2. 2
A study was done on the survival time of
patients with advanced cancer of the
stomach, bronchus, colon, ovary or breast
when treated with ascorbate1
. In this study,
the authors wanted to determine if the
survival times differ based on the affected
organ.
1
Cameron, E. and Pauling, L. (1978) Supplemental ascorbate in the supportive
treatment of cancer: re-evaluation of prolongation of survival time in terminal human
cancer. Proceedings of the National Academy of Science, USA, 75, 4538-4542.
A Problem
3. 3
A comparative dotplot of the survival times is
shown below.
A Problem
3000200010000
Survival Time (in days)
Dotplot for Survival Time
Cancer Type
Breast
Bronchus
Colon
Ovary
Stomach
4. 4
H0: µstomach = µbronchus = µcolon = µovary= µbreast
Ha: At least two of the µ’s are different
A Problem
The hypotheses used to answer the question
of interest are
The question is similar to ones encountered in
chapter 11 where we looked at tests for the
difference of means of two different variables. In
this case we are interested in looking a more than
two variable.
5. 5
A single-factor analysis of variance
(ANOVA) problems involves a comparison of
k population or treatment means µ1, µ2, … , µk.
The objective is to test the hypotheses:
H0: µ1 = µ2 = µ3 = …
= µk
Ha: At least two of the µ’s are different
Single-factor Analysis of Variance
(ANOVA)
6. 6
The analysis is based on k independently
selected samples, one from each population
or for each treatment.
In the case of populations, a random
sample from each population is selected
independently of that from any other
population.
When comparing treatments, the
experimental units (subjects or objects)
that receive any particular treatment are
chosen at random from those available
for the experiment.
Single-factor Analysis of Variance
(ANOVA)
7. 7
A comparison of treatments based on
independently selected experimental units is
often referred to as a completely randomized
design.
Single-factor Analysis of Variance
(ANOVA)
8. 8
70
60
50
40
Fertilizer
Yield
Dotplots of Yield by Fertilizer
(group means are indicated by lines)
Type 1 Type 2 Type 3
Notice that in the above comparative dotplot, the
differences in the treatment means is large relative to
the variability within the samples.
Single-factor Analysis of Variance
(ANOVA)
9. 9
Statistics
Psychology
Economics
Business
85
75
65
SubjectPrice
Dotplots of Price by Subject
(group means are indicated by lines)
Notice that in the above comparative dotplot, the
differences in the treatment means is not easily
understood relative to the sample variability.
ANOVA techniques will allow us to determined if those
differences are significant.
Single-factor Analysis of Variance
(ANOVA)
10. 10
ANOVA Notation
k = number of populations or treatments being compared
Population or treatment 1 2 … k
Population or treatment mean µ1 µ2 … µk
Sample mean …1x 2x kx
Population or treatment variance …2
1σ 2
2σ 2
kσ
Sample variance …
2
1s 2
2s 2
ks
Sample size n1 n2 … nk
11. 11
N = n1 + n2 + … + nk
(Total number of observations in the data set)
ANOVA Notation
T
x grand mean
N
= =
T = grand total = sum of all N observations
1 1 2 2 k kn x n x n x= + + +L
12. 12
Assumptions for ANOVA
1. Each of the k populations or treatments,
the response distribution is normal.
2. σ1 = σ2 = … = σk (The k normal
distributions have identical standard
deviations.
3. The observations in the sample from any
particular one of the k populations or treatments
are independent of one another.
4. When comparing population means, k random
samples are selected independently of one
another. When comparing treatment means,
treatments are assigned at random to subjects or
objects.
13. 13
Definitions
( ) ( ) ( )
2 2 2
1 1 2 2 k kSSTr n x x n x x n x x= − + − + + −L
A measure of disparity among the sample
means is the treatment sum of squares,
denoted by SSTr is given by
A measure of variation within the k samples, called
error sum of squares and denoted by SSE is
given by
( ) ( ) ( )2 2 2
1 1 2 2 k kSSE n 1 s n 1 s n 1 s= − + − + + −L
14. 14
Definitions
A mean square is a sum of squares divided
by its df. In particular,
The error df comes from adding the df’s associated
with each of the sample variances:
(n1 - 1) + (n2 - 1) + …+ (nk - 1)
= n1 + n2 … + nk - 1 - 1 - … - 1 = N - k
mean square for
treatments = MSTr =
SSTr
k 1−
mean square for error = MSE =
SSE
N k−
15. 15
Example
Three filling machines are used by a bottler to
fill 12 oz cans of soda. In an attempt to
determine if the three machines are filling the
cans to the same (mean) level, independent
samples of cans filled by each were selected
and the amounts of soda in the cans measured.
