2.16 Analysis of Variance (ANOVA) Rev DD 20100604

8/6/2019 2.16 Analysis of Variance (ANOVA) Rev DD 20100604

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2.16 Analysis of Variance (ANOVA)

Six Sigma Black Belt and Green Belt

Week 2

Revised 4th

June 2010


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2010-06-04 SKF Group Slide 1 SKF (Group Six Sigma) 2.16 Analysis of Variance (ANOVA)

Objectives

To introduce ANOVA hypothesis testing

Graphical method for analysing differences between means obtained

from two or more samples

Analysis of Variance (ANOVA) methods for analysing the differencesbetween means

To understand the relationship of

"within" subgroup estimates of variation and

"between" subgroup estimates of variation

To understand the measuring effect size

To practice examples

To introduce the Post Hoc test


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SKF Six Sigma roadmap

Six Sigma methodology and roadmap for common tool usage


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By knowing and controlling the Xs, we reduce thevariability in Y.

We validate Xs and Ys with hypothesis testing.

Variable

(Continuous)

Variables with categories

(Attribute)

Validating key process inputs and outputs with ANOVA

Y = f(X1

, X2

, X3

, ..., Xn

)


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6/712010-06-04 SKF Group Slide 5 SKF (Group Six Sigma) 2.16 Analysis of Variance (ANOVA)

Method

Uses sums of squared differences, just like a standard deviation, to

evaluate the total variability of the system

Calculates "standard deviations" for each source and subtracts their

variability from the total


7/712010-06-04 SKF Group Slide 6 SKF (Group Six Sigma) 2.16 Analysis of Variance (ANOVA)

ANOVA graphical

Between subgroup variation (signal)

Within subgroup variation (error)


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Degree of freedom introduction

Degrees of freedom (df) is the number of independent comparisons

available to estimate a specific statistic.

In ANOVA, the degrees of freedom are based on the total number of

responses and the number of levels at which factors are tested.

What is the minimum number of comparisons it would take todetermine which person is the shortest?


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The degree of freedom concept?

Example:

Consider a sample ofn

= 3 scores with a mean of X-bar = 5. The first

score in the sample can be selected without any restrictions; all

scores are independent of each other and they can have any value.

For this demonstration assume X = 2 is obtained for the first scoreand X = 9 for the second.

At this point, however, the third score can be determined.


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The degree of freedom concept?

In this case the third score must be X = 4.

The reason that the third score has to be X = 4 is, the entire sampleofn

= 3 scores has a mean of: X-bar = 5, which means that the sum

of the total must be: X = 15. The first two scores add up to 11

(= 9 + 2), so the third score must be X = 4.

In this case the first two out of three scores were free to have

any

value, but the final score was dependent on the values chosen for thefirst two. With a sample ofn

scores, the first n-1 scores are free to

vary, but the final score can be determined.

As a result, the sample is said to have n-1 degrees of freedom (df).The degrees of freedom determine the number of scores in thesample which are independent and free to vary.


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Thus in other words ...

Degrees of freedom are "statistical cash" ...

We "earn" a degree of freedom for every data point we collect

We "spend" a degree of freedom for every parameter we estimate

Degrees of freedom (within groups):

Earn a degree of freedom for each observation within each group

Spend one degree of freedom to calculate the average for each group

dfW

= n

1, where n

= sample size / treatment

Degrees of freedom (between groups):

Earn a degree of freedom for each group

Spend one degree of freedom to calculate the overall average

dfB

= k

1, where k

= # of group averages or number of treatments


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Degree of freedom and ANOVA

The n-1 degrees of freedom for a sample is the same n-1 that isused in the formulas for sample variance and sample standarddeviation.

