Post on 31-Dec-2015
description
2
One-Way Analysis of Variance
Y= DEPENDENT VARIABLE
(“yield”)
(“response variable”)
(“quality indicator”)
X = INDEPENDENT VARIABLE
(A possibly influential FACTOR)
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)
1-Way ANOVA 4
Completely Randomized Design (CRD)
4
• 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)
1-Way ANOVA 55
Example: Y = LIFETIME (HOURS)
BRAND3 replications
per level 1 2 3 4 5 6 7 8
1.8 4.2 8.6 7.0 4.2 4.2 7.8 9.0
5.0 5.4 4.6 5.0 7.8 4.2 7.0 7.4
1.0 4.2 4.2 9.0 6.6 5.4 9.8 5.8
2.6 4.6 5.8 7.0 6.2 4.6 8.2 7.4 5.8
1-Way ANOVA 77
Statistical ModelC “levels” OF BRAND
R observations for each level
Y11 Y12 • • • • • • •Y1R
Yij
Y21
•
•
•
•
•
•
YcI
•
•
•
•
•
1
2
•
•
•
•
C
1 2 • • • • • • • • R
Yij = + i + ij
i = 1, . . . . . , C
j = 1, . . . . . , R
YcR• • • • • • • •
1-Way ANOVA 88
Where
= OVERALL AVERAGE
i = index for FACTOR (Brand) LEVEL
j= index for “replication”
i = Differential effect associated with
ith level of X (Brand i) = i –
and ij = “noise” or “error” due to other factors associated with the (i,j)th data value.
i = AVERAGE associated with ith level of X (brand i)
= AVERAGE of i ’s.
1-Way ANOVA 99
Yij = + i + ij
By definition, i = 0C
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).
1-Way ANOVA 1010
Y • = Y i /C = “GRAND MEAN”
(assuming same # data points in each column)
(otherwise, Y • = mean of all the data)
i=1
c
Let Y1, Y2, etc., be level means
1-Way ANOVA 1111
MODEL: Yij = + i + ij
Y• estimates
Yi - Y • estimatesi (= i – ) (for all i)
These estimates are based on Gauss’ (1796)
PRINCIPLE OF LEAST SQUARES
and on COMMON SENSE
1-Way ANOVA 1212
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
<
<
1-Way ANOVA 1313
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
+
1-Way ANOVA 1414
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”]
(Yij - Y• )2 = R • (Yi - Y•)
2 + (Yij - Yi)
2C R
i=1 j=1 { { {i=1
C C R
i=1 j=1
TSS
TOTAL SUM OF SQUARES
=
=
SSB
SUM OF SQUARES BETWEEN SAMPLES
+
+
SSW (SSE)
SUM OF SQUARES WITHIN SAMPLES( ( (
( ((
1-Way ANOVA 1515
ANOVA TABLE
SOURCE OF VARIABILITY
SSQ DFMeansquare
(M.S.)
Between samples (due to brand)
Within samples (due to error)
SSB C - 1 MSBSSBC - 1
SSW (R - 1) • CSSW
(R-1)•C= MSW
=
TOTAL TSS RC -1
1-Way ANOVA 1616
Example: Y = LIFETIME (HOURS)
BRAND3 replications
per level 1 2 3 4 5 6 7 8
1.8 4.2 8.6 7.0 4.2 4.2 7.8 9.0
5.0 5.4 4.6 5.0 7.8 4.2 7.0 7.4
1.0 4.2 4.2 9.0 6.6 5.4 9.8 5.8
2.6 4.6 5.8 7.0 6.2 4.6 8.2 7.4 5.8
SSB = 3 ( [2.6 - 5.8]2 + [4.6 - 5.8]
2 + • • • + [7.4 - 5.8]2)
= 3 (23.04)
= 69.12
1-Way ANOVA 1717
(1.8 - 2.6)2 = .64 (4.2 - 4.6)2 =.16 (9.0 -7.4)2 = 2.56
(5.0 - 2.6)2 = 5.76 (5.4 - 4.6)2= .64 • • • • (7.4 - 7.4)2 = 0
(1.0 - 2.6)2 = 2.56 (4.2 - 4.6)2= .16 (5.8 - 7.4)2 = 2.56
8.96 .96 5.12
Total of (8.96 + .96 + • • • + 5.12),
SSW = 46.72
SSW =?
1-Way ANOVA 1818
ANOVA TABLE
Source of Variability
SSQ df M.S.
BRAND
ERROR
69.12
46.72
7
= 8 - 1
16
= 2 (8)
9.87
2.92
TOTAL 115.84 23
= (3 • 8) -1
1-Way ANOVA 1919
We can show:
E (MSB) = 2 +
“VCOL”{
MEASURE OF DIFFERENCES AMONG LEVEL
MEANS
RC-1
• (i - )2
{
i
((
E (MSW) = 2
(Assuming Yij follows N(j 2) and they are independent)
1-Way ANOVA 2020
E ( MSBC ) = 2 + VCOL
E ( MSW ) = 2
This suggests that
if MSBC
MSW > 1 ,
There’s some evidence of non-zero VCOL, or “level of X affects Y”
if MSBC
MSW< 1 ,
No evidence that VCOL > 0, or that “level of X affects Y”
1-Way ANOVA 2121
With HO: Level of X has no impact on Y
HI: Level of X does have impact on Y,
We need
MSBC
MSW> > 1
to reject HO.
