Upload
richard-ferreria
View
212
Download
1
Tags:
Embed Size (px)
DESCRIPTION
Chapter 15
Citation preview
1 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Chapter 15
The Analysis of Variance
2 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
70
60
50
40
FertilizerY
ield
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Sta
tistic
s
Psy
cho
log
y
Eco
nom
ics
Bus
ine
ss
85
75
65
SubjectP
rice
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 …21 2
2 2k
Sample variance …21s 2
2s 2ks
Sample size n1 n2 … nk
11 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
N = n1 + n2 + … + nk (Total number of observations in the data set)
ANOVA Notation
Tx grand mean
N
T = grand total = sum of all N observations
1 1 2 2 k kn x n x n x
12 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Assumptions for ANOVA
1. Each of the k populations or treatments, the response distribution is normal.
2. 1 = 2 = … = k (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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Definitions
2 2 2
1 1 2 2 k kSSTr n x x n x x n x x
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 21 1 2 2 k kSSE n 1 s n 1 s n 1 s
14 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 = SSTrk 1
mean square for error = MSE = SSEN k
15 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
ExampleThree 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 112.033 11.985 12.009 12.00912.033 12.025 12.054 12.050
Machine 212.031 11.985 11.998 11.99211.985 12.027 11.987
Machine 312.034 12.021 12.038 12.05812.001 12.020 12.029 12.01112.021
16 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example
2 2 21 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
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example
SSTrk 1
mean square for treatments = MSTr =
SSTr 0.00301552MSTr 0.0015078
k 1 3 1
mean square for error = MSE = SSEN k
SSE 0.0082579MSE 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 ’s, the larger µMSTr will be relative to µMSE.
20 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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: MSTrF
MSE
21 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Consider the earlier example involving the three filling machines.
Machine 112.033 11.985 12.009 12.009 12.03312.025 12.054 12.050
Machine 212.031 11.985 11.998 11.992 11.98512.027 11.987
Machine 312.034 12.021 12.038 12.058 12.00112.020 12.029 12.011 12.021
23 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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: = 0.01
5. Test statistic:MSTr
FMSE
25 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example
6. Looking at the comparative dotplot, it seems reasonable to assume that the distributions have the same ’s. We shall look at the normality assumption on the next slide.*
12.0612.0512.0412.0312.0212.0112.0011.99
Fill
Dotplot for FillMachine
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example6. 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: 0.692A-Squared: 0.235
Anderson-Darling Normality Test
N: 8StDev: 0.0230078Average: 12.0248
12.05512.04512.03512.02512.01512.00511.99511.985
.999
.99
.95
.80
.50
.20
.05
.01
.001
Pro
bab
ility
Machine 1
Normal Probability Plot
P-Value: 0.031A-Squared: 0.729
Anderson-Darling Normality Test
N: 7StDev: 0.0198890Average: 12.0007
12.0312.0212.0112.0011.99
.999
.99
.95
.80
.50
.20
.05
.01
.001
Pro
bab
ility
Machine 2
Normal Probability Plot
P-Value: 0.702A-Squared: 0.237
Anderson-Darling Normality Test
N: 9StDev: 0.0164958Average: 12.0259
12.0612.0512.0412.0312.0212.0112.00
.999
.99
.95
.80
.50
.20
.05
.01
.001
Pro
bab
ility
Machine 3
Normal Probability Plot
27 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example7. 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 0.0015078F 3.835
MSE 0.00039313df treatment df k 1 3 1 2
df error df N k 24 3 21
28 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example8. P-value:
3.835
dfden / dfnum 2
21 0.100 2.570.050 3.470.025 4.420.010 5.780.001 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 0.0015078F 3.835
MSE 0.00039313df treatment df k 1 3 1 2
df error df N k 24 3 21
Recall
1
2
MSTr 0.0015078F 3.835
MSE 0.00039313df treatment df k 1 3 1 2
df error df N k 24 3 21
Recall
29 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example
9. Conclusion:
Since P-value > = 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Total Sum of Squares
The relationship between the three sums of squares is SSTo = SSTr + SSEwhich 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.
