Analysis of Variance (ANOVA)

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

2
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
3
Comparing Multiple Means - Type I Error

• Suppose a 0.05 P(Type 1 error) 0.05
• (1 a) P (accept H0 H0 is true) 0.95
• Conducting multiple t-tests increases the
  probability of a Type 1 error
• The 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 0.599
• Making the comparisons simultaneously (as in an
  ANOVA) reduces the error back to 0.05

4
Analysis of Variance (ANOVA) Terms

• Independent variable that which is varied
• Treatment
• Factor
• Level the selected categories of the factor
• In a singlefactor experiment there are a levels
• Dependent variable the measured result
• Observations
• Replicates
• (N observations in the total experiment)
• Randomization performing experimental runs in
  random order so that other factors dont
  influence results.

5
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

6
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 Observations Observations Observations Observations 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
7
To determine if there is a difference in the
response at the 4 levels

• Calculate sums of squares
• Calculate degrees of freedom
• Calculate mean squares
• Calculate the F statistic
• Organize the results in the ANOVA table
• Conduct the hypothesis test

8
Calculate the sums of squares
9
Additional Calculations

• Calculate Degrees of Freedom
• dftreat 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

10
Organizing the Results

• Build the ANOVA table
• Determine significance
• fixed a-level ? compare to Fa,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        
11
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

12
Hypothesis Test Results

• 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
• Fcrit less than Fcalc
• Draw the graphic
• Reject the null hypothesis
• Conclusion The mean tensile strength differs for
  at least one hardwood concentration.

13
Hypothesis Test Results

• 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
• Fcrit less than Fcalc
• Draw the graphic
• Reject the null hypothesis
• Conclusion The mean tensile strength differs for
  at least one hardwood concentration.

14
Post-hoc Analysis Hand Calculations

• Calculate and check residuals, eij Oi - Ei
• plot residuals vs treatments
• normal probability plot
• Perform ANOVA and determine if there is a
  difference in the means
• If the decision is to reject the null hypothesis,
  identify which means are different using Tukeys
  procedure
• Model yij µ ai eij

15
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

16
Numerical Methods - Computer

• Tukeys test
• Duncans 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

17
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 a i ßj eij

18
Example

• Robins Air Force Base uses CO2 to strip paint
  from F-15s. 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)

19
ANOVA One-Way with Blocking

• Construct the ANOVA table
• Where,

20
Blocking Example

• Your turn fill in the blanks in the following
  ANOVA table (from Excel)
• Make decision and draw conclusions

ANOVA
Source of Variation SS df MS F P-value F crit
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 ___        
21
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 a i ßj (a ß)ijk eij
• a is the main effect of Treatment A
• ß is the main effect of Treatment B
• The a ß component is the interaction effect

22
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

23
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

24
Two-Way ANOVA
25
Your Turn

• Fill in the blanks
• What does this mean?

ANOVA
Source of Variation SS df MS F P-value F crit
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        
26
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

ANOVA
Source of Variation SS df MS F P-value F crit
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        
27
Understanding Interaction Effects

• Graphical methods
• graph means vs factors
• identify where the effect will change the result
  for one factor based on the value of the other.