1-Way ANOVA 2222
More Formally,
HO: 1 = 2 = • • • c = 0
HI: not all j = 0
OR
HO: 1 = 2 = • • • • c
HI: not all j are EQUAL
(All level means are equal)
1-Way ANOVA 2323
The distribution of
MSB
MSW= “Fcalc” , is
The F - distribution with (C-1, (R-1)C)degrees of freedom
Assuming
HO true.
C = Table Value
1-Way ANOVA 2424
In our problem:
ANOVA TABLE
Source of Variability
SSQ df M.S.
BRAND
ERROR
69.12
46.72
7
16
9.87
2.92 = 9.87 2.92
Fcalc
3.38
1-Way ANOVA 2626
Hence, at = .05, Reject Ho .
(i.e., Conclude that level of BRAND does have an impact on battery lifetime.)
1-Way ANOVA 2727
MINITAB INPUT life brand
1.8 1
5.0 1
1.0 1
4.2 2
5.4 2
4.2 2
. .
. .
. .
9.0 8
7.4 8
5.8 8
1-Way ANOVA 2828
ONE FACTOR ANOVA (MINITAB)
Analysis of Variance for life
Source DF SS MS F P
brand 7 69.12 9.87 3.38 0.021
Error 16 46.72 2.92
Total 23 115.84
MINITAB: STAT>>ANOVA>>ONE-WAY
Estimate of the common variance ^2
1-Way ANOVA 2929
1 2 3 4 5 6 7 8
0
1
2
3
4
5
6
7
8
9
10
brand
lifeBoxplots of life by brand
(means are indicated by solid circles)
1-Way ANOVA 3030
Assumptions
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
Normality plot& test
Residual plot& test
Run order plot
1-Way ANOVA 3131
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
1-Way ANOVA 3232
Minitab: stat>>basic statistics>>normality test
RESI1
Perc
ent
43210-1-2-3-4
99
95
90
80
70
605040
30
20
10
5
1
Mean -1.48030E-16StDev 1.425N 24AD 0.481P-Value 0.212
Probability Plot of RESI1Normal
1-Way ANOVA 3333
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
1-Way ANOVA 3434
Minitab: Stat >> Anova >> One-way
Fitted Value
Resi
dual
8765432
3
2
1
0
-1
-2
Residuals Versus the Fitted Values(response is life)
1-Way ANOVA 3535
Minitab: Stat>> Anova>> Test for Equal variancesbra
nd
95% Bonferroni Confidence Intervals for StDevs
8
7
6
5
4
3
2
1
403020100
Test Statistic 4.20P-Value 0.757
Test Statistic 0.31P-Value 0.938
Bartlett's Test
Levene's Test
Test for Equal Variances for life
1-Way ANOVA 3636
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
1-Way ANOVA 3737
Observation Order
Resi
dual
24222018161412108642
3
2
1
0
-1
-2
Residuals Versus the Order of the Data(response is life)
Minitab: Stat >> Anova >> One-way
1-Way ANOVA 3838
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
1-Way ANOVA 3939
Brand
A B C
32 32 28
30 32 21
30 26 15
29 26 15
26 22 14
23 20 14
20 19 14
19 16 11
18 14 9
12 14 8
BATTERY LIFETIME (hours)
(each column rank ordered, for simplicity)
Mean: 23.9 22.1 14.9 (here, irrelevant!!)
1-Way ANOVA 4040
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
1-Way ANOVA 4141
Brand
A B C
32 (29) 32 (29) 28 (24)
30 (26.5) 32 (29) 21 (18)
30 (26.5) 26 (22) 15 (10.5)
29 (25) 26 (22) 15 (10.5)
26 (22) 22 (19) 14 (7)
23 (20) 20 (16.5) 14 (7)
20 (16.5) 19 (14.5) 14 (7)
19 (14.5) 16 (12) 11 (3)
18 (13) 14 (7) 9 (2)
12 (4) 14 (7) 8 (1)T1 = 197 T2 = 178 T3 = 90
n1 = 10 n2 = 10 n3 = 10
Ranks in ( )
1-Way ANOVA 4242
TEST STATISTIC:
H =12
N (N + 1)• (Tj
2/nj ) - 3 (N + 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
K
j = 1
1-Way ANOVA 4343
H =
[ 12 197 2 178 2 902
30 (31) 10 10 10+ +
[ - 3 (31)
= 8.41
(with adjustment for ties, we get 8.46)
1-Way ANOVA 4444
We can show that, under HO , H is well
approximated by a 2 distribution with df = K - 1.
What do we do with H?
Here, df = 2, and at = .05, the critical value = 5.99
2
df
dfFdf,=
5.99 8.41 = H
= .05
Reject HO; conclude that mean lifetime NOT the same for all 3 BRANDS
8
1-Way ANOVA 4545
• Kruskal-Wallis Test: life versus brand
• Kruskal-Wallis Test on life
• brand N Median AveRank Z• 1 3 1.800 4.5 -2.09• 2 3 4.200 7.8 -1.22• 3 3 4.600 11.8 -0.17• 4 3 7.000 16.5 1.05• 5 3 6.600 13.3 0.22• 6 3 4.200 7.8 -1.22• 7 3 7.800 20.0 1.96• 8 3 7.400 18.2 1.48• Overall 24 12.5
• H = 12.78 DF = 7 P = 0.078• H = 13.01 DF = 7 P = 0.072 (adjusted for ties)
Minitab: Stat >> Nonparametrics >> Kruskal-Wallis