Transcript One-way Anova: Inferences about More than Two Population Means

Always be mindful of the kindness and not the faults of others.
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One-way Anova: Inferences about
More than Two Population Means
Model
and test for oneway anova
Assumption checking
Nonparamateric
alternative
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Analysis of Variance & One Factor
Designs (One-Way ANOVA)
Y= RESPONSE VARIABLE (of numerical type)
(e.g. battery lifetime)
X = EXPLANATORY VARIABLE (of categorical type)
(A possibly influential FACTOR)
(e.g. brand of battery)
OBJECTIVE: To determine the impact of X on Y
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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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Example: Y = LIFETIME (HOURS)
3 replications
per level
1
BRAND
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
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Analysis of Variance
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Statistical Model
C “levels” OF BRAND
R observations for each level
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2 ••• • • •••R
1 Y11 Y12
• • • • • • •Y1R
2 Y21
•
•
•
•
•
•
•
Yij
•
•
•
•
•
•
•
•
C YcI • • • • • • • •YcR
Yij = + i + ij
i = 1, . . . . . , C
j = 1, . . . . . , R
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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.
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Yij =  + i + ij
C
By definition, 
i=1
i = 0
The experiment produces
RxC
Yij data values.
The analysis produces estimates of
, ,,c . (We can then get estimates of
the ij by subtraction).
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Let Y1, Y2, etc., be level means
c
Y • = Y i C = “GRAND MEAN”
i=1
/
(assuming same # data points in each column)
(otherwise, Y • = mean of all the data)
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MODEL:
Y•
Yi - Y •
Yij =  + i + ij
estimates 
estimates i (= i – )
(for all i)
These estimates are based on Gauss’ (1796)
PRINCIPLE OF LEAST SQUARES
and on COMMON SENSE
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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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Then, Yij = Y• + (Yi - Y• ) + ( Yij - Yi)
{
{
{
or, (Yij - Y• ) = (Yi - Y•) + (Yij - Yi )
(3)
VARIABILITY
in Y
Variability
Variability
TOTAL
=
in Y
+
in Y
associated
associated
with X
with all other
factors
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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
2
C R
2
(Yij - Y• ) = R •  (Yi - Y•) + (Yij - Yi)
i=1
i=1 j=1
{
{
{
i=1 j=1
2
(
(
TSS
TOTAL SUM OF
SQUARES
=
SSB
+
=
SUM OF
SQUARES
BETWEEN
SAMPLES
+
(
SSW (SSE)
(
SUM OF SQUARES
WITHIN SAMPLES
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ANOVA TABLE
SOURCE OF
VARIABILITY
Between samples
(due to brand)
Within samples
(due to error)
TOTAL
SSQ
SSB
SSW
TSS
DF
C-1
Mean
(M.S.)
square
SSB
= MSB
C-1
SSW
= MSW
(R - 1) • C
(R-1)•C
RC -1
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Example: Y = LIFETIME (HOURS)
BRAND
3 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
2
2
2
=
3 ( [2.6 - 5.8] + [4.6 - 5.8] + • • • + [7.4 - 5.8] )
=
3 (23.04)
=
69.12
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SSW =?
2
(4.2 - 4.6)2 =.16
2
(5.4 - 4.6)2= .64 • • • • (7.4 - 7.4)2 = 0
2
(4.2 - 4.6)2= .16
(5.8 - 7.4)2 = 2.56
.96
5.12
(1.8 - 2.6) = .64
(5.0 - 2.6) = 5.76
(1.0 - 2.6) = 2.56
8.96
(9.0 -7.4)2 = 2.56
Total of (8.96 + .96 + • • • + 5.12),
SSW = 46.72
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ANOVA TABLE
Source of
Variability
SSQ
df
M.S.
BRAND
69.12
7
9.87
= 8-1
ERROR
46.72
16
2.92
= 2 (8)
TOTAL
115.84
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= (3 • 8) -1
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We can show:
“VCOL”
E (MSB) = 2
(
+
MEASURE OF
DIFFERENCES
AMONG LEVEL
MEANS
R
C-1
•
(i - )2
i
E (MSW) = 2
(Assuming Yij follows N(j ,2) and they are independent)
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E ( MSBC ) = 2 + VCOL
E ( MSW ) = 2
This suggests that
if
if
MSBC
MSW
MSBC
MSW
>
<
There’s some
evidence of non1 , zero V , or “level
COL
of X affects Y”
No evidence that
1,
VCOL > 0, or that
“level of X affects Y”
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With HO:
HI:
Level of X
has no impact
on Y
Level of X
does have
impact on Y,
We need
MSBC
MSW
>>1
to reject HO.