Remember, variance is defined as the mean square deviation. This

mean it is computed by finding the sum and dividing by the numberof scores:

Mean = Sum / Number of scores

To calculate sample variance (mean squared deviation), we find thesum of the squared deviations (SS) and divide by the number ofscores that are free to vary. This number is n-1 = df.

df

Sum of squared deviations

Number of scores free to vary

SS==s2

i i


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Calculate the F statistic Example of seal life by shift (1)Shift 1 Shift 2 Shift 3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

Mean 24.93 22.61 20.59

df = (5-1) (5-1) (5-1)

Overall average = 22.71

Data collection !

dftotal

= (4) + (4) + (4) = 12

C l l h F i i


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Calculate the F statistic Example of seal life by shift (2)

Mean shift 1

(24.93) Mean shift 2(22.61)

Mean shift 3(20.59)

Overall average = 22.71


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C l l h F i i


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Calculate the F statistic Example of seal life by shift (3)between

between2between

dfSSs =

within

within2

withindf

SSs =

25.600.921123.582

ssF 2

within

2

betweendf,df 21

===

23.58213

47.164dfSS

between

between=

=

0.9211315

11.0532

df

SS

within

within=

=

The F-distribution depends on two sets of degrees of freedom:

the df from each variance: df1

for the between

and df2

for the within.

Number of shifts

Total data available


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What is the distribution of the F-ratio?

This is the distribution of F-ratios that would occur if there was nodifference in group means.

For example, say Im willing to take a 5% chance of being wrong by

saying there is more between than within variation.

Fcritical

at 5%

5% of the total area is fromthis F value, Fcrit

to the right

The curvechanges as afunction of thenumerator df

anddenominator df

Represents theamount of risk I'mwilling to take ofbeing wrong when Isay that Ive foundthis factor to be asignificant effect.

A calculated F-ratio > Fcrit

gives me less than a 5%chance that the largerbetween variationoccurred by chance alone.

Remember you choose the amount of risk to take, then find a corresponding Fcritical


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F di t ib ti t bl


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F-distribution table Probability points of the F-distributionDegrees of

Freedom for

Denominator

Degrees of Freedom for Numerator (df)1 2 3 4 5 6 7 8 9 10 15 20

1161.4 199.5 315.7 224.6 230.2 234.0 236.8 238.9 240.5 241.9 245.9 248.0

4052 5000 5403 5625 5764 5859 5928 5981 6022 6056 6157 6209

218.51 19.00 19.16 19.25 19.30 19.33 19.35 19.37 19.38 19.40 19.43 19.45

98.50 99.00 99.17 99.25 99.30 99.33 99.36 99.37 99.39 99.40 99.43 99.45

3

10.13 9.55 9.28 9.12 9.01 8.94 8.89 8.85 8.81 8.79 8.70 8.66

34.12 30.82 29.46 28.71 28.24 27.91 27.67 27.49 27.35 27.23 26.87 26.69

47.71 6.94 6.59 6.39 6.26 6.16 6.09 6.04 6.00 5.96 5.86 5.80

21.20 18.00 16.69 15.98 15.52 15.21 14.98 14.80 14.66 14.55 14.20 14.02

56.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 4.74 4.62 4.56

16.26 13.27 12.06 11.39 10.97 10.67 10.46 10.29 10.16 10.05 9.72 9.55

65.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10 4.06 3.94 3.87

13.75 10.92 9.78 9.15 8.75 8.47 8.26 8.10 7.98 7.87 7.56 7.40

75.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68 3.64 3.51 3.44

12.25 9.55 8.45 7.85 7.46 7.19 6.99 6.84 6.72 6.62 6.31 6.16

85.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39 3.35 3.22 3.15

11.26 8.65 7.59 7.01 6.63 6.37 6.18 6.03 5.91 5.81 5.52 5.36

95.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.14 3.01 2.94

10.56 8.02 6.99 6.42 6.06 5.80 5.61 5.47 5.35 5.26 4.96 4.81

10 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 2.98 2.85 2.7710.04 7.56 6.55 5.99 5.64 5.39 5.20 5.06 4.94 4.85 4.56 4.41

154.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 2.54 2.40 2.33

8.68 6.36 5.42 4.89 5.56 4.32 4.14 4.00 3.89 3.80 3.52 3.37

204.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 2.35 2.20 2.12

8.10 5.85 4.94 4.43 4.10 3.87 3.70 3.56 3.46 3.37 3.09 2.94

= 0.05 ... first row

= 0.01 ... second rowNumerator

Denominator


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Mean sum of squares (MS)

In ANOVA, we use the term Mean Square, or simply

MS, in stead of the

term variance.