The samples are given below.
Machine 1
12.033 11.985 12.009 12.009
12.033 12.025 12.054 12.050
Machine 2
12.031 11.985 11.998 11.992
11.985 12.027 11.987
Machine 3
12.034 12.021 12.038 12.058
12.001 12.020 12.029 12.011
12.021
16. 16
Example
( ) ( ) ( )
2 2 2
1 1 2 2 k k
2 2 2
SSTr n x x n x x n x x
8(0.0065833) 7(-0.0174524) 9(0.0077222)
0.000334672+0.00213210+0.00053669
0.00301552
= − + − + + −
= + +
=
=
L
1 1 1n 8, x 12.0248, s 0.02301= = =
3 3 3n 9, x 12.0259, s 0.01650= = =
2 2 2n 7, x 12.0007, s 0.01989= = =
x 12.018167=
17. 17
Example
( ) ( ) ( )2 2 2
1 1 2 2 k k
2 2 2
SSE n 1 s n 1 s n 1 s
7(0.0230078) 6(0.0198890) 8(0.01649579)
0.0037055 0.0023734 0.0021769
0.00825582
= − + − + + −
= + +
= + +
=
L
1 1 1n 8, x 12.0248, s 0.02301= = =
3 3 3n 9, x 12.0259, s 0.01650= = =
2 2 2n 7, x 12.0007, s 0.01989= = =
x 12.018167=
18. 18
Example
SSTr
k 1−
mean square for treatments = MSTr =
SSTr 0.00301552
MSTr 0.0015078
k 1 3 1
= = =
− −
mean square for error = MSE =
SSE
N k−
SSE 0.0082579
MSE 0.00039313
N k 24 3
= = =
− −
1 1 1n 8, x 12.0248, s 0.02301= = =
3 3 3n 9, x 12.0259, s 0.01650= = =
2 2 2n 7, x 12.0007, s 0.01989= = =
x 12.018167=
19. 19
Comments
Both MSTr and MSE are quantities that are
calculated from sample data.
As such, both MSTr and MSE are statistics
and have sampling distributions.
More specifically, when H0 is true, µMSTr = µMSE.
However, when H0 is false, µMSTr = µMSE and the greater
the differences among the µ’s, the larger µMSTr will be
relative to µMSE.
20. 20
The Single-Factor ANOVA F Test
Null hypothesis: H0: µ1 = µ2 = µ3 = …
= µk
Alternate hypothesis: At least two of the µ’s
are different
Test Statistic: MSTr
F
MSE
=
21. 21
The Single-Factor ANOVA F Test
When H0 is true and the ANOVA assumptions
are reasonable, F has an F distribution with
df1 = k - 1 and df2 = N - k.
Values of F more contradictory to H0 than what was
calculated are values even farther out in the upper tail,
so the P-value is the area captured in the upper tail of
the corresponding F curve.
22. 22
Example
Consider the earlier example involving the
three filling machines.
Machine 1
12.033 11.985 12.009 12.009 12.033
12.025 12.054 12.050
Machine 2
12.031 11.985 11.998 11.992 11.985
12.027 11.987
Machine 3
12.034 12.021 12.038 12.058 12.001
12.020 12.029 12.011 12.021
23. 23
Example
SSTr 0.00301552= SSE 0.00825582=
MSTr 0.0015078= MSE 0.00039313=
1 1 1n 8, x 12.0248, s 0.02301= = =
3 3 3n 9, x 12.0259, s 0.01650= = =
2 2 2n 7, x 12.0007, s 0.01989= = =
x 12.018167=
24. 24
Example
1. Let µ1, µ2 and µ3 denote the true mean
amount of soda in the cans filled by
machines 1, 2 and 3, respectively.
