1. Chapter 3 Experiments with a Single Factor: The Analysis of Variance

2. 3.1 An Example
Chapter 2: A signal-factor experiment with two levels of the factor
Consider signal-factor experiments with a levels of the factor, a  2
Example:
The tensile strength of a new synthetic fiber.
The weight percent of cotton
Five levels: 15%, 20%, 25%, 30%, 35%
a = 5 and n = 5

3. Does changing the cotton weight percent change the mean tensile strength?
Is there an optimum level for cotton content?

4. 3.2 The Analysis of Variance
a levels (treatments) of a factor and n replicates for each level.
yij: the jth observation taken under factor level or treatment i.

5. Models for the Data
Means model:
yij is the ijth observation,
i is the mean of the ith factor level
ij is a random error with mean zero
Let i =  + i ,  is the overall mean and i is the ith treatment effect
Effects model:

6. Linear statistical model
One-way or Signal-factor analysis of variance model
Completely randomized design: the experiments are performed in random order so that the environment in which the treatment are applied is as uniform as possible.
For hypothesis testing, the model errors are assumed to be normally and independently distributed random variables with mean zero and variance, 2, i.e. yij ~ N(+i, 2)
Fixed effect model: a levels have been specifically chosen by the experimenter.

7. 3.3 Analysis of the Fixed Effects Model
Interested in testing the equality of the a treatment means, and E(yij) = i =  + i, i = 1,2, …, a
H0: 1 = … = a v.s.
H1: i  j, for at least one pair (i, j)
Constraint:
H0: 1 = … = a =0 v.s. H1: i  0, for at least one i

8. Notations:
3.3.1 Decomposition of the Total Sum of Squares
Total variability into its component parts.
The total sum of squares (a measure of overall variability in the data)
Degree of freedom: an – 1 = N – 1

9. SSTreatment: sum of squares of the differences between the treatment averages (sum of squares due to treatments) and the grand average, and a – 1 degree of freedom
SSE: sum of squares of the differences of observations within treatments from the treatment average (sum of squares due to error), and N – a degrees of freedom.

10. A large value of SSTreatments reflects large differences in treatment means
A small value of SSTreatments likely indicates no differences in treatment means
dfTotal = dfTreatment + dfError
If there are no differences between a treatment means,

11. Mean squares:
3.3.2 Statistical Analysis
Assumption: ij are normally and independently distributed with mean zero and variance 2

12. SST/2 ~ Chi-square (N – 1), SSE/2 ~ Chi-square (N – a), SSTreatments/2 ~ Chi-square (a – 1), and SSE/2 and SSTreatments/2 are independent (Theorem 3.1)
H0: 1 = … = a =0 v.s. H1: i  0, for at least one i

13. Reject H0 if F0 > F, a-1, N-a
Rewrite the sum of squares:
See page 71

14. Response:. Strength. ANOVA for Selected Factorial Model
Response: Strength ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares]
Sum of Mean F Source Squares DF Square Value Prob > F Model < A < Pure Error Cor Total
Std. Dev R-Squared Mean Adj R-Squared C.V Pred R-Squared
PRESS Adeq Precision 9.294

15. 3.3.3 Estimation of the Model Parameters
Model: yij =  + i + ij
Estimators:
Confidence intervals:

16. Example 3.3 (page 75)
Simultaneous Confidence Intervals (Bonferroni method): Construct a set of r simultaneous confidence intervals on treatment means which is at least 100(1-): 100(1-/r) C.I.’s
3.3.4 Unbalanced Data
Let ni observations be taken under treatment i, i=1,2,…,a, N = i ni,

17. 1. The test statistic is relatively insensitive to small departures from the assumption of equal variance for the a treatments if the sample sizes are equal.
2. The power of the test is maximized if the samples are of equal size.

18. 3.4 Model Adequacy Checking
Assumptions: yij ~ N(+i, 2)
The examination of residuals
Definition of residual:
The residuals should be structureless.

19. 3.4.1 The Normality Assumption
Plot a histogram of the residuals
Plot a normal probability plot of the residuals
See Table 3-6

20. May be
Slightly skewed (right tail is longer than left tail)
Light tail (the left tail of error is thinner than the tail part of standard normal)
Outliers
The possible causes of outliers: calculations, data coding, copy error,….
Sometimes outliers are more informative than the rest of the data.

21. Detect outliers: Examine the standardized residuals,
3.4.2 Plot of Residuals in Time Sequence
Plotting the residuals in time order of data collection is helpful in detecting correlation between the residuals.
Independence assumption

23. 3.4.3 Plot of Residuals Versus Fitted Values
Plot the residuals versus the fitted values
Structureless

24. Variance-stabilizing transformation
Nonconstant variance: the variance of the observations increases as the magnitude of the observation increase, i.e. yij  2
If the factor levels having the larger variance also have small sample sizes, the actual type I error rate is larger than anticipated.
Variance-stabilizing transformation
Poisson
Square root transformation
Lognormal
Logarithmic transformation
Binomial
Arcsin transformation

25. Statistical Tests for Equality Variance:
Bartlett’s test:
Reject null hypothesis if

26. Example 3.4: the test statistic is
Bartlett’s test is sensitive to the normality assumption
The modified Levene test:
Use the absolute deviation of the observation in each treatment from the treatment median.
Mean deviations are equal => the variance of the observations in all treatments will be the same.
The test statistic for Levene’s test is the ANOVA F statistic for testing equality of means.

