1. The greatest blessing in life is in giving and not taking.
1-Way Anova
1-Way ANOVA 1 1

2. One-Way Analysis of Variance
Y= DEPENDENT VARIABLE
(“yield”)
(“response variable”)
(“quality indicator”)
X = INDEPENDENT VARIABLE
(A possibly influential FACTOR) 2 2

3. = Many other factors (possibly, some we’re unaware of)
OBJECTIVE: To determine the impact of X on Y
Mathematical Model:
Y = f (x, ) , where  = (impact of) all factors other than X
Ex: Y = Battery Life
(hours)
X = Brand of Battery
= Many other factors (possibly, some we’re unaware of)

4. Completely Randomized Design (CRD)
Goal: to study the effect of Factor X
The same # of observations are taken randomly and independently from the individuals at each level of Factor X
i.e. n1=n2=…nc (c levels)
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5. Example: Y = LIFETIME (HOURS) BRAND
3 replications per level
5.8
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6. Analysis of Variance
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7. R observations for each level
Statistical Model
C “levels” OF BRAND
R observations for each level
• • •  •  •  • • • R 1 2 • C
Y11 Y12 • • • • • • •Y1R
Yij = + i + ij
i = 1, , C
j = 1, , R
Y21 •
YcI •
Yij
YcR
•   •  •   •    •   •    •    • 
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8. mi = AVERAGE associated with ith level of X (brand i)
Where
= OVERALL AVERAGE
i = index for FACTOR (Brand) LEVEL
j= index for “replication”
i = Differential effect associated with
ith level of X (Brand i) = mi – m
and ij = “noise” or “error” due to other factors associated with the (i,j)th data value.
mi = AVERAGE associated with ith level of X (brand i)
m = AVERAGE of mi ’s.
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9. The experiment produces
Yij =  + i + ij
By definition, i = 0 C
i=1
The experiment produces
R x C Yij data values.
The analysis produces estimates of ,c. (We can then get estimates of the ij by subtraction).
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10. Let Y1, Y2, etc., be level means
Y • = Y i /C = “GRAND MEAN”
(assuming same # data points in each column)
(otherwise, Y • = mean of all the data) c
i=1
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11. These estimates are based on Gauss’ (1796) PRINCIPLE OF LEAST SQUARES
MODEL: Yij =  + i + ij
Y• estimates 
Yi - Y • estimatesi (= mi – m)
(for all i)
These estimates are based on Gauss’ (1796)
PRINCIPLE OF LEAST SQUARES
and on COMMON SENSE
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12. If you insert the estimates into the MODEL,
MODEL: Yij =  + j + ij
If you insert the estimates into the MODEL,
(1) Yij = Y • + (Yj - Y • ) + ij.
<
it follows that our estimate of ij is
(2) ij = Yij – Yj, called residual
<
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13. { { { Then, Yij = Y• + (Yi - Y• ) + ( Yij - Yi)
or, (Yij - Y• ) = (Yi - Y•) + (Yij - Yi ) { { {
(3)
TOTAL
VARIABILITY
in Y
Variability
in Y
associated
with X
Variability
in Y
associated
with all other factors + =
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14. SUM OF SQUARES BETWEEN SAMPLES SUM OF SQUARES WITHIN SAMPLES
If you square both sides of (3), and double sum both sides (over i and j), you get, [after some unpleasant algebra, but lots of terms which “cancel”] {{
C R C
C R
(Yij - Y• )2 = R •  (Yi - Y•)2 + (Yij - Yi)2 {
i=1 j=1
i=1
i=1 j=1
TSS
TOTAL SUM OF SQUARES (
SSB
SUM OF SQUARES BETWEEN SAMPLES = +
SSW (SSE)
SUM OF SQUARES WITHIN SAMPLES ( ( ( ( (
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15. ANOVA TABLE SSB SSB C - 1 = MSB C - 1 SSW MSW SSW (R - 1) • C =
SOURCE OF VARIABILITY
SSQ DF
Mean
square
(M.S.)
Between samples (due to brand)
SSB
SSB
C - 1 =
MSB
C - 1
Within samples (due to error)
SSW
MSW
SSW
(R - 1) • C =
(R-1)•C
TOTAL TSS RC -1
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16. Example: Y = LIFETIME (HOURS) BRAND
3 replications per level
5.8
SSB = 3 ( [ ]2 + [ ] 2 + • • • + [ ]2)
= 3 (23.04) =
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17. SSW =?
( )2 = ( )2 = ( )2 = 2.56
( )2 = ( )2= .64 • • • • ( )2 = 0
( )2 = ( )2= ( )2 = 2.56
Total of ( • • • ),
SSW = 46.72
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18. ANOVA TABLE Source of Variability df SSQ M.S. BRAND 69.12 7 9.87 ERROR=
9.87
ERROR
46.72 16
= 2 (8)
2.92
TOTAL
= (3 • 8) -1
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19. { ( ( { We can show: i “VCOL” E (MSB) = 2 + R C-1 E (MSW) = 2
MEASURE OF DIFFERENCES AMONG LEVEL MEANS ( R (
• (i - )2 {
C-1 i
E (MSW) = 2
(Assuming Yij follows N(j , 2) and they are independent)
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20. > 1 , < 1 , E ( MSBC ) = 2 + VCOL E ( MSW ) = 2
This suggests that
There’s some evidence of non-zero VCOL, or “level of X affects Y”
if MSBC
> 1 ,
MSW
if MSBC
No evidence that VCOL > 0, or that “level of X affects Y”
< 1 ,
MSW
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21. With HO: Level of X has no impact on Y
HI: Level of X does have impact on Y,
We need
MSBC
> > 1
MSW
to reject HO.
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22. (All level means are equal) HO: 1 = 2 = • • • • c
More Formally,
HO: 1 = 2 = • • • c = 0
HI: not all j = 0 OR
(All level means are equal)
HO: 1 = 2 = • • • • c
HI: not all j are EQUAL
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23.  The distribution of MSB = “Fcalc” , is MSW
The F - distribution with (C-1, (R-1)C)
degrees of freedom 
Assuming
HO true.
C = Table Value
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24. In our problem: ANOVA TABLE Source of Variability SSQ df M.S. Fcalc
BRAND
69.12 7
9.87
3.38
ERROR
46.72 16 =
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25. F table: table 8
 = .05
C =
(7,16 DF)
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26. Hence, at  = .05, Reject Ho .
(i.e., Conclude that level of BRAND does have an impact on battery lifetime.)
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27. MINITAB INPUT life brand 1.8 1 5.0 1 1.0 1 4.2 2 5.4 2 . . 9.0 8 7.4 8
1.8 1
5.0 1
1.0 1
4.2 2
5.4 2
. .
9.0 8
7.4 8
5.8 8
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28. ONE FACTOR ANOVA (MINITAB)
MINITAB: STAT>>ANOVA>>ONE-WAY
Analysis of Variance for life
Source DF SS MS F P
brand
Error
Total
Estimate of the common variance s^2
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29. 1-Way ANOVA 29