Remember that variance is defined as the mean of the

squared deviations. In the same way that we use SS

to stand for the

sum of the squared deviations, we now will use MS

to stand for the

mean of the squared deviations. For the final F-ratio we will need anMSbetween

treatments for the numerator and MSwithin

treatments for the

denominator.

within

between

MS

MSratio-F =

MSbetween

= SSbetween/ dfbetween

MSwithin

= SSwithin/ dfwithin

P titi f i d F ti


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Partition of variance and F-ratio OverviewTotal variability

Between treatmentsvariance

Within treatmentsvariance

Measures differences due to:

Treatment effects and

Chance

Measures differences due to:

Chance

Signal Error

Variance (MSbetween

) = SSbetweendfbetween

Variance (MSwithin

) = SSwithindfwithin

F-ratio =MSbetween

MSwithin


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The p-value and ANOVA

Assumptions ...

H0 : There are no differences between subgroups meansHA

: There are differences between subgroups means

Low p-values suggest that there ARE differences betweensubgroups means.

Tip: P-value is low, H0

must go !


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The R-squared value and ANOVA

To provide an indication of how large the effect actually is, we

check

p-value but also the R2

value to take the decision if the result is robust

or not.

For Analysis of Variance, the simplest and most direct way to measureeffect size is to compute R2, the percentage of variance accounted for.

In simpler terms, R2

measures how much of the difference between

scores is accounted for by the differences between treatments.

SSbetween

measures the variability accounted for by the treatment

differences, and SStotal

measures the total variability.

total

between2

SS

SSR =


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The R-squared value and ANOVA - examples

R2

= 90%

Sum

ofvariance

Variance explained by thefactor (treatment)

Error, part ofvariance not

explained by thefactor (xs)

R2

= 50 %

50%

90%

Sum

ofvariance

Which model is more robust? A or B?

A B


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ANOVA assumptions

1.

Normality

2.

Homogeneity of variance (equal variances)

3.

Independence of error


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Independence of error

Errors should be independent for each value and over

time

If not, then do not assume test is valid

Identify why error is not independent and correct

We use control charts to check the stability and detectthe special cause


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Normality

The values in each group are Normally distributed

While the ANOVA method is robust against departures

from normality as in the t-test, especially with largesample sizes, non-normal distributions where normality

would be expected may indicate an area of investigation

Master Black Belt may be consulted when non-normal

data is being analysed (non-parametric tests)


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Homogeneity of variance

The variance within each group is equal

However, if the sample sizes are equal between groups,

the F-test is robust enough for unequal variances

Always try to have equal sample sizes

If both normality and equal variances are violated,Master Black Belt may be consulted


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The p-value

For a classical hypothesis test, use the p-value to evaluatethe probability that the calculated F-ratio (or test statistic)was due to within

subgroup noise.

Low p-values

suggest that there ARE differences between

subgroups means:

H0

: There are no differences between subgroups means

HA

: There are differences between subgroups means


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Examples to practice !


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One-way ANOVA

Stat > ANOVA > One-Way

Data must be in one column and the subscripts in another

Can be used with balanced and unbalanced designs

A one-way analysis of variance (ANOVA) tests the hypothesis that

the means of several populations are equal

The method is an extension of the two-sample t-test, specifically

for the case were the population variances are assumed to be

equal. A one-way analysis of variance requires the following: A response, or measurement taken from the units sampled

A factor, or discrete variable which is altered systematically


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One-way ANOVA example: Tire brand test

Four cars:

1 2 3 4

Four brands of tires:

A B C D

Objective: To determine tread wear of tires after 30,000 km of driving.

Problem: How do we assign 16 tires to the 4 cars?

Assign each of the 16 tires at random to a wheel. (Large variability

within brands.)

Ref.: "Fundamental Concepts in the Designof Experiments" by Hicks and Turner

Cars 1 2 3 4C (12) A (14) C (10) A (13)

A (17) A (13) D (11) D (9)

D (13) B (14) B (14) B (8)D (11) C (12) B (13) C (9)

Difference in tread thickness in mm.Model

Tread wear = Overall mean + Brand effect + error

Data of tread wear of tires


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Data of tread wear of tires Each of the 16 tires assigned at random to wheelCar Brand Tread

One

C 12

A 17

D 13

D 11

Two

A 14

A 13

B 14

C 12

Three

D 10

C 11

B 14

B 13

Four

A 13

D 9

B 8

C 9

Open the file

and

check the different assumptions:

Stability

Normality


(equal variances)

ANOVA - Tire Brand.MTW

One-way ANOVA


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One way ANOVA Stat > ANOVA > One-way

You wish to compare the mean tread wear

for the different types of brands of tires.H0 is that

the tread wear are all the same.

