1. CHAPTER 13 Design and Analysis of Single-Factor Experiments:
The Analysis of Variance

2. Learning Objectives Design and conduct engineering experiments
Understand how the analysis of variance is used to analyze the data
Use multiple comparison procedures
Make decisions about sample size
Understand the difference between fixed and random factors
Estimate variance components
Understand the blocking principle
Design and conduct experiments involving the randomized complete block design

3. Engineering Experiments
Experiments are a natural part of the engineering decision-making process
Designed to improve the performance of a subset of processes
Processes can be described in terms of controllable variables
Determine which subset has the greatest influence
Such analysis can lead to
Improved process yield
Reduced variability in the process and closer conformance to nominal or target requirements
Reduced cost of operation

4. Steps In Experimental Design
Usually designed sequentially
Determine which variables are most important
Used to refine the information to determine the critical variables for improving the process
Determine which variables result in the best process performance

5. Single Factor Experiment
Assume a parameter of interest
Consist of making up several specimens in two samples
Analyzed them using the statistical hypotheses methods
Can say an experiment with single factor
Has two levels of investigations
Levels are called treatments
Treatment has n observations or replicates

6. Designing Engineering Experiments
More than two levels of the factor
This chapter shows
ANalysis Of VAriance (ANOVA)
Discuss randomization of the experimental runs
Design and analyze experiments with several factors

7. Linear Statistical Model
Following linear model
Yij= µ+i+ij
i=1, 2,…,a, and j=1, 2,…,n
Yij is (ij)th observation
µ called the overall mean
i called the ith treatment effect
ij is a random error component with mean zero and variance 2
Each treatment defines a population
Mean µi consisting of the overall mean µ
Plus an effect i
Pg. 471 Fig 13-1b

8. Completely Randomized Design
Table shows the underlying model
Following observations are taken in random order
Treatments are used as uniform as possible
Called completely randomized design

9. Fixed-effects and Random Models
Chosen in two different ways
Experimenter chooses the a treatments
Called the fixed-effect model
Experimenter chooses the treatments from a larger population
Called random-effect model

10. Development of ANOVA
Total of the observations and the average of the observations under the ith treatment
Grand total of all observations and the grand mean
N=an is the total number of observations
“dot” subscript notation implies summation

11. Hypothesis Testing H0: 1=2=…=a=0
Interested in testing the equality of the following a treatment means
1,2 ….. a
Equivalent
H0: 1=2=…=a=0
H1: a#0 for at least one i
If the null hypothesis is true, changing the levels of the factor has no effect on the mean response

12. Components of Total Variability
Total variability in data is described by the total sum of squares
Partitions this total variability into two parts
Measure the differences between treatments
Measure the random error effect

13. Computational Formulas
Mean square for treatments
MSTreatments=SSTreatments /(a-1)
Error mean square
MSE=SSE /[a(n-1)]
Efficient formulas
Total sum of squares
Treatment sum of squares
Error sum of squares
SSE=SST - SSTreatment

14. ANOVA TABLE

15. Using Computer Software
Packages have the capability to analyze data from designed experiments
Presents the output from the Minitab one-way analysis of variance routine

16. Example
The tensile strength of a synthetic fiber is of interest to the manufacturer. It is suspected that strength is related to the percentage of cotton in the fiber. Five levels of cotton percentage are used, and five replicates are run in random order, resulting in the data below. Use α=0.05.
a) Does cotton percentage affect breaking strength?

17. Solution Use the general steps in hypothesis testing
Parameter of interest is the cotton percentage
H0: 1=2= 3=4=5=0
H1: i #0 for at least one I
α = 0.05
Test statistic
Fo = MSTR /MSE
6. Reject Ho if fo > fα,(a-1)n(a-1)
Computations

18. Initial calculations Compute the last two columns Conc 1 2 3 4 5 15 711 9 49
9.8 20 12 17 18 77
15.4 25 14 19 88
17.6 30 22 23
108
21.6 35 10 54
10.8
=376

19. Solution - Cont. 8. Since fo=14.75> f0.05,4,20= 2.87, reject Ho
Compute SST, SSTR, SSE , MSTR, and MSE
=(7)2 +(7)2 + ….+(376)2/25=
=((49)2 +(77)2 +..+(54)2)/5 -376/ =475.7
SSE = =
MSTR= SSTR/a-1 = /4 = 118.9
MSE = SSE/a(n-1)=161.20/5(5-1) = 8.0
Hence, the test statistic
Fo = MSTR /MSE = /8.06 = 14.75
8. Since fo=14.75> f0.05,4,20= 2.87, reject Ho

20. Solution ANOVA results Source DF SS MS F P
COTTON
Error
Total
Reject H0 and conclude that cotton percentage affects breaking strength