2. H0: µ1 = µ2 = µ3
3. Ha: At least two among are µ1, µ2 and µ3
different
4. Significance level: α = 0.01
5. Test statistic:
MSTr
F
MSE
=
25. 25
Example
6. Looking at the comparative dotplot, it
seems reasonable to assume that the
distributions have the same σ’s. We shall
look at the normality assumption on the
next slide.*
12.0612.0512.0412.0312.0212.0112.0011.99
Fill
Dotplot for Fill
Machine
Machine 1
Machine 2
Machine 3
*When the sample sizes are large, we can make judgments about
both the equality of the standard deviations and the normality of the
underlying populations with a comparative boxplot.
26. 26
Example
6. Looking at normal plots for the samples, it
certainly appears reasonable to assume that
the samples from Machine’s 1 and 2 are
samples from normal distributions.
Unfortunately, the normal plot for the sample
from Machine 2 does not appear to be a
sample from a normal population. So as to
have a computational example, we shall
continue and finish the test, treating the
result with a “grain of salt.”
P-Value: 0.692
A-Squared: 0.235
Anderson-Darling Normality Test
N: 8
StDev: 0.0230078
Average: 12.0248
12.05512.04512.03512.02512.01512.00511.99511.985
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Machine 1
Normal Probability Plot
P-Value: 0.031
A-Squared: 0.729
Anderson-Darling Normality Test
N: 7
StDev: 0.0198890
Average: 12.0007
12.0312.0212.0112.0011.99
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Machine 2
Normal Probability Plot
P-Value: 0.702
A-Squared: 0.237
Anderson-Darling Normality Te
N: 9
StDev: 0.0164958
Average: 12.0259
12.0612.0512.0412.0312.0212.0112.00
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Machine 3
Normal Probability Plot
27. 27
Example
7. Computation:
SSTr 0.00301552= SSE 0.00825582=
MSTr 0.0015078= MSE 0.00039313=
1 1 1n 8, x 12.0248, s 0.02301= = =
3 3 3n 9, x 12.0259, s 0.01650= = =
2 2 2n 7, x 12.0007, s 0.01989= = =
x 12.018167=
1 2 3N n n n 8 7 9 24, k 3= + + = + + = =
1
2
MSTr 0.0015078
F 3.835
MSE 0.00039313
df treatment df k 1 3 1 2
df error df N k 24 3 21
= = =
= = − = − =
= = − = − =
28. 28
Example
8. P-value:
3.835
dfden / dfnum α 2
21 0.100 2.57
0.050 3.47
0.025 4.42
0.010 5.78
0.001 9.77
From the F table with
numerator df1 = 2 and
denominator df2 = 21 we
can see that
0.025 < P-value < 0.05
(Minitab reports this value
to be 0.038
1
2
MSTr 0.0015078
F 3.835
MSE 0.00039313
df treatment df k 1 3 1 2
df error df N k 24 3 21
= = =
= = − = − =
= = − = − =
Recall
1
2
MSTr 0.0015078
F 3.835
MSE 0.00039313
df treatment df k 1 3 1 2
df error df N k 24 3 21
= = =
= = − = − =
= = − = − =
Recall
29. 29
Example
9. Conclusion:
Since P-value > α = 0.01, we fail to reject
H0. We are unable to show that the mean
fills are different and conclude that the
differences in the mean fills of the
machines show no statistically significant
differences.
30. 30
Total Sum of Squares
The relationship between the three sums of
squares is SSTo = SSTr + SSE
which is often called the fundamental identity
for single-factor ANOVA.
Informally this relation is expressed as
Total variation = Explained variation + Unexplained variation
Total sum of squares, denoted by SSTo,
is given by
with associated df = N - 1.
all N obs.
2
SSTo (x x)= −∑
31. 31
Single-factor ANOVA Table
The following is a fairly standard way of
presenting the important calculations from
an single-factor ANOVA. The output from
most statistical packages will contain an
additional column giving the P-value.