Any variation is caused by random variationfound in each brand. The HA is that differentbrands have different tread wear.

Normality and stability


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5.02.50.0-2.5-5.0

99

90

50

10

1

Residual

Perc

ent

14131211

2

0

-2

-4

Fitted V alue

Resi

dual

3210-1-2-3-4

3

2

1

0

Residual

Frequency

16151413121110987654321

2

0

-2

-4

Observation Order

Res

idual

Normal Probability Plot Versus Fits

Histogram Versus Order

Residual Plots for Tread

Normality and stability One-way ANOVA Residual plots

Normal ?

Stable ?

V i


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D

C

B

A

181614121086420

B

rand

95% Bonferroni Confidence Intervals for StDevs

Test St at istic 1.52

P-V alue 0.677

Test St at istic 0.15

P-V alue 0.926

Bart lett 's Test

Levene's Test

Test for Equal Variances for Tread

Variance

Variances are equal ?

O ANOVA l


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One-way ANOVA example

1.

Open the file 2.

Select Stat > ANOVA > One-Way

3.

Select Tread for the Response

and Brand for the Factor

4.

Click on OK

ANOVA - Tire Brand.MTW

Interpreting the One-way ANOVA


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One-way ANOVA: Tread versus Brand

Source DF SS MS F P

Brand 3 30.69 10.23 2.44 0.115

Error 12 50.25 4.19

Total 15 80.94

S = 2.046 R-Sq = 37.92% R-Sq(adj) = 22.39%

Individual 95% CIs For Mean Based on

Pooled StDev

Level N Mean StDev -------+---------+---------+---------+--

A 4 14.250 1.893 (----------*----------)

B 4 12.250 2.872 (----------*----------)

C 4 11.000 1.414 (----------*----------)

D 4 10.750 1.708 (----------*----------)

-------+---------+---------+---------+--

10.0 12.0 14.0 16.0

Pooled StDev = 2.046

MINIT

AB

Interpreting the One way ANOVA Output from the session windowThe 1st

row "Brand" gives the stats for

the variation between the means of thefactor levels.The 2nd

row "Error" gives the stats for

the variation due to random error.The 3rd

row "Total" gives the stats for

the overall variability in the data.

Interpreting the One-way ANOVA


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One-way ANOVA: Tread versus Brand

Source DF SS MS F P

Brand 3 30.69 10.23 2.44 0.115

Error 12 50.25 4.19

Total 15 80.94

S = 2.046 R-Sq = 37.92% R-Sq(adj) = 22.39%

Individual 95% CIs For Mean Based on

Pooled StDev

Level N Mean StDev -------+---------+---------+---------+--

A 4 14.250 1.893 (----------*----------)

B 4 12.250 2.872 (----------*----------)

C 4 11.000 1.414 (----------*----------)

D 4 10.750 1.708 (----------*----------)

-------+---------+---------+---------+--

10.0 12.0 14.0 16.0

Pooled StDev = 2.046

MINIT

AB

Interpreting the One way ANOVA Output from the session window1.

What is your decision?

2.

The result is robust or not and why?

3. Which Brand is best?

Two-way ANOVA


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Two way ANOVAUsing a 2nd

variable to block Car variation

Assign each tire at random but under the condition thateach tire occurs exactly once on each car.

Reduces

unexplained variability.

Cars 1 2 3 4B (14) D (11) A (13) C (9)

C (12) C (12) B (13) D (9)

A (17) B (14) D (11) B (8)D (13) A (14) C (10) A (13)

Model

Tread wear = Overall mean + Brand effect + Car effect + errorDifference in tread thickness in mm.



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Data of tread wear of tires Each tire occurs exactly once on each carCar Brand Tread

One

B 14

C 12

A 17

D 13

Two

D 11

C 12

B 14

A 14

Three

A 13

B 13

D 11

C 10

Four

C 9

D 9

B 8

A 13

Open the file

and check the assumptions:

Stability

Normality


(equal variances)

ANOVA - Tire Brand Car.MTW

T ANOVA l Ti b d t t


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Two-way ANOVA example: Tire brand test

1.

