦𝑖𝑗 = 𝑦𝑖. + 𝑦.𝑗 − 𝑦..

33. 8/6/2022 33
Considering blocking for example 13.1
with blocking
SSB = 53.70833 SSE = 76.45833 SST= 512.9583 SSt = 382.7917
F0block = 0.702452 s2= 5.097222 F0= 25.0327
f0.01,5,15=2.27 f0.01,3,15=2.49
f0.01,3,20=2.38
The hypothesis on the blocking factor is not rejected.

34. An experiment is conducted to determine the effect of air voids on percentage retained
strength of asphalt. For purposes of the experiment, air voids are controlled at three levels;
low (2–4%), medium (4–6%), and high (6–8%). The data are shown in the following below.
(a) Do the different levels of air voids significantly affect mean retained strength? Use 0.01.
(b) Find the P-value for the F-statistic in part (a).
(c) Find a 95% confidence interval on mean retained strength where there is a high level of
air voids.
(d) Find a 95% confidence interval on the difference in mean retained strength at the low
and high levels of air voids.
8/6/2022 34
Problem 13.13
Air Voids Retained Strength %
Low 106 90 103 90 79 88 92 95
Medium 80 69 94 91 70 83 87 83
High 78 80 62 69 76 85 69 85

35. Analysis of Variance for STRENGTH
Source DF SS MS F P
AIRVOIDS 2 1230.3 615.1 8.30 0.002
Error 21 1555.8 74.1
Total 23 2786.0 Reject H0
b) P-value = 0.002
c) 95% Confidence interval on the mean of retained strength where there is a high level of
air voids
8/6/2022 35
𝑦3 − 𝑡0.025,21
𝑀𝑆𝐸
𝑛
≤ 𝜇𝑖 ≤ 𝑦3 + 𝑡0.025,21
𝑀𝑆𝐸
𝑛
75.5 − 2.08
74.1
8
≤ 𝜇𝑖 ≤ 75.5 + 2.08
74.1
8
69.17 ≤ 𝜇𝑖 ≤ 81.83
d) 95% confidence interval on the difference between the means of retained strength at the high
level and the low levels of air voids.
𝑦1 − 𝑦3 − 𝑡0.025,21
2𝑀𝑆𝐸
𝑛
≤ 𝜇𝑖 ≤ 𝑦1 − 𝑦3 + 𝑡0.025,21
2𝑀𝑆𝐸
𝑛
92.875 − 75.5 − 2.08
2𝑀𝑆𝐸
𝑛
≤ 𝜇1 − 𝜇4 ≤ (92.875 − 75.5) + 2.08
2(74.1)
8
8.42 ≤ 𝜇1 − 𝜇4 ≤ 26.33

36. 8/6/2022 36
Problem 13.41
Subject
Test 1 2 3 4
Trainee 0.05 0.05 0.04 0.15
Trainer 0.05 0.05 0.04 0.17
Lab 0.04 0.04 0.03 0.10
A field test for detecting the presence of arsenic in urine samples is conducted. The
experiment compared the test as performed by both a trainee and an experienced trainer to an
analysis at a remote laboratory.
Four subjects were selected for testing and are considered as blocks. The response variable is
arsenic content (in ppm) in the subject’s urine. The data are as follows:
Is there any difference in the arsenic test procedure?
Analysis of Variance
Source DF SS MS F P
TEST 2 0.0014000 0.0007000 3.00 0.125
SUBJECT 3 0.0212250 0.0070750 30.32 0.001
Error 6 0.0014000 0.0002333
Total 11 0.0240250
Fail to reject H0, there is no evidence of differences between the tests.
Solution