Always be mindful of the kindness and not the faults of others.

1

Always be mindful of the kindness and not the faults of others.

2

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

Model and test for one-way anovaAssumption checkingNonparamateric alternative

3

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

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)

5

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

Analysis of Variance

6

7

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•   •  •   •    •   •    •    • 

8

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

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).

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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

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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

12

MODEL: Yij = + i + ij

If you insert the estimates into the MODEL,

(1) Yij = Y • + (Yi - Y • ) + ij.

it follows that our estimate of ij is

(2) ij = Yij – Yi, called residual

<

<

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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

+

14

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

TSSTOTAL 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

SSQ DF Meansquare

(M.S.)

Between samples (due to brand)

Within samples (due to error)

SSB C - 1 MSBSSBC - 1

SSW (R - 1) • C SSW(R-1)•C = MSW

=

TOTAL TSS RC -1

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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

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(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 = 2.568.96 .96 5.12

Total of (8.96 + .96 + • • • + 5.12),

SSW = 46.72

SSW =?

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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

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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)

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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”

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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.

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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)

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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 TABLESource 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

25

= .05

C = 2.66 3.38

F table: table 8

(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 brand

1.8 15.0 11.0 14.2 25.4 24.2 2. .. .. .9.0 87.4 85.8 8

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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

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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)

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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

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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

RESI1

Perc

ent

43210-1-2-3-4

99

9590

80706050403020

105

1

Mean -1.48030E-16StDev 1.425N 24AD 0.481P-Value 0.212

Probability Plot of RESI1Normal

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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

Fitted Value

Resid

ual

8765432

3

2

1

0

-1

-2

Residuals Versus the Fitted Values(response is life)

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Minitab: Stat>> Anova>> Test for Equal variancesbr

and

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

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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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Observation Order

Resid

ual

24222018161412108642

3

2

1

0

-1

-2

Residuals Versus the Order of the Data(response is life)

Minitab: Stat >> Anova >> One-way

What to do if the equal variances assumption fails

• Transform response Y (Table 8.17, page 423)

• Example 8.4

• If the problem cannot be fixed, do the non-parametric procedure (next slide).

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39

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

BATTERY LIFETIME (hours)

(each column rank ordered, for simplicity)

Mean: 23.9 22.1 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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Brand

A B C32 (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 ( )

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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

44

H =

[ 12 197 2 178 2 902

30 (31) 10 10 10+ +

[ - 3 (31)

= 8.41

(with adjustment for ties, we get 8.46)

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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

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• 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