32. 32
Single-factor ANOVA Table
The ANOVA table supplied by Minitab
One-way ANOVA: Fills versus Machine
Analysis of Variance for Fills
Source DF SS MS F P
Machine 2 0.003016 0.001508 3.84 0.038
Error 21 0.008256 0.000393
Total 23 0.011271
33. 33
Another Example
A food company produces 4 different
brands of salsa. In order to determine if the
four brands had the same sodium levels, 10
bottles of each Brand were randomly (and
independently) obtained and the sodium
content in milligrams (mg) per tablespoon
serving was measured.
The sample data are given on the next
slide.
Use the data to perform an appropriate
hypothesis test at the 0.05 level of
significance.
34. 34
Another Example
Brand A
43.85 44.30 45.69 47.13 43.35
45.59 45.92 44.89 43.69 44.59
Brand B
42.50 45.63 44.98 43.74 44.95
42.99 44.95 45.93 45.54 44.70
Brand C
45.84 48.74 49.25 47.30 46.41
46.35 46.31 46.93 48.30 45.13
Brand D
43.81 44.77 43.52 44.63 44.84
46.30 46.68 47.55 44.24 45.46
35. 35
Another Example
1. Let µ1, µ2 , µ3 and µ4 denote the true
mean sodium content per tablespoon in
each of the brands respectively.
2. H0: µ1 = µ2 = µ3 = µ4
3. Ha: At least two among are µ1, µ2, µ3 and
µ4 are different
4. Significance level: α = 0.05
5. Test statistic:
MSTr
F
MSE
=
36. 36
6. Looking at the following comparative
boxplot, it seems reasonable to assume
that the distributions have the equal σ’s as
well as the samples being samples from
normal distributions.
Another Example
BrandD
BrandC
BrandB
BrandA
49
48
47
46
45
44
43
42
Boxplots of Brand A - Brand D
(means are indicated by solid circles)
37. 37
Example
Treatment df = k - 1 = 4 - 1 = 3
7. Computation:
Brand k si
Brand A 10 44.900 1.180
Brand B 10 44.591 1.148
Brand C 10 47.056 1.331
Brand D 10 45.180 1.304
xi
= − + − + − + −
= − + −
+ − + −
=
2 2 2 2
1 1 2 2 3 3 4 4
2 2
2 2
SSTr n (x x) n (x x) n (x x) n (x x)
10(44.900 45.432) 10(44.591 45.432)
10(47.056 45.432) 10(45.180 45.432)
36.912
=x 45.432
38. 38
Example
7. Computation (continued):
Error df = N - k = 40 - 4 = 36
( ) ( ) ( ) ( )= − + − + − + −
= + + +
=
2 2 2 2
1 1 2 2 3 3 4 4
2 2 2 2
SSE n 1 s n 1 s n 1 s n 1 s
9(1.180) 9(1.148) 9(1.331) 9(1.304)
55.627
= = = = =SSTr
SSE
SSTr 36.912
MSTr 12.304df 3F 7.963
SSE 55.627MSE 1.5452
df 36
40. 40
Example
9. Conclusion:
Since P-value < α = 0.001, we reject H0.
We can conclude that the mean sodium
content is different for at least two of the
Brands.
We need to learn how to interpret the results and
will spend some time on developing techniques to
describe the differences among the µ’s.
41. 41
Multiple Comparisons
A multiple comparison procedure is a
method for identifying differences among the
µ’s once the hypothesis of overall equality
(H0) has been rejected.
The technique we will present is based on
computing confidence intervals for difference
of means for the pairs.
Specifically, if k populations or treatments are studied, we
would create k(k-1)/2 differences. (i.e., with 3 treatments one
would generate confidence intervals for µ1 - µ2, µ1 - µ3 and µ2 -
µ3.) Notice that it is only necessary to look at a confidence
interval for µ1 - µ2 to see if µ1 and µ2 differ.
42. 42
The Tukey-Kramer Multiple
Comparison Procedure
When there are k populations or treatments
being compared, k(k-1)/2 confidence
intervals must be computed. If we denote the
relevant Studentized range critical value by
q, the intervals are as follows:
For µi - µj:
Two means are judged to differ significantly
if the corresponding interval does not include
zero.
i j
i j
MSE 1 1
( ) q
2 n n
µ − µ ± +
43. 43
The Tukey-Kramer Multiple
Comparison Procedure
When all of the sample sizes are the same,
we denote n by n = n1 = n2 = n3 = … = nk,
and the confidence intervals (for µi - µj)
simplify to
i j
MSE
( ) q
n
µ − µ ±
44. 44
Example (continued)
Continuing with example dealing with the
sodium content for the four Brands of salsa we
shall compute the Tukey-Kramer 95% Tukey-
Kramer confidence intervals for µA - µB, µA - µC,
µA - µD, µB - µC, µB - µD and µC - µD.
