Analysis of Variance (ANOVA)

EGR 252 Spring 2009 - Ch.13 Part 1 1

Analysis of Variance (ANOVA)

A single-factor ANOVA can be used to compare more than two means.For example, suppose a manufacturer of paper used for grocery bags is concerned about the tensile strength of the paper. Product engineers believe that tensile strength is a function of the hardwood concentration and want to test several concentrations for the effect on tensile strength.

If there are 2 different hardwood concentrations (say, 5% and 15%), then a z-test or t-test is appropriate:

H0: μ1 = μ2

H1: μ1 ≠ μ2


Comparing More Than Two Means

What if there are 3 different hardwood concentrations (say, 5%, 10%, and 15%)?

H0: μ1 = μ2 H0: μ1 = μ3 H0: μ2 = μ3

H1: μ1 ≠ μ2 H1: μ1 ≠ μ3 H1: μ2 ≠ μ3

How about 4 different concentrations (say, 5%, 10%, 15%, and 20%)?

All of the above, PLUS

H0: μ1 = μ4 H0: μ2 = μ4 H0: μ3 = μ4

H1: μ1 ≠ μ4 H1: μ2 ≠ μ4 H1: μ3 ≠ μ4

What about 5 concentrations? 10?

and and

and and


Comparing Multiple Means - Type I Error Suppose α = 0.05 P(Type 1 error) = 0.05

(1 – α) = P (accept H0 | H0 is true) = 0.95Conducting multiple t-tests increases the

probability of a Type 1 errorThe greater the number of t-tests, the

greater the error probability 4 concentrations: (0.95)4 = 0.814

5 concentrations: (0.95)5 = 0.774

10 concentrations: (0.95)10 = 0.599Making the comparisons simultaneously

(as in an ANOVA) reduces the error back to 0.05


Analysis of Variance (ANOVA) Terms

Independent variable: that which is variedTreatment Factor

Level: the selected categories of the factor In a single–factor experiment there are a levels

Dependent variable: the measured resultObservations Replicates (N observations in the total experiment)

Randomization: performing experimental runs in random order so that other factors don’t influence results.


The Experimental Design

Suppose a manufacturer is concerned about the tensile strength of the paper used to produce grocery bags. Product engineers believe that tensile strength is a function of the hardwood concentration and want to test several concentrations for the effect on tensile strength. Six specimens were made at each of the 4 hardwood concentrations (5%, 10%, 15%, and 20%). The 24 specimens were tested in random order on a tensile test machine.

Terms Factor: Hardwood Concentration Levels: 5%, 10%, 15%, 20% a = 4 N = 24


The Results and Partial Analysis

The experimental results consist of 6 observations at each of 4 levels for a total of N = 24 items. To begin the analysis, we calculate the average and total for each level.

Hardwood Observations

Concentration 1 2 3 4 5 6 Totals Averages

5% 7 8 15 11 9 10 60 10.00

10% 12 17 13 18 19 15 94 15.67

15% 14 18 19 17 16 18 102 17.00

20% 19 25 22 23 18 20 127 21.17

383 15.96


To determine if there is a difference in the response at the 4 levels …

1. Calculate sums of squares

2. Calculate degrees of freedom

3. Calculate mean squares

4. Calculate the F statistic

5. Organize the results in the ANOVA table

6. Conduct the hypothesis test


Calculate the sums of squares

TreatmentTotalError

a

i

iTreatment

a

i

n

jijTotal

SSSSSS

N

y

n

ySS

N

yySS

2

1

2

2

1 1

2

96.51224

383109111587

2222222 TotalSS

79.382

24

383

6

1271029460 22222

TreatmentSS

1667.130 TreatmentTotalError SSSSSS


Additional Calculations

Calculate Degrees of Freedomdftreat = a – 1 = 3

df error = a(n – 1) = 20

dftotal = an – 1 = 23

Mean Square, MS = SS/df

MStreat = 382.7917/3 = 127.5972

MSE = 130.1667 /20 = 6.508333

Calculate F = MStreat / MSError = 127.58 / 6.51 = 19.61


Organizing the Results

Build the ANOVA table

Determine significance

fixed α-level compare to Fα,a-1, a(n-1)

p – value find p associated with this F with degrees of freedom a-1, a(n-1)

ANOVA

Source of Variation SS df MS F P-value F crit

Treatment 382.79 3 127.6 19.6 3.6E-06 3.1

Error 130.17 20 6.5083

Total 512.96 23


Conduct the Hypothesis Test

Null Hypothesis: The mean tensile strength is the same for each hardwood concentration.

Alternate Hypothesis: The mean tensile strength differs for at least one hardwood concentration

Compare Fcrit to Fcalc

Draw the graphic

State your decision with respect to the null hypothesis

State your conclusion based on the problem statement


Hypothesis Test Results



Fcrit less than Fcalc

Draw the graphic

Reject the null hypothesis

Conclusion: The mean tensile strength differs for at least one hardwood concentration.


