Lecture 8: Analysis of Variance
Formal Analysis
Partitioning Variability
The Analysis of Variance –
ANOVA
Quantify the total amount
of variability and split it
among the sources.
Total Variability
Variability due to
treatments effects.
+
Variability due to
chance (random) error.
Measurement
Experimental Material
Planned Systematic
1
Total Variability
2
Total Variability
∑
How much variability is
there for the entire set of
experimental data?
Compute a sample variance.
4140.95
19
1
3
4
Partitioning Variability
Partitioning Variability
Sum of Squares
C Total
Degrees of Freedom
C Total
Sum of Squares
Treatments
+
Degrees of Freedom
Treatments
Sum of Squares
Error
5
+
Degrees of Freedom
Error
6
1
Lecture 8: Analysis of Variance
Treatments
SSTreatment
Estimated Treatment Effects
̂
50 mmol Na/day:
143.10 – 153.05 = –9.95
= 10(–9.95)2 + 10(+9.95)2
= 1980.05
200 mmol Na/day:
163.00 – 153.05 = +9.95
7
8
Random Error
SSError
Pool the sums of squares
within each treatment by
creating the weighted sum
of treatment sample
variances.
1
= 9(11.75)2 + 9(10.10)2
= 2160.9
9
Degrees of Freedom
10
Analysis of Variance
Source
Treatment: number of
treatments minus 1, (k – 1)
Error: total number of
observations minus the number
of treatments, (N – k)
11
df
Sum of Mean F Ratio Prob > F
Squares Square
Diet
1
1980.05 1980.05 16.4935
Error
18
2160.90
C. Total
19
4140.95
0.0007*
120.05
12
2
Lecture 8: Analysis of Variance
Test of Hypothesis
F Ratio
F Ratio is the mean square
for treatment divided by the
mean square error.
:
:
or
:
:
13
14
F Ratio
Prob > F
A large F Ratio indicates
that differences between
treatment sample means are
larger than random error.
This is the P-value for the F
Ratio.
15
16
Prob > F
Decision
Small P-values (<0.05)
indicate that difference in
treatment sample means are
statistically significant.
The F Ratio=16.4935 and Pvalue=0.0007.
The small P-value indicates
that the difference in treatment
sample means is statistically
significant.
17
18
3
Lecture 8: Analysis of Variance
Generalization
JMP
For men with Stage 1
hypertension, a low sodium
diet (50 mmol Na/day) will
result in lower average blood
pressure than a high sodium
diet (200 mmol Na/day).
Analyze
Fit Y by X
Y, Response: Systolic BP
X, Factor: Diet
19
20
Menu next to Oneway
Oneway Analysis of Systolic BP (mmHg) By Diet
200
Means and Std Dev
Means/ANOVA/pooled t
150
100
050 mmol/day
200 mmol/day
Diet
21
22
23
24
Means and Std Dev
4