Open the file 2.

Select Stat > ANOVA > Two-Way

3.

Select Tread for the Response

and Brand for the Row factorand Car for Column factor.Check Display means.

4.

Click on OK

ANOVA - Tire Brand Car.MTW

Interpreting the Two-way ANOVA


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Two-way ANOVA: Tread versus Brand, Car

Source DF SS MS F P

Brand 3 30.6875 10.2292 7.96 0.007

Car 3 38.6875 12.8958 10.04 0.003Error 9 11.5625 1.2847

Total 15 80.9375

S = 1.133 R-Sq = 85.71% R-Sq(adj) = 76.19%

MINIT

AB

Interpreting the Two way ANOVA Output from the session windowp-values are low for Car and Brand,therefore: Brands are not the same, andTread loss for Cars is not the same.

Lets look at the residuals plots ...

T ANOVA R id l l t


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Two-way ANOVA Residual plots

210-1-2

99

90

50

10

1

Residual

Perc

ent

161412108

1

0

-1

-2

Fit t ed Value

Residual

1.00.50.0-0.5-1.0-1.5-2.0

4

3

2

1

0

Residual

Freq

uency

16151413121110987654321

1

0

-1

-2

Observat ion Order

Res

idual




The residuals plots show no unusual observations.The Histogram is not bell shaped (only 16 observations) so it is hard to interpret.

Interpreting the Two-way ANOVA


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Two-way ANOVA: Tread versus Brand, Car

Individual 95% CIs For Mean Based on Pooled StDev

Brand Mean -+---------+---------+---------+--------

A 14.25 (-------*-------)B 12.25 (-------*-------)

C 10.75 (-------*-------)

D 11.00 (-------*-------)

-+---------+---------+---------+--------

9.6 11.2 12.8 14.4

Individual 95% CIs For Mean Based on Pooled StDev

Car Mean --------+---------+---------+---------+-

Four 9.75 (------*-----)

One 14.00 (-----*-----)

Three 11.75 (------*-----)Two 12.75 (------*-----)

--------+---------+---------+---------+-

10.0 12.0 14.0 16.0

MINITA

B

p g y Output from the session window

The confidence intervals show:Brands are not the same, andTread loss for Cars is not the same.

1.


2.

The result is robust or not and why?

3.

Which factor is significant?

Lets look at the data graphically


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Let s look at the data graphically

Graph > Chart > Values from a table

Lets look at the data graphically


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Let s look at the data graphically

Graph > Chart > Values from a table > Data View

Displaying the Two way ANOVA design


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Displaying the Two-way ANOVA design

Car

Brand

FourThreeTwoOne

DACBDACBDACBDACB

18

16

14

12

10

8

6

4

2

0

Tread

B

C

A

D

Brand

Chart of Tread

All 4 Brands performed better in Car One. This is an assignable difference due to Car.

Also it appears that Brand A performs better at each Car than the other Brands.

Displaying the Two way ANOVA design


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Displaying the Two-way ANOVA design

Brand

Car

DACB

Four

Three

Two

One

Four

Three

Two

One

Four

Three

Two

One

Four

Three

Two

One

18

16

14

12

10

8

6

4

2

0

Tread

One

Two

Three

Four

Car

Chart of Tread

Here we are trying to discover which Brand of Tires had the best

Tread Wear characteristics.

We included a blocking variable to explain some of the variability. Based on a comparison

of the bar chart and the ANOVA table which Brand should be selected?

Three-way ANOVA


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y Using a Latin Square designEach brand appears once in each position and only once oneach car (2 restrictions on randomisation).

Minimises

variability.

Model

Tread wear = Overall mean + Brand effect + Car effect+ Position

effect

+ errorDifference in tread thickness in mm.