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More Formally,
HO: 1 = 2 = • • • c = 0
HI: not all j = 0
OR
HO: 1 = 2 = • • • • c
(All level means
are equal)
HI: not all j are EQUAL
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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
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In our problem:
ANOVA TABLE
Source of
Variability
SSQ
df
M.S.
BRAND
69.12
7
9.87
3.38
ERROR
46.72
16
2.92
= 9.87
2.92
Fcalc
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F table: table 8
 = .05
C = 2.66
3.38
(7,16 DF)
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Hence, at  = .05, Reject Ho .
(i.e., Conclude that level of
BRAND does have an impact on
battery lifetime.)
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MINITAB INPUT
life
1.8
5.0
1.0
4.2
5.4
4.2
.
.
.
9.0
7.4
5.8
brand
1
1
1
2
2
2
.
.
.
8
8
8
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ONE FACTOR ANOVA
(MINITAB)
MINITAB: STAT>>ANOVA>>ONE-WAY
Analysis of Variance for life
Source
DF
SS
MS
F
P
3.38
0.021
brand
7
69.12
9.87
Error
16
46.72
2.92
Total
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115.84
Estimate of the common variance ^2
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Boxplots of life by brand
(means are indicated by solid circles)
10
9
8
7
life
6
5
4
3
2
1
8
7
6
5
4
3
2
brand
1
0
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Assumptions
MODEL:
Yij = + i + ij
Run order plot
Normality plot
& test
Residual plot
& test
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
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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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Minitab: stat>>basic statistics>>normality test
Probability Plot of RESI1
Normal
99
Mean
StDev
N
AD
P-Value
95
90
-1.48030E-16
1.425
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0.481
0.212
Percent
80
70
60
50
40
30
20
10
5
1
-4
-3
-2
-1
0
RESI1
1
2
3
4
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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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Minitab: Stat >> Anova >> One-way
Residuals Versus the Fitted Values
(response is life)
3
Residual
2
1
0
-1
-2
2
3
4
5
Fitted Value
6
7
8
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Minitab: Stat>> Anova>> Test for Equal variances
Test for Equal Variances for life
1
Bartlett's Test
Test Statistic
P-Value
2
Lev ene's Test
Test Statistic
P-Value
3
brand
4.20
0.757
0.31
0.938
4
5
6
7
8
0
10
20
30
40
95% Bonferroni Confidence Intervals for StDevs
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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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Minitab: Stat >> Anova >> One-way
Residuals Versus the Order of the Data
(response is life)
3
Residual
2
1
0
-1
-2
2
4
6
8
10
12
14
16
Observation Order
18
20
22
24
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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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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
Mean: 23.9
22.1
BATTERY LIFETIME
(hours)
(each column rank
ordered, for
simplicity)
14.9 (here, irrelevant!!)
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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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Ranks in ( )
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
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TEST STATISTIC:
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H=
N (N + 1)
K
•  (Tj2/nj ) - 3 (N + 1)
j=1
nj = # data values in column j
K
N = nj
j=1
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.)
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H=
30 (31)
[
197 2 178 2 902
+
+
10
10
10
[
12
- 3 (31)
= 8.41
(with adjustment for ties, we get 8.46)
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What do we do with H?
We can show that, under HO , H is well
2
approximated by a  distribution with
df = K - 1.
Here, df = 2, and at = .05, the critical value = 5.99
-,df
 = .05
= F-,df,
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
2
df
5.99
8.41 = H
Reject HO; conclude that mean lifetime NOT
the same for all 3 BRANDS
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Minitab: Stat >> Nonparametrics >> KruskalWallis
•
Kruskal-Wallis Test: life versus brand
•
Kruskal-Wallis Test on life
•
•
•
•
•
•
•
•
•
•
brand
1
2
3
4
5
6
7
8
Overall
•
•
H = 12.78 DF = 7 P = 0.078
H = 13.01 DF = 7 P = 0.072 (adjusted for ties)
N
3
3
3
3
3
3
3
3
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Median
1.800
4.200
4.600
7.000
6.600
4.200
7.800
7.400
AveRank
4.5
7.8
11.8
16.5
13.3
7.8
20.0
18.2
12.5
Z
-2.09
-1.22
-0.17
1.05
0.22
-1.22
1.96
1.48
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