2SSTo (x x)
31 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 PMachine 2 0.003016 0.001508 3.84 0.038Error 21 0.008256 0.000393Total 23 0.011271
33 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Another Example
Brand A43.85 44.30 45.69 47.13 43.3545.59 45.92 44.89 43.69 44.59
Brand B42.50 45.63 44.98 43.74 44.9542.99 44.95 45.93 45.54 44.70
Brand C45.84 48.74 49.25 47.30 46.4146.35 46.31 46.93 48.30 45.13
Brand D43.81 44.77 43.52 44.63 44.8446.30 46.68 47.55 44.24 45.46
35 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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: = 0.05
5. Test statistic:MSTr
FMSE
36 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
6. Looking at the following comparative boxplot, it seems reasonable to assume that the distributions have the equal ’s as well as the samples being samples from normal distributions.
Another Example
Bra
nd
D
Bra
nd
C
Bra
nd
B
Bra
nd
A
49
48
47
46
45
44
43
42
Boxplots of Brand A - Brand D(means are indicated by solid circles)
37 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Treatment df = k - 1 = 4 - 1 = 3
7. Computation:Brand k si
Brand A 10 44.9001.180Brand B 10 44.5911.148Brand C 10 47.0561.331Brand D 10 45.1801.304
xi
2 2 2 21 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example7. Computation (continued):
Error df = N - k = 40 - 4 = 36
2 2 2 21 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 36.912MSTr 12.304df 3F 7.963
SSE 55.627MSE 1.5452df 36
39 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example8.P-value:
F = 7.96 with dfnumerator= 3 and dfdenominator= 36
Using df = 30 we find
P-value < 0.001
7.96
40 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example
9. Conclusion:
Since P-value < = 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 i - j:
Two means are judged to differ significantly if the corresponding interval does not include zero.
i ji j
MSE 1 1( ) q
2 n n
43 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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
44 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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
55.627MSE 1.5452, n n n n n 10
36Interpolating from the table
q 3.81 i.e. 60% of the way from 3.85 to 3.79
MSE 1.5452q 3.81 1.498
n 10
45 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Example (continued)
Difference95% Confidence
Limits95% Confidence
Interval
A - B 0.309 ± 1.498 (-1.189, 1.807)
A - C -2.156 ± 1.498 (-3.654, -0.658)
A - D -0.280 ± 1.498 (-1.778, 1.218)
B - C -2.465 ± 1.498 (-3.963, -0.967)
B - D -0.589 ± 1.498 (-2.087, 0.909)
C - D 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Minitab Output for Example
One-way ANOVA: Sodium versus Brand
Analysis of Variance for Sodium Source DF SS MS F PBrand 3 36.91 12.30 7.96 0.000Error 36 55.63 1.55Total 39 92.54 Individual 95% CIs For Mean Based on Pooled StDevLevel 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Minitab Output for Example
Tukey's pairwise comparisons
Family error rate = 0.0500Individual 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
ExampleWhen 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 2, 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Summary of the Randomized Block F Test
Notation:
Letk = 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 21 2 kSSTr l[(x x) (x x) ... (x x) ]
SSE SSTo SSTr SSBl
2 2 21 2 lSSBl k[(b x) (b x) ... (b x) ]
2
all x
x x
55 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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
FMSE
whereSSTr SSE
MSTr and MSEk 1 (k 1)(l 1)
56 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
The ANOVA Table for a Randomized Block Experiment
S o u r c e o f V a r i a t i o n d f
S u m o f S q u a r e s M e a n S q u a r e F