Hypothesis Test Results



Fcrit less than Fcalc

Draw the graphic

Reject the null hypothesis

Conclusion: The mean tensile strength differs for at least one hardwood concentration.


Post-hoc Analysis: “Hand Calculations”

1. Calculate and check residuals, eij = Oi - Ei

plot residuals vs treatments normal probability plot

2. Perform ANOVA and determine if there is a difference in the means

3. If the decision is to reject the null hypothesis, identify which means are different using Tukey’s procedure:

4. Model: yij = μ + αi + εij

nMSkqyy Esr /1),,()(


Graphical Methods - Computer Individual 95% CIs For Mean Based on

Pooled StDev

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

5% 6 10.000 2.828 (----*----)

10% 6 15.667 2.805 (----*-----)

15% 6 17.000 1.789 (----*-----)

20% 6 21.167 2.639 (-----*----)

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

8.0 12.0 16.0 20.0

Data

20%15%10%5%

25

20

15

10

5

Boxplot of 5% , 10% , 15% , 20%


Numerical Methods - Computer

Tukey’s test Duncan’s Multiple Range test Easily performed in Minitab

Tukey 95% Simultaneous Confidence Intervals (partial results)

10% subtracted from:

Lower Center Upper ----+---------+---------+---------+-----15% -2.791 1.333 5.458 (-----*-----)

20% 1.376 5.500 9.624 (-----*-----)

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

-7.0 0.0 7.0 14.0


Blocking

Creating a group of one or more people, machines, processes, etc. in such a manner that the entities within the block are more similar to each other than to entities outside the block.

Balanced design: n = 1 for each treatment/block category

Model: yij = μ + α i + βj + εij


Example:Robins Air Force Base uses CO2 to strip paint from F-15’s. You have been asked to design a test to determine the optimal pressure for spraying the CO2. You realize that there are five machines that are being used in the paint stripping operation. Therefore, you have designed an experiment that uses the machines as blocking variables. You emphasized the importance of balanced design and a random order of testing. The test has been run with these results (values are minutes to strip one fighter):


ANOVA: One-Way with Blocking

1. Construct the ANOVA table

Where,

SSBSSASSTSSE

yykSSB

yybSSA

yySST

b

jj

k

ii

k

i

b

jij

1

2

1

2

1 1

2

)(

)(

)(


Blocking Example

Your turn: fill in the blanks in the following ANOVA table (from Excel):

2. Make decision and draw conclusions:

ANOVA


Rows 89.733 2 44.867 8.492 0.0105 4.458968

Columns 77.733 ___ _____ ____ 0.0553 _______

Error 42.267 8 5.2833

Total 209.73 ___


Two-Way ANOVA

Blocking is used to keep extraneous factors from masking the effects of the one treatment you are interested in studying.

A two-way ANOVA is used when you are interested in determining the effect of two treatments.

Model: yijk = μ + α i + βj + (α β)ijk + εij

α is the main effect of Treatment Aβ is the main effect of Treatment BThe α β component is the interaction effect


Two-Way ANOVA w/ Replication Your fame as an experimental design expert grows. You

have been called in as a consultant to help the Pratt and Whitney plant in Columbus determine the best method of applying the reflective stripe that is used to guide the Automated Guided Vehicles (AGVs) along their path. There are two ways of applying the stripe (paint and coated adhesive tape) and three types of flooring (linoleum and two types of concrete) in the facilities using the AGVs. You have set up two identical “test tracks” on each type of flooring and applied the stripe using the two methods under study. You run 3 replications in random order and count the number of tracking errors per 1000 ft of track. The results are as follows:


Two-Way ANOVA Example

Analysis is the similar to the one-way ANOVA; however we are now concerned with interaction effects

The two-way ANOVA table displays three calculated F values


Two-Way ANOVA


Your Turn

Fill in the blanks …

What does this mean?

ANOVA


Sample 0.4356 1

_____ 2.3976 0.14748 4.74722

Columns 4.48 2 2.24 12.33 0.00123 3.88529

Interaction 0.9644 ___ 0.4822 _____ 0.11104 3.88529

Within 2.18 ___ 0.1817

Total 8.06 17


What if Interaction Effects are Significant?

For example, suppose a new test was run using different types of paint and adhesive, with the following results:

Linoleum Concrete I Concrete IIPaint 10.7 10.8 12.2

10.9 11.1 12.311.3 10.7 12.5

Adhesive 11.2 11.9 10.911.6 12.2 11.610.9 11.7 11.9

ANOVA


Sample 0.109 1 0.1089 1.071 0.3211 4.7472

Columns 1.96 2 0.98 9.639 0.0032 3.8853

Interaction 2.831 2 1.4156 13.92 0.0007 3.8853

Within 1.22 12 0.1017

Total 6.12 17


Understanding Interaction EffectsGraphical methods:

graph means vs factors identify where the effect will change the result for one factor

based on the value of the other.

Interaction

10.5

11

11.5

12

12.5

0 1 2 3 4

Floor Type

Tra

ckin

g E

rro

rs

Paint

Adhesive

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Analysis of Variance (ANOVA)

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