Position 1 2 3 4I C (12) D (11) A (13) B (8)

I I B (14) C (12) D (11) A (13)

I I I A (17) B (14) C (10) D (9)I V D (13) A (14) B (13) C (9)



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Each brand appears once in each position and each carCar Position Brand Tread

One

Left Front B 14

Right Front C 12

Left Back A 17

Right Back D 13

Two

Left Front D 11

Right Front C 12

Left Back B 14

Right Back A 14

Three

Left Front A 13

Right Front B 13

Left Back D 11

Right Back C 10

Four

Left Front C 9

Right Front D 9

Left Back B 8

Right Back A 13

Open the file

and check the

assumptions:

Stability

Normality


(equal variances)

ANOVA - Tire Brand Car Position.MTW

Three-way ANOVA


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y Stat > ANOVA > General Linear ModelFill out the dialog box as shown.

Click OK.

Interpreting the General Linear Model


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General Linear Model: Tread versus Car, Position, Brand

Factor Type Levels Values

Car fixed 4 Four, One, Three, Two

Position fixed 4 Left Back, Left Front, Right Back, Right FrontBrand fixed 4 A, B, C, D

Analysis of Variance for Tread, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P

Car 3 38.6875 38.6875 12.8958 14.40 0.004

Position 3 6.1875 6.1875 2.0625 2.30 0.177

Brand 3 30.6875 30.6875 10.2292 11.42 0.007

Error 6 5.3750 5.3750 0.8958

Total 15 80.9375

S = 0.946485 R-Sq = 93.36% R-Sq(adj) = 83.40%

MINITA

B

p g Output from the session windowThe 1st

half of the table lists the

value for each level of each factor.The 2nd

half is the ANOVA table.

Two factors are statisticallysignificant at the

= 0.05 level:

Car, Brand. Factor Position

doesnt

appear to be a significant effect.The residual plots will confirm

whether the basic assumptionsabout the error have been met.

Lets look at the residuals plots ...

General Linear Model Residual plots


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10-1

99

90

50

10

1

Residual

Pe

rcent

16141210

1.0

0.5

0.0

-0.5

-1.0

Fit t ed Value

Re

sidual

1.00.50.0-0.5-1.0

4

3

2

1

0

Residual

Frequency

16151413121110987654321

1.0

0.5

0.0

-0.5

-1.0

Observat ion Order

Residual




Review the residual plots and state the conclusions about the assumptions

regarding error, i.e. that the errors for each treatment level areindependent, normally distributed with a mean = 0 and a constant

variance.

ANOVA example with GLM


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ANOVA example with GLM

How to include the interaction in the model?

2.

Select Tread for Response

Car and Brand

for Model.

For the interaction we create Car*Brand.

3. Click on OK

1.

Select Stat > ANOVA > General Linear Model

GLM with an unbalanced and nested design


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Four chemical companies produce insecticides that can be used to

kill

mosquitoes, but the composition of the insecticides differs from

company to

company.

An experiment is conducted to test the efficacy of the insecticides by placing

400 mosquitoes inside a glass container treated with a single insecticide andcounting the live mosquitoes 4 hours later.

Three replications are performed for each product.

The goal is to compare the product effectiveness of the different companies.The factors

are fixed

because you are interested in comparing the particular

brands.

The factors are nested

because each insecticide for each company is unique.

You use GLM to analyse your data because the design is unbalanced:

Company A: 3 type of products

Company B: 2 type of products

Company C: 2 type of products

Company D: 4 type of products



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For the

Nested design add (Company)

2.

Select NMosquito for Response

Company and Product for Model.

1.

Select Stat > ANOVA > General Linear Model

3.

Click on OK



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ANOVA table in session window

1.


2.

Which parameter is significant?

Multi-way ANOVA


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Multi way ANOVA

Two-way, Balanced, General Linear Model.

Two-way ANOVA may also be used to analyse a design where thereare two controllable factors, both of which are of interest.

More than two factors can be analysed using Balanced ANOVA orGeneral Linear Model.

There may be more than one factor that has an effect on theresponse variable.

This commonly occurs in manufacturing processes. It is often wise toinclude more than one factor in the analysis.

Valuable resources can be used more efficiently by investigating

several

factors at one time.

More error can be explained by including additional factors in the model.

By including more factors interactions can be studied.

What about the other ANOVA options?Wh h i ?


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When are they appropriate?One-wayANOVA

Studies the effect of one factor at various levels on a response

variable.Two-wayANOVA

Studies the effect of two factors and their interaction at variouslevels on a response variable.