21. Multiple Comparisons Following the ANOVA
When H0:1=2=…=a=0 is rejected
Know that some of the treatment are different
Doesn’t identify which means are different
Called multiple comparisons methods
Called Fisher’s least significant difference (LSD) method

22. Fisher’s Least Significant Difference (LSD) Method
Compares all pairs of means with the H0: = for all i # j
Test statistic
Pair of means i and j would be different
Least significant difference, LSD, is

23. Example
Use Fisher’s LSD method with α = 0.05 test to analyze the means of five different levels of cotton percentage content in the previous example
Recall H0 was rejected and concluded that cotton percentage affects the breaking strength
Apply the Fisher’s LSD method to determine which treatment means are different

24. Solution Summarize a =5 means, n=5, MSE = 8.06, and t0.025,20=2.086
Treatment means are
9.8
15.4
17.6
21.6
10.8

25. Solution –Cont. Value of LSD Comparisons
5 Vs. 1=I10.8–9.8I=1
5 Vs. 2=I I=4.6>3.74
5 Vs. 3=I I=6.8>3.74
5 Vs. 4=I I=10.8>3.74
4 Vs. 1=I I=11.8>3.74
4 Vs. 2=I I=6.2 > 3.74
4 Vs. 3=I21.6 –17.6I=4>3.74
3 Vs. 1=I I=7.8>3.74
3 Vs. 2=I I=2.2
2 Vs. 1=I I=5.6>3.74
From this analysis, we see that there are significant differences between all pairs of means except 5 vs. 1 and 3 vs. 2

26. C.I. on Treatment Means
Confidence interval on the mean of the ith treatment µi
Confidence interval on the difference in two treatment means

27. Determining Sample Size
Choice of the sample size to use is important
OC curves provide guidance in making this selection
Power of the ANOVA test is
1-β=P( Reject H0 | H0 is false)
=P(F0> fα, a-1, a(n-1) | H0 is false)
Plot β against a parameter 

28. Sample OC Curves

29. Example
Suppose that four normal populations have common variance 2=25 and means µ1 =50, µ 2=60, µ3=50, and µ4=60. How many observations should be taken on each population so that the probability of rejecting the hypothesis of equality of means is at least 0.90? Use α=0.05

30. Solution Average mean 1 = -5, 2 = 5, 3 = -5, 4 = 5
Various choices:
Therefore, n = 5 is needed

31. The Random-effects Model
A large number of possible levels
Experimenter randomly selects a of these levels from the population of factor levels
Called random-effect model
Valid for the entire population of factor levels

32. Linear Statistical Model
Following linear model
Yij= µ+i+ij
=1,2,….a, j=1,2,…n
Yij is the (ij)th observation
i and ij are independent random variables
Identical in structure to the fixed-effects case
Parameters have a different interpretation
ij are with mean 0 and variance 2
i are with mean zero and variance 2

33. Testing the Hypothesis
Testing the hypothesis that the individual treatment effects are zero is meaningless
Appropriate to test a hypothesis about the variance of the treatment effect
H0: 2 =0 vs. H1: 2 >0
2 =0, all treatments are identical
There is variability between them
Total variability
SST=SSTreatments +SSE
Expected values of the MS
E(MSTreatments)= 2 + n2 and E(MSE)= 2
Computational procedure and construction of the ANOVA table are identical to the fixed-effects case
Eq ,22

34. Randomized Complete Block Design
Desired to design an experiment so that the variability arising from a nuisance factor can be controlled
Recall about the paired t-test
When all experimental runs cannot be made under homogeneous conditions
See the paired t-test as a method for reducing the noise in the experiment by blocking out a nuisance factor effect
Randomized block design can be viewed as an extension of the paired t-test
Factor of interest has more than two levels
More than two treatments must be compared

35. Randomized Complete Block Design
General procedure for a randomized complete block design consists of selecting b blocks
Data that result from running a randomized complete block design for investigating a single factor with a levels and b blocks

36. Linear Statistical Model
Following linear model
Yij is the (ij)th observation
j is the effect of the jth block
µ called the overall mean
i called the ith treatment effect
ij is a random error component with mean zero and variance 2

37. Hypothesis Testing H0: 1=2=…=a=0
Interested in testing the equality of the a treatment means
1,2 ….. a
Equivalent
H0: 1=2=…=a=0
H1: a#0 for at least one i
If the null hypothesis is true, changing the levels of the factor has no effect on the mean response

38. Displaying Data

39. Components of Total Variability
Total variability in data
Partitions this total variability into three parts
Or symbolically,
SST=SStreatments+SSblocks+SSerrors

40. Computational Formulas
Computing formulas for the sums of squares
Error sum of squares
SSerrors=SST-SStreatments-SSblocks
Computer software package will be used to perform the analysis of variance

41. Analysis of Variance

42. Next Agenda
Ends our discussion with the analysis of variance when there are more then two levels of a single factor
In the next chapter, we will show how to design and analyze experiments with several factors with more than two levels