T r e a t m e n t s k – 1 S S T r
S S T rM S T r
k 1
M S T rF
M S E
B l o c k s l - 1 S S B l
S S B lM S B l
l 1
E r r o r ( k – 1 ) ( l – 1 ) S S E
S S EM S E
( k 1) ( l 1)
T o t a l k l - 1 S S T o
57 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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
58 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 CTide (100 oz liquid detergent) 6.39 5.59 5.241 lb Land O'Lakes Butter 3.99 3.49 2.981 dozen Large Grade AA eggs 1.49 1.49 0.72Tropicana (no pulp, non-conc) OJ (64 oz) 3.99 2.99 2.502 Liter Diet Coke 1.39 1.50 1.041 loaf Wonderbread 2.09 2.09 1.4318 oz jar Skippy Peanut Butter 2.49 2.49 1.77
61 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Calculations
Treatments: k 3 Blocks: l 7
57.15x 2.7214
21
2 2 21 2 3
2 2 2
SSTr l[(x x) (x x) (x x) ]
7[(3.1186 2.7214) (2.8057 2.7214) (2.2400 2.7214) ]
7[0.15772 0.00710 0.23177] 7[0.39660] 2.7762
SSTr 2.7762MSTr 1.3881
k 1 3 1
62 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Calculations
2 2 21 2 7
2 2 2
2 2 2
2
SSBl k[(b x) (b x) ... (b x) ]
3[(5.7400 2.7214) (3.4867 2.7214) (1.2333 2.7214)
(3.1600 2.7214) (1.3100 2.7214) (1.8700 2.7214)
(2.2500 2.7214) ]
3[9.1118+0.58559+2.21443+0.19234+1.9921+0.
72493+0.22224]
3[15.04344] 45.1303
SSTr 2.7762MSTr 1.3881
k 1 3 1
63 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Calculations
SSE SSTo SSTr SSBl
48.6356 2.7762 45.1303
0.72893
SSE 0.72893MSE
(k 1)(l 1) (3 1)(7 1)
0.06074
MSTr 1.3881F 22.85
MSE 0.060744
den
num
df (k 1)(l 1) (3 1)(7 1) 12
df k 1 3 1 2
64 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 1.3881F 22.85
MSE 0.060744
65 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Conclusions
We find q = 4.34 for the 95% Tukey confidence intervals. The confidence intervals are
Difference95% Confidence
Limits95% Confidence
Interval
A - B 0.313 ± 0.404 (-0.091, 0.717)
A - C 0.879 ± 0.404 (0.474, 1.283)
B - C 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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.
68 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Two-Factor ANOVA Example
Supervisor Day Swing Night495 547 481 457 500 578 504 496 485607 517 515 428 518 497481 520 498 508 471 560 572 550 583533 507 518 578 625 598
Shift
Fred
Wilma
69 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Two-Factor ANOVA Example
Supervisor Day Swing NightFred 529.40 495.60 500.00 508.33Wilma 507.80 527.00 585.60 540.13Mean 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Two-Factor ANOVA ExampleTypically, 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
Mea
n
Interaction Plot - Data Means for Cases
Day Night Swing
WilmaFred
590
580
570
560
550
540
530
520
510
500
Supervisor
Shift
Mea
n
Interaction Plot - Data Means for Cases
72 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Basic Assumptions for Two-Factor ANOVA
The observations on any particular treatment are independently selected from a normal distribution with variance 2 (the same variance for each treatment), and samples from different treatments are independent of one another.
74 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Two-Factor ANOVA TableThe 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 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
Two-Factor ANOVA Example
Source dfSum of Squares
Mean Square F
Shift 2 5437 2719 1.82Supervisor 1 7584 7584 5.07Interaction 2 14365 7183 4.80Error 24 35878 1495Total 29 63265
76 © 2008 Brooks/Cole, a division of Thomson Learning, Inc.
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 PShift 2 5437 2719 1.82 0.184Supervis 1 7584 7584 5.07 0.034Interaction 2 14365 7183 4.80 0.018Error 24 35878 1495Total 29 63265
1. Test of H0: no interaction between supervisor and Shift
There is evidence of an interaction.