Outline One-Factor Analysis of Variance (11.1) Randomized Block - - PDF document
2/8/2007 219323 Probability and Statistics for Software Statistics for Software and Knowledge Engineers Lecture 11: The Analysis of Variance Th A l i f V i Monchai Sopitkamon, Ph.D. Outline One-Factor Analysis of Variance (11.1)
Monchai Sopitkamon, Ph.D.
One-Factor Analysis of Variance (11.1) Randomized Block Designs (11.2)
One-Factor Layouts (11.1.1) Partitioning the Total Sum of Squares Partitioning the Total Sum of Squares
The Analysis of Variance Table (11.1.3) Pairwise Comparisons of the Factor Level
Sample Size Determination (11.1.5)
Model Assumptions (11.1.6)
Comparing three or more population means Objectives: Objectives:
– To determine if the population means are unequal – To determine which population means are different and by how much
Completely randomized design – a set of
Uses the analysis of variance (ANOVA)
Randomizer
Randomizer
K populations w/ unknown population
If k = 1, one-sample inference problems
If k = 2, two-sample comparison
If k ≥ 3, one-factor ANOVA problems (this
xij jth observation from the ith population One factor layout If n1 = n2 = … = nk balanced data set, else unbalanced data set
Estim ating the population ( factor level) m eans
Hypothesis testing
– Null hypothesis: Null hypothesis: H0 : μ1 = … = μk population means are all equal – Alternative hypothesis: HA : μi ≠ μj for some i and j at least two of the population means are not equal Acceptance of H0 means that no evidence that any
Rejection of H0 means that there is evidence that there is some of the population means are unequal, and so it is not plausible to assume that the population means are all equal
Partition of total sum of squares for com pletely random ized one factor layout
Total Sum of Squares (SST) – a measure of
2 1 1 ..
= =
k i n j ij
i
1 1 k i n j ij
i
1 1 .. T i j
= =
ij
i
= k i i
1
Levels: A B C and D
A B C D 11 12 18 11 13 14 16 12 17 17 18 16 Page Replacement Algorithm
programs under each CPU configuration .
Number of Processors 17 17 18 16 17 19 20 15 15 21 22 14 16 18 15 17 14 19 17 13 10 18 21 16 12 16 16 17 14 18 20 18
A B C D 11 12 18 11 Page Replacement Algorithm Number of Processors 13 14 16 12 17 17 18 16 17 19 20 15 15 21 22 14 16 18 15 17 14 19 17 13 10 18 21 16 12 16 16 17
14 18 20 18 Grand Mean 16.075
Partition of total sum of squares for com pletely random ized one factor layout
Treatment Sum of Squares (SSTr) – a
=
k i i i
1 2 ..
where
k n ij
i
⋅ i
sample mean corresponding to group or level i
i
number of observations in group or level i
1 1 .. T i j ij
= =
A B C D 11 12 18 11 13 14 16 12 Page Replacement Algorithm
Number of Processors
13 14 16 12 17 17 18 16 17 19 20 15 15 21 22 14 16 18 15 17 14 19 17 13 10 18 21 16 12 16 16 17 14 18 20 18 Mean 13.9 17.2 18.3 14.9 Grand Mean 16.075
Partition of total sum of squares for com pletely random ized one factor layout
Error Sum of Squares (SSE) – a measure of
2 1 1
= = ⋅
k i n j i ij
i
⋅ i
sample mean corresponding to group or level i
ij
j-th observation in group or level i
A B C D 11 12 18 11 13 14 16 12 Page Replacement Algorithm Number of Processors 13 14 16 12 17 17 18 16 17 19 20 15 15 21 22 14 16 18 15 17 14 19 17 13 10 18 21 16 12 16 16 17 14 18 20 18 14 18 20 18 Mean 13.9 17.2 18.3 14.9
SSE = (11-13.9)2 + …+(14-13.9)2 + (12-17.2)2 + … + (18-17.2)2 + (18-18.3)2 + … + (20-18.3)2 + (11-14.9)2 + … + (18-14.9)2 = 213.5
SST = SSTr + SSE = 123.275 + 213.5 = 336.775
I nterpretation of th f the sum of squares for treatm ents and the sum of squares for error
Dependence of p-value on the sum of squares for treatm ents and the sum of squares for error
T −
are all equal is p-value = P(X ≥ F) where F-statistic is and RV X has an distribution
MSE MSTr F =
k n k
T
F
− − , 1
P-value calculation for
analysis of variance table variance table
A B C D 11 12 18 11 13 14 16 12 17 17 18 16 Page Replacement Algorithm
Number of Processors
17 17 18 16 17 19 20 15 15 21 22 14 16 18 15 17 14 19 17 13 10 18 21 16 12 16 16 17 14 18 20 18 Mean 13.9 17.2 18.3 14.9 Grand Mean 16.075 SSA 47 30625 12 65625 49 50625 13 80625 123 275
SST
SSA 47.30625 12.65625 49.50625 13.80625 123.275 SSW 213.5 SST 336.775 MSA 41.091667 MSW 5.9305556 F 6.93 df numer. 3 df denom. 36 Fu 2.87 (from table)