BalancedANOVA

Studies the impact of 2 or more factors and their interactions atvarious levels on a response variable. The levels of factors are

structured such that there are an equal number of levels andobservations within each level for each factor.GeneralLinearModel

Studies the impact of 2 or more factors and their interactions atvarious levels on a response variable. The number of levels andobservations may vary. The factors may be a mixture nested andcrossed relationship. User must specify factors, interactions andnested/crossed relationships of interest.

FullyNestedANOVA

Studies the impact of 2 or more factors. The factors arestructured in a hierarchical structure such that one factor isnested (or unique to) the factor above it. No interactions areobtained.

Partitioning of sums of squares


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Partitioning of sums of squares

SS Within

Brands

Total

SS

SS Between

Brands

SS Within

Cars

SS Between

Cars

SS Within

(Error)

SS Between

Positions

Summary ANOVA


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Summary ANOVA

One Way ANOVA

To analyse the difference between means from 2 or more samples

Balanced ANOVA

To compare the means of populations that are classified in two ormore ways (two or more factors)

General Linear Model

Similar to above

Last words


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Last words

We reviewed:

Graphical methods for analysing differences between means obtainedfrom 2 or more samples.

Analysis of Variance (ANOVA) methods for analysing the differencesbetween means.

Methods for determining whether or not significant differences invariance exist between two or more samples.


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Appendix

Post Hoc tests


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Pos Hoc es s

Definition: Post hoc tests are additional hypothesis tests that are doneafter an ANOVA to determine exactly which mean differences aresignificant and which are not. These tests are done when:

You reject H0

and there are three or more treatments.

Rejecting H0

indicates that at least one difference exists among the

treatments.

With k

= 3 or more, the problem is to find where the differences are.

Note that when you have two treatments, rejecting H0

indicates that the

two means are not equal, in this case there is no question about

which

means are different, and there is no need to do Post Hoc Tests.

The first test we consider is Tukeys HSD test. Tukeys test allows youto compute a single value that determines the minimum difference

between treatment mean that is necessary for significance.

Post Hoc tests


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This value, called the Honestly Significant Difference (HSD) is then usedto compare any two treatments (Xs). If the mean difference exceedTukeys HSD you conclude that there is significant difference betweentreatments. The formula is:

N: number of data for each treatment

Where the value ofq

is found in the table (next slide). To locate the

appropriate value ofq, you must know the number of treatments in theoverall experiment (k) and the degree of freedom for the Error andselect the Alpha-risk (0.05) q

value used in this test is called a

Studentised range statistic.

Tukeys test requires that the sample size must be the same for all

treatments.

n

MSqHSD within=

Post Hoc tests


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Tukeys HSD test example


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y p

Example of seal life by shift

Shift 1 Shift 2 Shift 3

25.40 23.40 20.00

26.31 21.80 22.20

24.10 23.50 19.75

23.74 22.75 20.60

25.10 21.60 20.40

Mean 24.93 22.61 20.59

ANOVA result:

P-value is low, the difference issignificant between the shifts.

Now the question is:

Which mean differences are

significant and which are not?

Tukeys HSD test example


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y p

Tukeys HSD calculation step 1:Determine the q value, in this example k=3 and dffor Error = 12. Checkthe value in the table, we get = 3.77 with Alpha-risk = 0.05

Tukeys HSD calculation step 2:

Determine the HSD value

Tukeys HSD calculation step 3:

The mean difference between any two samples must be at least 1,618to be significant. Using this value, we can make the followingconclusions :

Shift 1 is significantly different from Shift 2 (Mean S1

Mean S2

= 2.32)


Mean S3

= 4.34)


Mean S3

= 2.02)

1.6185

0.9213.77HSD ==

Summary


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y

ANOVA is used as a hypothesis test and we also use it forcomponents of variation studies

The X is attribute and Y is variable

very common data sets

ANOVA introduced us to 3 preliminary tests before concluding to

accept or reject the null:

Stability

Normality


All hypothesis tests require these or similar tests of assumptions

Use the appropriate design before to calculate ANOVA

Use Tukeys HSD test to adjust the conclusion if needed

2.16 Analysis of Variance (ANOVA) Rev DD 20100604

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Transcript of 2.16 Analysis of Variance (ANOVA) Rev DD 20100604