= = = = = = =
= ÷
= =
A B C D
55.627
MSE 1.5452, n n n n n 10
36
Interpolating from the table
q 3.81
i.e. 60% of the way from 3.85 to 3.79
MSE 1.5452
q 3.81 1.498
n 10
45. 45
Example (continued)
Difference
95% Confidence
Limits
95% Confidence
Interval
µA - µB 0.309 ± 1.498 (-1.189, 1.807)
µA - µC -2.156 ± 1.498 (-3.654, -0.658)
µA - µD -0.280 ± 1.498 (-1.778, 1.218)
µB - µC -2.465 ± 1.498 (-3.963, -0.967)
µB - µD -0.589 ± 1.498 (-2.087, 0.909)
µC - µD 1.876 ± 1.498 (0.378, 3.374)
Notice that the confidence intervals for µA – µB, µA – µC
and µC – µD do not contain 0 so we can infer that the mean
sodium content for Brands C is different from Brands A, B
and D.
46. 46
Example (continued)
We also illustrate the differences with the
following listing of the sample means in
increasing order with lines underneath those
blocks of means that are indistinguishable.
Brand B Brand A Brand D Brand C
44.591 44.900 45.180 47.056
Notice that the confidence interval for µA – µC, µB – µC, and
µC – µD do not contain 0 so we can infer that the mean
sodium content for Brand C and all others differ.
47. 47
Minitab Output for Example
One-way ANOVA: Sodium versus Brand
Analysis of Variance for Sodium
Source DF SS MS F P
Brand 3 36.91 12.30 7.96 0.000
Error 36 55.63 1.55
Total 39 92.54
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev ------+---------+---------+---------+
Brand A 10 44.900 1.180 (-----*------)
Brand B 10 44.591 1.148 (------*-----)
Brand C 10 47.056 1.331 (------*------)
Brand D 10 45.180 1.304 (------*-----)
------+---------+---------+---------+
Pooled StDev = 1.243 44.4 45.6 46.8 48.0
48. 48
Minitab Output for Example
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0107
Critical value = 3.81
Intervals for (column level mean) - (row level mean)
Brand A Brand B Brand C
Brand B -1.189
1.807
Brand C -3.654 -3.963
-0.658 -0.967
Brand D -1.778 -2.087 0.378
1.218 0.909 3.374
49. 49
Simultaneous Confidence Level
The Tukey-Kramer intervals are created in a
manner that controls the simultaneous
confidence level.
For example at the 95% level, if the procedure is used
repeatedly on many different data sets, in the long run only
about 5% of the time would at least one of the intervals not
include that value of what it is estimating.
We then talk about the family error rate being 5% which is
the maximum probability of one or more of the confidence
intervals of the differences of mean not containing the true
difference of mean.
50. 50
Randomized Block Experiment
Suppose that experimental units (individuals
or objects to which the treatments are
applied) are first separated into groups
consisting of k units in such a way that the
units within each group are as similar as
possible. Within any particular group, the
treatments are then randomly allocated so
that each unit in a group receives a different
treatment. The groups are often called
blocks and the experimental design is
referred to as a randomized block design.
This slide as
well as all
remaining
slides in this
show refer to
materials that
are available
on the
Testbook
website.
51. 51
Example
When choosing a variety of melon to plant, one
thing that a farmer might be interested in is the
length of time (in days) for the variety to bear
harvestable fruit. Since the growing conditions
(soil, temperature, humidity) also affect this, a
farmer might experiment with three hybrid
melons (denoted hybrid A, hybrid B and hybrid
C) by taking each of the four fields that he
wants to use for growing melons and
subdividing each field into 3 subplots (1, 2 and
3) and then planting each hybrid in one subplot
of each field. The blocks are the fields and the
treatments are the hybrid that is planted. The
question of interest would be “Are the mean
times to bring harvestable fruit the same for all
three hybrids?”
