Conducting ANOVA’s
Conducting ANOVA’s
I.
Why?
A. more than two groups to compare
Conducting ANOVA’s
I.
Why?
A. more than two groups to compare
What’s the prob?
D. putrida low density
D. putrida high density
D. putrida with D. tripuncatata
Conducting ANOVA’s
I.
Why?
A. more than two groups to compare
What’s the prob?
D. putrida low density
D. putrida high density
D. putrida with D. tripuncatata
What was our solution?
Conducting ANOVA’s
I.
Why?
A. more than two groups to compare
What’s the prob?
D. putrida low density
U
D. putrida high density
D. putrida with D. tripuncatata
What was our solution?
U
U
What’s the prob?
D. putrida low density
U
D. putrida high density
D. putrida with D. tripuncatata
U
U
Tested each contrast at p = 0.05
Probability of being correct in rejecting each Ho:
1
0.95
2
0.95
3
0.95
What’s the prob?
D. putrida low density
U
D. putrida high density
D. putrida with D. tripuncatata
U
U
Tested each contrast at p = 0.05
Probability of being correct in rejecting all Ho:
1
2
3
0.95 x
0.95 x
0.95 = 0.86
So, Type I error rate has increased from 0.05 to 0.14
Probability of being correct in rejecting all Ho:
1
2
3
0.95 x
0.95 x
0.95 = 0.86
So, Type I error rate has increased from 0.05 to 0.14
Hmmmm….. What can we do to maintain a 0.05
level across all contrasts?
Probability of being correct in rejecting all Ho:
1
2
3
0.95 x
0.95 x
0.95 = 0.86
So, Type I error rate has increased from 0.05 to 0.14
Hmmmm….. What can we do to maintain a 0.05
level across all contrasts?
Right. Adjust the comparison-wise error rate.
Probability of being correct in rejecting all Ho:
1
2
3
0.95 x
0.95 x
0.95 = 0.86
So, Type I error rate has increased from 0.05 to 0.14
Simplest: Bonferroni correction:
Comparison-wise p = experiment-wise p/n
Where n = number of contrasts.
Experient-wise = 0.05
Comparison-wise = 0.05/3 = 0.0167
Probability of being correct in rejecting all Ho:
1
2
3
0.95 x
0.95 x
0.95 = 0.86
So, Type I error rate has increased from 0.05 to 0.14
Simplest: Bonferroni correction:
Comparison-wise p = experiment-wise p/n
Where n = number of contrasts.
Experient-wise = 0.05
Comparison-wise = 0.05/3 = 0.0167
So, confidence = 0.983
Probability of being correct in rejecting all Ho:
1
2
3
0.983 x
0.983 x
0.983 = 0.95
So, Type I error rate is now 0.95
Simplest: Bonferroni correction:
Comparison-wise p = experiment-wise p/n
Where n = number of contrasts.
Experient-wise = 0.05
Comparison-wise = 0.05/3 = 0.0167
So, confidence = 0.983
Conducting ANOVA’s
I.
Why?
A. more than two groups to compare
What’s the prob?
- multiple comparisons reduce
experiment-wide alpha level.
- Bonferroni adjustments assume
contrasts as independent; but they are both
part of the same experiment…
- Bonferroni adjustments assume
contrasts as independent; but they are both
part of the same experiment…
Our consideration of 1 vs. 2 might consider
the variation in all treatments that were part
of this experiment; especially if we are
interpreting the differences between mean
comparisons as meaningful.
1 vs. 2 – not significant
1 vs. 3 – significant. So, interspecific
competition is more important than
intraspecific competition
- Bonferroni adjustments assume
contrasts as independent; but they are both
part of the same experiment…
Our consideration of 1 vs. 2 might consider
the variation in all treatments that were part
of this experiment; especially if we are
interpreting the differences between mean
comparisons as meaningful.
Conducting ANOVA’s
I.
Why?
A. more than two groups to compare
B. complex design with multiple factors
- blocks
- nested terms
- interaction effects
- correlated variables (covariates)
- multiple responses
Conducting ANOVA’s
I. Why?
II. How?
A. Variance Redux
Of a population
Of a sample
S2 =
Sum of squares
n-1
S2 =
“Sum of squares” = SS
n-1
2
(Sx)
=
SS
2
S(x ) n
S2 =
MS =
“Sum of squares” = SS
n-1
2
(Sx)
