1. ANOVA
11/3

2. Comparing Several Groups
Section:
M 1-3
M 3-5
T 9-11
W 1-3
R 9-11
Exam 2 Mean:
82.7%
80.1%
85.4%
77.7%
Do the group means differ?
Naive approach
t-tests of all pairs
Each test doesn't use all data  Less power
10 total tests  Greater chance of Type I error
Analysis of Variance (ANOVA)
Single test for any group differences
Null hypothesis: All means are equal
Works using variance of the sample means
Also based on separating explained and unexplained variance

3. Variance of Sample Means
Say we have k groups of n subjects each
Want to test whether mi are all equal for i = 1,…,k
Even if all population means are equal, sample means will vary
Central limit theorem tells how much:
Compare actual variance of sample means to amount expected by chance or
Estimate s2 using MSE or MSresidual
Test statistic:
If F is large enough
Sample means vary more than expected by chance
Reject null hypothesis m M1 M2 M3 Mk

4. Partitioning Sum of Squares
General approach to ANOVA uses sums of squares
Works with unequal sample sizes and other complications
Breaks total variability into explained and unexplained parts
SStotal
Sum of squares for all data, treated as a single sample
Based on differences from grand mean,
SSresidual (SSwithin)
Variability within groups, just like with independent-samples t-test
Not explainable by group differences
SStreatment (SSbetween)
Variability explainable by difference among groups
Treat each datum as its group mean and take difference from grand mean

5. Magic of Squares M1 SStotal = S(X – M)2 ×5 M M3 ×5
SSresidual = Si(S(Xi – Mi)2) ×6 ×4 M4 M2
SStreatment = Si(ni(Mi – M)2)

6. Test Statistic for ANOVA
Goal: Determine whether group differences account for more variability than expected by chance
Same approach as with regression
Calculate mean squares:
H0: MStreatment ≈ s2
Estimate s2 using
Test statistic:

7. Two Views of ANOVA Simple: Equal sample sizes
General: Using sums of squares m M1 M2 M3 Mk
If sample sizes equal:

8. Degrees of Freedom dfregression = k – 1 SStotal dftotal = Sni – 1
dfresidual = S(ni – 1)
= Sni – k .