52. 52
Assumptions and Hypotheses
The single observation made on any
particular treatment in a given block is
assumed to be selected from a normal
distribution. The variance of this distribution
is σ2
, the same for each block-treatment
combinations. However, the mean value
may depend separately both on the
treatment applied and on the block. The
hypotheses of interest are as follows:
H0: The mean value does not depend on
which treatment is applied
Ha: The mean value does depend on
which treatment is applied
53. 53
Summary of the Randomized
Block F Test
Notation:
Let
k = number of treatments
l = number of blocks
= average of all observations in
block I
ib
ix = average if all observations for
treatment i
= average of all kl observations in
the experiment (the grand mean)
x
54. 54
Summary of the Randomized
Block F Test
Sums of squares and associated df’s are as
follows.
Sum of Squares Symbol df Formula
Treatments SSTr k - 1
Blocks SSBl l - 1
Error SSE (k - 1)(l - 1)
Total SSTo kl - 1
2 2 2
1 2 kSSTr l[(x x) (x x) ... (x x) ]= −+−++−
SSE SSTo SSTr SSBl= − −
2 2 2
1 2 lSSBl k[(b x) (b x) ... (b x) ]= −+−++−
( )
2
all x
x x−∑
55. 55
Summary of the Randomized
Block F Test
SSE is obtained by subtraction through the
use of the fundamental identity
SSTo = SSTr + SSBl + SSE
The test is based on df1 = k - 1 and df2 = (k - 1)(l - 1)
Test statistic:
MSTr
F
MSE
=
where
SSTr SSE
MSTr and MSE
k 1 (k 1)(l 1)
= =
− − −
56. 56
The ANOVA Table for a
Randomized Block Experiment
Source of
Variation df
Sum of
Squares Mean Square F
Treatments k –1 SSTr
SSTr
MSTr
k 1
=
−
MSTr
F
MSE
=
Blocks l -1 SSBl
SSBl
MSBl
l 1
=
−
Error (k–1)(l–1) SSE
SSE
MSE
(k 1)(l 1)
=
− −
Total kl - 1 SSTo
57. 57
Multiple Comparisons
As before, in single-factor ANOVA, once
H0 has been rejected, declare that
treatments I and j differ significantly if the
interval
does not include zero, where q is based
on a comparison of k treatments and
error df = (k - 1)(l - 1).
i j
MSE
( ) q
l
µ − µ ±
58. 58
Example (Food Prices)
In an attempt to measure which of 3 grocery
chains has the best overall prices, it was felt
that there would be a great deal of variability
of prices if items were randomly selected
from each of the chains, so a randomized
block experiment was devised to answer the
question.
A list of standard items was developed
(typically a fairly large representative list
would be used, but do to a problem with
insufficient planning, only 7 items were left
“in the shopping cart.” and the price
recorded for each of these items in each of
the stores.
59. 59
Example (Food Prices)
Because of the problem that the Blocking
variable (the item) wasn’t set up with a well
designed, representative sample of the
items in a typical shopping basket, the
results should be taken with a “grain of salt.”
For the purposes of showing the
calculations, we shall treat this as being the
contents of a “representative” shopping
basket.
The data appear in the next slide along with
the hypotheses.
60. 60
Example (Food Prices)
H0: µA = µB = µC
Ha: At least two among are µA, µB and µC are
different
Product Store A Store B Store C
Tide (100 oz liquid detergent) 6.39 5.59 5.24
1 lb Land O'Lakes Butter 3.99 3.49 2.98
1 dozen Large Grade AA eggs 1.49 1.49 0.72
Tropicana (no pulp, non-conc) OJ (64 oz) 3.99 2.99 2.50
2 Liter Diet Coke 1.39 1.50 1.04
1 loaf Wonderbread 2.09 2.09 1.43
18 oz jar Skippy Peanut Butter 2.49 2.49 1.77
64. 64
den
num
df (k 1)(l 1) (3 1)(7 1) 12
df k 1 3 1 2
= − − = − − =
= − = − =
Conclusions
We can reject the hypothesis that
the mean prices are the same in
all three stores.