=
SS
2
S(x ) n
n-1
Conducting ANOVA’s
I. Why?
II. How?
A. Variance Redux
B. The ANOVA Table
Source of Variation df
SS
MS
F
p
Weight gain in mice fed different diets
Group A
(Control)
10.8
11
9.7
10.1
11.2
9.8
10.5
9.5
10
10.2
Group sums
Sx
Sx2
(Sx)2/n
Group B
(Junk Food)
12.7
13.9
11.8
13
11
10.9
13.6
10.9
11.5
12.8
Group C
(Health Food)
9.8
8.6
8
7.5
9
10
8.1
7.8
7.9
9.1
Weight gain in mice fed different diets
Group A
(Control)
10.8
11
9.7
10.1
11.2
9.8
10.5
9.5
10
10.2
Group B
(Junk Food)
12.7
13.9
11.8
13
11
10.9
13.6
10.9
11.5
12.8
Group C
(Health Food)
9.8
8.6
8
7.5
9
10
8.1
7.8
7.9
9.1
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
85.8
742.92
736.164
310.7
3305.09
3283.789
Weight gain in mice fed different diets
Group A
(Control)
10.8
11
9.7
10.1
11.2
9.8
10.5
9.5
10
10.2
Group B
(Junk Food)
12.7
13.9
11.8
13
11
10.9
13.6
10.9
11.5
12.8
Group C
(Health Food)
9.8
8.6
8
7.5
9
10
8.1
7.8
7.9
9.1
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
85.8
742.92
736.164
310.7
3305.09
3283.789
Correction term = (SSx)2/N = (310.7)2/30
Weight gain in mice fed different diets
Group A
(Control)
10.8
11
9.7
10.1
11.2
9.8
10.5
9.5
10
10.2
Group B
(Junk Food)
12.7
13.9
11.8
13
11
10.9
13.6
10.9
11.5
12.8
Group C
(Health Food)
9.8
8.6
8
7.5
9
10
8.1
7.8
7.9
9.1
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
Source of Variation df
TOTAL
SS
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
p
29
2
(Sx)
=
SS
2
S(x ) n
SStotal = 3305.09 – 3217.816 = 87.274
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
Source of Variation df
TOTAL
29
SS
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
p
87.274
2
(Sx)
=
SS
2
S(x ) n
SStotal = 3305.09 – 3217.816 = 87.274
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
2
(Sx)
= SS
S(x2) n
Source of Variation df
TOTAL
GROUP
29
2
SS
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
p
87.274
65.973
SSgroup = 3283.789 – 3217.816 = 65.973
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
Source of Variation df
TOTAL
GROUP
29
2
SS
87.274
65.973
MS =
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
32.986
p
2
(Sx)
S(x2) n
n-1
MSgroup = 65.973/2 = 32.986
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
Source of Variation df
TOTAL
GROUP
“ERROR” (within)
29
2
27
SS
87.274
65.973
21.301
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
32.986
p
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
Source of Variation df
TOTAL
GROUP
“ERROR” (within)
29
2
27
SS
87.274
65.973
21.301
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
32.986
0.789
MSerror = 21.301/27 = 0.789
p
GOOD
GRIEF !!!
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
Source of Variation df
TOTAL
GROUP
“ERROR” (within)
F=
29
2
27
SS
87.274
65.973
21.301
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
p
32.986
0.789
Variance (MS) between groups
Variance (MS) within groups
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
Source of Variation df
TOTAL
GROUP
“ERROR” (within)
F=
29
2
27
SS
87.274
65.973
21.301
32.986
= 41.81
0.789
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
32.986
0.789
41.81
p
Weight gain in mice fed different diets
Group sums
Sx
Sx2
(Sx)2/n
Totals
102.8
1059.76
1056.784
122.1
1502.41
1490.841
Source of Variation df
TOTAL
GROUP
“ERROR” (within)
F=
29
2
27
SS
87.274
65.973
21.301
32.986
= 41.81
0.789
85.8
742.92
736.164
310.7
3305.09
3283.789
3217.816 = CT
MS
F
32.986
0.789
p
41.81 < 0.05
Conducting ANOVA’s
I. Why?
II. How?
III. Comparing Means
“post-hoc mean comparison tests – after ANOVA
TUKEY – CV = q
MSerror
n
Q from table A.7 = 3.53
n = n per group (10)
= 0.9915
Means:
Health Food
Control
Junk Food
8.58
10.29
12.25
H – C = 1.70
J – C = 1.93
H – J = 3.67
All greater than 0.9915, so all mean
comparisions are significantly
different at an experiment-wide
error rate of 0.05.
Means:
Health Food
Control
Junk Food
8.58 a
10.29 b
12.25 c
H – C = 1.70
J – C = 1.93
H – J = 3.67
All greater than 0.9915, so all mean
comparisions are significantly
different at an experiment-wide
error rate of 0.05.