MSTr
F-crit MSE SSTr SSE
Analysis of variance table for one factor layout Reject H0 if F-statistic > F-critical (from
Excel sheet
If H0 is rejected (not all population means
Need to do Tukey multiple comparisons to
– By computing the differences μI – μj for 1 ≤ i < j ≤ k among all k(k – 1)/2 pairs of factor level means means
Compute confidence intervals
− ⋅ ⋅ j i k n k j i j i
T
, , α
If the CI for the difference μI – μj contains 0,
If the CI for the difference μI – μj does not
The CI indicates by how much the factor
Pairw ise confidence intervals for the carpet fiber blends experim ent
Schem atic presentation of the results of the carpet fiber blends experim ent
Conclusions
– Fiber types 1, 3, and 5 are indistinguishable from one Fiber types 1, 3, and 5 are indistinguishable from one another – These three fibers are not significantly different – Fiber types 2 and 6 and indistinguishable, as are fiber types 4 and 6, but type 2 have average scores larger than that of type 4 – The best fiber type is either type 2 or type 6
The power of the test of H0 that the factor
This power is the prob that the H0 is rejected
An increase in sample sizes also results in a
The minimum number of sample sizes, n, to
2 2 , ,
k n k
T −
α
Ex.63 pg.507: Roadway Base Aggregates
3564 10 1479 81 . 3 2 2
, ,
= × × = =
−
n MSE q L
k n k
T
α
2 2 2 2 2 , ,
−
k n k
T
α
The ANOVA for a one-factor layout is based on
– The observations are independent and have normal distribution with the same variance.
Independence is ensured by randomization of
Normality: ANOVA F test is robust as long as
Homogeneity of variance computed from
– If variances are much different, make pairwise comparisons (using the general procedure from Chapter 9) between the k factor levels, one pair at a time, and use a small CI (90%).
2 i
One-Factor Layouts with Blocks (11.2.1) Partitioning the Total Sum of Squares Partitioning the Total Sum of Squares
The Analysis of Variance Table (11.2.3) Pairwise Comparisons of the Factor Level
Model Assumptions (11.2.5)
A complete randomized block design – an
Randomizer
Randomizer
A random ized block design data set
Number of Processors
Partition of the total sum of squares for a random ized block design
Total Sum of Squares (SST):
2 1 1 ..
= =
k i b j ij
k b ij
1 1 ..
i j j
= =
Number of Processors
Partition of the total sum of squares for a random ized block design
Treatment Sum of Squares (SSTr)
2 1 ..
= ⋅ −
k i i
b
1 .
j ij i
=
Number of Processors
SSTr = 5 * [ (14.0-15.61)2+(15.1-15.61)2+…+(13.1-15.61)2] = 155.018
Partition of the total sum of squares for a random ized block design
Block Sum of Squares (SSBl)
2 1 .. .
=
b j j
k
1 .
i ij j
=
Number of Processors
SSBl = 4 * [(13.0-15.61)2+(14.5-15.61)2+…+(16.4-15.61)2] = 76.133
Partition of the total sum of squares for a random ized block design
Error Sum of Squares (SSE)
2 1 1 .. . .
= =
k i b j j i ij
…. (11.0-13.1-13.0+15.61)2+(13.5-13.1-16.4+15.61)2 = 1.287
Mean Squares for Treatments (MSTr)
Mean Squares for Blocks (MSBl)
Mean Square Error (MSE)
A p-value for the null hypothesis that the factor
MSE MSTr F
T = ) 1 )( 1 ( , 1 − − − b k k
F
MSE MSBl FB =
) 1 )( 1 ( , 1 − − − b k b
F
Analysis of variance table for a random ized block design Reject H0 if F-statistic > F-critical (from
Excel sheet
Number of Processors
SSTr
MSTr F-crit
=MSTr/MSE
If H0 is rejected (not all population means
Need to do Tukey multiple comparisons to
– By computing the differences μI – μj for 1 ≤ i < j ≤ k among all k(k – 1)/2 pairs of factor level means means
Compute confidence intervals
b k k j i j i ) 1 )( 1 ( , , − − ⋅ ⋅
α
Ex.55 pg.525: Heart Rate Reduction
) 1 )( 1 ( , ,
− −
b k k α
CIs of Diff lower upper μ1 − μ2
heart rate reduction between 2-14.6%
μ1 − μ3
μ2 − μ3
3.81
compared to low dose
reduction between 4.5-17.1% compared to low dose No significant difference in reduction between high and med doses!
Schem atic presentation of the results of the heart rate reductions exam ple
Modeling assumption
xij = μ + αI + βj + εij xij μ + αI + βj + εij where εij are error terms w/ independently distributed, or εij ∼ N(0, σ2) – normal distribution w/ common variance assumption of error terms are the same as for the completely randomized design
No interaction between the factor levels and