The actual differences can be
estimated with confidence
intervals.
MSTr 1.3881
F 22.85
MSE 0.060744
= = =
65. 65
Conclusions
We find q = 4.34 for the 95% Tukey confidence
intervals. The confidence intervals are
Difference
95% Confidence
Limits
95% Confidence
Interval
µA - µB 0.313 ± 0.404 (-0.091, 0.717)
µA - µC 0.879 ± 0.404 (0.474, 1.283)
µB - µC 0.566 ± 0.404 (0.161, 0.970)
Store C Store B Store A
$2.24 $2.81 $3.20
We therefore conclude that Store A is cheaper on the average
than Store B and Store C.
66. 66
Two-Factor ANOVA
Notation:
k = number of levels of factor A
l = number of levels of factor B
kl = number of treatments (each one a
combination of a factor A level and
a factor B level)
m = number of observations on each
treatment
67. 67
Two-Factor ANOVA Example
A grocery store has two stocking
supervisors, Fred & Wilma. The store is
open 24 hours a day and would like to
schedule these two individuals in a manner
that is most effective. To help determine how
to schedule them, a sample of their work
was obtained by scheduling each of them for
5 times in each of the three shifts and then
tracked the number of cases of groceries
that were emptied and stacked during the
shift. The data follows on the next slide.
69. 69
Interactions
There is said to be an interaction between
the factors, if the change in true average
response when the level of one factor
changes depend on the level of the other
factor.
One can look at the possible interaction
between two factors by drawing an
interactions plot, which is a graph of the
means of the response for one factor plotted
against the values of the other factor.
70. 70
Two-Factor ANOVA Example
Supervisor Day Swing Night
Fred 529.40 495.60 500.00 508.33
Wilma 507.80 527.00 585.60 540.13
Mean Output
for Each Shift
518.60 511.30 542.80 524.23
Mean Output
for Each
Supervisor
Shift
A table of the sample means for the 30
observations.
71. 71
Two-Factor ANOVA Example
Typically, only one of these interactions plots will
be constructed. As you can see from these
diagrams, there is a suggestion that Fred does
better during the day and Wilma is better at night
or during the swing shift. The question to ask is
“Are these differences significant?” Specifically is
there an interaction between the supervisor and
the shift.
Fred
Wilma
SwingNightDay
590
580
570
560
550
540
530
520
510
500
Shift
Supervisor
Mean
Interaction Plot - Data Means for Cases
Day
Night
Swing
WilmaFred
590
580
570
560
550
540
530
520
510
500
Supervisor
Shift
Mean
Interaction Plot - Data Means for Cases
72. 72
Interactions
If the graphs of true average responses are
connected line segments that are parallel, there
is no interaction between the factors. In this
case, the change in true average response
when the level of one factor is changed is the
same for each level of the other factor.
Special cases of no interaction are as follows:
1.The true average response is the same for
each level of factor A (no factor A main
effects).
2.The true average response is the same for
each level of factor B (no factor B main
effects).
73. 73
Basic Assumptions for Two-
Factor ANOVA
The observations on any particular treatment
are independently selected from a normal
distribution with variance σ2
(the same
variance for each treatment), and samples
from different treatments are independent of
one another.
74. 74
Two-Factor ANOVA Table
The following is a fairly standard way of
presenting the important calculations for an two-
factor ANOVA.
The fundamental identity is SSTo = SSA + SSB + SSAB +SSE
75. 75
Two-Factor ANOVA Example
Source df
Sum of
Squares
Mean
Square F
Shift 2 5437 2719 1.82
Supervisor 1 7584 7584 5.07
Interaction 2 14365 7183 4.80
Error 24 35878 1495
Total 29 63265
76. 76
Two-Factor ANOVA Example
Minitab output for the Two-Factor ANOVA
Two-way ANOVA: Cases versus Shift, Supervisor
Analysis of Variance for Cases
Source DF SS MS F P
Shift 2 5437 2719 1.82 0.184
Supervis 1 7584 7584 5.07 0.034
Interaction 2 14365 7183 4.80 0.018
Error 24 35878 1495
Total 29 63265
1. Test of H0: no interaction between
supervisor and Shift
There is evidence of an interaction.