27. Example 3.5:
Four methods of estimating flood flow frequency procedure (see Table 3.7)
ANOVA table (Table 3.8)
The plot of residuals v.s. fitted values (Figure 3.7)
Modified Levene’s test: F0 = 4.55 with P-value = Reject the null hypothesis of equal variances.

28. Find y* = y that yields a constant variance. *  +-1
Let E(y) =  and y  
Find y* = y that yields a constant variance.
*  +-1
Variance-Stabilizing Transformations
* and  
= 1 - 
Transformation
*constant 1
No transformation
*  1/2
Square root
*  
Log
*  3/2
3/2
-1/2
Reciprocal square root
*  2 2 -1
Reciprocal

29. How to find :
Use
See Figure 3.8, Table 3.10 and Figure 3.9

30. 3.5 Practical Interpretation of Results
Conduct the experiment => perform the statistical analysis => investigate the underlying assumptions => draw practical conclusion
3.5.1 A Regression Model
Qualitative factor: compare the difference between the levels of the factors.
Quantitative factor: develop an interpolation equation for the response variable.

31. Regression analysis See Figure 3.1
Final Equation in Terms of Actual Factors:
Strength = * Cotton Weight % * Cotton Weight %^ E-003 * Cotton Weight %^3
This is an empirical model of the experimental results

32. 3.5.2 Comparisons Among Treatment Means
If that hypothesis is rejected, we don’t know which specific means are different
Determining which specific means differ following an ANOVA is called the multiple comparisons problem
3.5.3 Graphical Comparisons of Means

33. 3.5.4 Contrast
A contrast: a linear combination of the parameters of the form
H0:  = 0 v.s. H1:   0
Two methods for this testing.

34. The first method:

35. The second method:

36. The C.I. for a contrast, 
Unequal Sample Size

37. 3.5.5 Orthogonal Contrast
Two contrasts with coefficients, {ci} and {di}, are orthogonal if ci di = 0
For a treatments, the set of a – 1 orthogonal contrasts partition the sum of squares due to treatments into a – 1 independent single-degree-of-freedom components. Thus, tests performed on orthogonal contrasts are independent.
See Example 3.6 (Page 94)

38. 3.5.6 Scheffe’s Method for Comparing All Contrasts
Scheffe (1953) proposed a method for comparing any and all possible contrasts between treatment means.
See Page 95 and 96

39. 3.5.7 Comparing Pairs of Treatment Means
Compare all pairs of a treatment means
Tukey’s Test:
The studentized range statistic:
See Example 3.7

40. Sometimes overall F test from ANOVA is significant, but the pairwise comparison of mean fails to reveal any significant differences.
The F test is simultaneously considering all possible contrasts involving the treatment means, not just pairwise comparisons.
The Fisher Least Significant Difference (LSD) Method
For H0: i = j

41. The least significant difference (LSD):
See Example 3.8
Duncan’s Multiple Range Test
The a treatment averages are arranged in ascending order, and the standard error of each average is determined as

42. Assume equal sample size, the significant ranges are
Total a(a-1)/2 pairs
Example 3.9
The Newman-Keuls Test
Similar as Duncan’s multiple range test
The critical values:

43. 3.5.8 Comparing Treatment Means with a Control
Assume one of the treatments is a control, and the analyst is interested in comparing each of the other a – 1 treatment means with the control.
Test H0: i = a v.s. H1: : i  a, i = 1,2,…, a – 1
Dunnett (1964)
Compute
Reject H0 if
Example 3.10

44. 3.7 Determining Sample Size
Determine the number of replicates to run
3.7.1 Operating Characteristic Curves (OC Curves)
OC curves: a plot of type II error probability of a statistical test,

45. F0 = MSTreatment / MSE ~ noncentral F
If H0 is false, then
F0 = MSTreatment / MSE ~ noncentral F
with degree of freedom a – 1 and N – a and noncentrality parameter 
Chart V of the Appendix
Determine
Let i be the specified treatments. Then estimates of i :
For 2, from prior experience, a previous experiment or a preliminary test or a judgment estimate.

46. Example 3.11
Difficulty: How to select a set of treatment means on which the sample size decision should be based.
Another approach: Select a sample size such that if the difference between any two treatment means exceeds a specified value the null hypothesis should be rejected.

47. 3.7.2 Specifying a Standard Deviation Increase
Let P be a percentage for increase in standard deviation of an observation. Then
For example (Page 110): If P = 20, then

48. 3.7.3 Confidence Interval Estimation Method
Use Confidence interval.
For example: we want 95% C.I. on the difference in mean tensile strength for any two cotton weight percentages to be  5 psi and  = 3. See Page 110.

49. 3.9 The Regression Approach to the Analysis of Variance
Model: yij =  + i + ij

50. The normal equations
Apply the constraint
Then estimations are
Regression sum of squares (the reduction due to fitting the full model)

51. The error sum of squares:
Find the sum of squares resulting from the treatment effects:

52. The testing statistic for H0: 1 = … = a