30. Assumptions Yij = + i + ij 1.) the ij are indep. random variables
MODEL:
Yij = + i + ij
1.) the ij are indep. random variables
2.) Each ij is Normally Distributed E(ij) = 0 for all i, j
3.) 2(ij) = constant for all i, j
Run order plot
Normality plot
& test
Residual plot
& test
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31. Normal probability plot & normality test of residuals
Diagnosis: Normality
The points on the normality plot must more or less follow a line to claim “normal distributed”.
There are statistic tests to verify it scientifically.
The ANOVA method we learn here is not sensitive to the normality assumption. That is, a mild departure from the normal distribution will not change our conclusions much.
Normal probability plot & normality test of residuals
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32. Minitab: stat>>basic statistics>>normality test
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33. Diagnosis: Constant Variances
The points on the residual plot must be more or less within a horizontal band to claim “constant variances”.
There are statistic tests to verify it scientifically.
The ANOVA method we learn here is not sensitive to the constant variances assumption. That is, slightly different variances within groups will not change our conclusions much.
Tests and Residual plot: fitted values vs. residuals
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34. Minitab: Stat >> Anova >> One-way
1-Way ANOVA 34

35. Minitab: Stat>> Anova>> Test for Equal variances
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36. Diagnosis: Randomness/Independence
The run order plot must show no “systematic” patterns to claim “randomness”.
There are statistic tests to verify it scientifically.
The ANOVA method is sensitive to the randomness assumption. That is, a little level of dependence between data points will change our conclusions a lot.
Run order plot: order vs. residuals
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37. Minitab: Stat >> Anova >> One-way
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38. KRUSKAL - WALLIS TEST (Non - Parametric Alternative)
HO: The probability distributions are identical for each level of the factor
HI: Not all the distributions are the same
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39. BATTERY LIFETIME (hours)
Brand
A B C
BATTERY LIFETIME (hours)
(each column rank ordered, for simplicity)
Mean: (here, irrelevant!!)
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40. HO: no difference in distribution. among the three brands with
HO: no difference in distribution among the three brands with respect to battery lifetime
HI: At least one of the 3 brands differs in distribution from the others with respect to lifetime
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41. Ranks in ( ) Brand A B C 32 (29) 32 (29) 28 (24)
32 (29) (29) (24)
30 (26.5) (29) (18)
30 (26.5) (22) (10.5)
29 (25) (22) (10.5)
26 (22) (19) (7)
23 (20) (16.5) 14 (7)
20 (16.5) (14.5) (7)
19 (14.5) (12) (3)
18 (13) (7) (2)
12 (4) (7) (1)
T1 = T2 = T3 = 90
n1 = n2 = n3 = 10
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42. TEST STATISTIC: 12 •  (Tj2/nj ) - 3 (N + 1) H = N (N + 1)K 12
H =
•  (Tj2/nj ) - 3 (N + 1)
N (N + 1)
j = 1
nj = # data values in column j
N = nj
K = # Columns (levels)
Tj = SUM OF RANKS OF DATA ON COL j When all DATA COMBINED
(There is a slight adjustment in the formula as a function of the number of ties in rank.) K
j = 1
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43. [ [ H = = 8.41 (with adjustment for ties, we get 8.46)
30 (31) [ + +
- 3 (31)
= 8.41
(with adjustment for ties, we get 8.46)
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44. What do we do with H?
We can show that, under HO , H is well approximated by a 2 distribution with df = K - 1.
Here, df = 2, and at = .05, the critical value = 5.99
= H
 = .05
c21-adf df
F1-adf, = 8
Reject HO; conclude that mean lifetime NOT the same for all 3 BRANDS
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45. Minitab: Stat >> Nonparametrics >> Kruskal-Wallis
Kruskal-Wallis Test: life versus brand
Kruskal-Wallis Test on life
brand N Median AveRank Z
Overall
H = DF = 7 P = 0.078
H = DF = 7 P = (adjusted for ties)
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