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
EGR 252 Spring Ch.13 Part 1

2. Comparing More Than Two Means
What if there are 3 different hardwood concentrations (say, 5%, 10%, and 15%)?
H0: μ1 = μ H0: μ1 = μ H0: μ2 = μ3
H1: μ1 ≠ μ H1: μ1 ≠ μ H1: μ2 ≠ μ3
How about 4 different concentrations (say, 5%, 10%, 15%, and 20%)?
All of the above, PLUS
H0: μ1 = μ H0: μ2 = μ H0: μ3 = μ4
H1: μ1 ≠ μ H1: μ2 ≠ μ H1: μ3 ≠ μ4
What about 5 concentrations? 10?
and
and
5 concentrations = 5!/(2!*3!) = 10 tests …
10 concentrations = 10!/(2!*8!) = 45 tests …
and
and
EGR 252 Spring Ch.13 Part 1

3. Comparing Multiple Means - Type I Error
Suppose α = P(Type 1 error) = 0.05
(1 – α) = 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 concentrations =
5 conc. =
10 conc. =
EGR 252 Spring Ch.13 Part 1

4. Analysis of Variance (ANOVA) Terms
Independent variable: that which is varied
Treatment
Factor
Level: the selected categories of the factor
In a single–factor 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 don’t influence results.
4 concentrations =
5 conc. =
10 conc. =
EGR 252 Spring Ch.13 Part 1

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
EGR 252 Spring Ch.13 Part 1

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
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 94
15.67
15% 14 16
102
17.00
20% 25 22 23 20
127
21.17
383
15.96
Overlay:
a = 4
n = 6
Hardwood Observations
Concentration (%) Totals Averages
y11 y12 y13 y14 y15 y16 y1• y1• (bar)
y21 y22 y23 y24 y25 y26 y2• y2• (bar)
y31 y32 y33 y34 y35 y36 y3• y3• (bar)
y41 y42 y43 y44 y45 y46 y4• y4• (bar)
y•• y•• (bar)
EGR 252 Spring Ch.13 Part 1

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
SStotal = ( … ) – 3832/24 =
SStreat = ( )/6 – 3832/24 =
SSE = SStotal - SStreat =
dftreat = 3
dfE = 20
dftotal = 23
EGR 252 Spring Ch.13 Part 1

8. Calculate the sums of squares
SStotal = ( … ) – 3832/24 =
SStreat = ( )/6 – 3832/24 =
SSE = SStotal - SStreat =
dftreat = 3
dfE = 20
dftotal = 23
EGR 252 Spring Ch.13 Part 1

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 = /3 =
MSE = /20 =
Calculate F = MStreat / MSError = / 6.51 = 19.61
MStreat = SStreat/dftreat = /3 =
MSE = SSE/dfE = /20 =
F = /6.51 = 19.61
EGR 252 Spring Ch.13 Part 1

10. 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
F =
F.05,3,20 = 3.10
p-value =3.6E-06
EGR 252 Spring Ch.13 Part 1

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
conclusion: there is a difference in tensile strength as a function of hardwood concentration
EGR 252 Spring Ch.13 Part 1

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.
conclusion: there is a difference in tensile strength as a function of hardwood concentration
EGR 252 Spring Ch.13 Part 1

13. 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 Tukey’s procedure:
Model: yij = μ + αi + εij
note: α in model refers to the treatment effect (not the significance level)
EGR 252 Spring Ch.13 Part 1

14. Graphical Methods - Computer
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev
5% (----*----)
10% (----*-----)
15% (----*-----)
20% (-----*----)
EGR 252 Spring Ch.13 Part 1

15. 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% (-----*-----)
20% (-----*-----)
EGR 252 Spring Ch.13 Part 1