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Conducting ANOVA’s
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 densityD. putrida high densityD. putrida with D. tripuncatata
Conducting ANOVA’s
I. Why?
A. more than two groups to compare
What’s the prob?
D. putrida low densityD. putrida high densityD. 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 densityD. putrida high densityD. putrida with D. tripuncatata
What was our solution?
U
U
U
What’s the prob?
D. putrida low densityD. putrida high densityD. putrida with D. tripuncatata
Tested each contrast at p = 0.05
Probability of being correct in rejecting each Ho:
1 2 30.95 0.95 0.95
U
U
U
What’s the prob?
D. putrida low densityD. putrida high densityD. putrida with D. tripuncatata
Tested each contrast at p = 0.05
Probability of being correct in rejecting all Ho:
1 2 30.95 x 0.95 x 0.95 = 0.86
So, Type I error rate has increased from 0.05 to 0.14
U
U
U
Probability of being correct in rejecting all Ho:
1 2 30.95 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 30.95 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 30.95 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/nWhere n = number of contrasts.Experient-wise = 0.05Comparison-wise = 0.05/3 = 0.0167
Probability of being correct in rejecting all Ho:
1 2 30.95 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/nWhere n = number of contrasts.Experient-wise = 0.05Comparison-wise = 0.05/3 = 0.0167So, confidence = 0.983
Probability of being correct in rejecting all Ho:
1 2 30.983 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/nWhere n = number of contrasts.Experient-wise = 0.05Comparison-wise = 0.05/3 = 0.0167So, 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 significant1 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 compareB. 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
Sum of squaresn - 1S2 =
“Sum of squares” = SSn - 1S2 =
= SSS(x2) - (Sx)2 n
“Sum of squares” = SSn - 1S2 =
= SSS(x2) - (Sx)2 n
n - 1MS =
Conducting ANOVA’s
I. Why?II. How?
A. Variance ReduxB. The ANOVA Table
Source of Variation df SS MS F p
Group A Group B Group C(Control) (Junk Food) (Health Food)10.8 12.7 9.811 13.9 8.69.7 11.8 810.1 13 7.511.2 11 99.8 10.9 1010.5 13.6 8.19.5 10.9 7.810 11.5 7.910.2 12.8 9.1
Weight gain in mice fed different diets
Group sumsSxSx2
(Sx)2/n
Group A Group B Group C(Control) (Junk Food) (Health Food)10.8 12.7 9.811 13.9 8.69.7 11.8 810.1 13 7.511.2 11 99.8 10.9 1010.5 13.6 8.19.5 10.9 7.810 11.5 7.910.2 12.8 9.1
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
Group A Group B Group C(Control) (Junk Food) (Health Food)10.8 12.7 9.811 13.9 8.69.7 11.8 810.1 13 7.511.2 11 99.8 10.9 1010.5 13.6 8.19.5 10.9 7.810 11.5 7.910.2 12.8 9.1
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
Correction term = (SSx)2/N = (310.7)2/30
Group A Group B Group C(Control) (Junk Food) (Health Food)10.8 12.7 9.811 13.9 8.69.7 11.8 810.1 13 7.511.2 11 99.8 10.9 1010.5 13.6 8.19.5 10.9 7.810 11.5 7.910.2 12.8 9.1
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29
SStotal = 3305.09 – 3217.816 = 87.274
= SSS(x2) - (Sx)2 n
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29 87.274
SStotal = 3305.09 – 3217.816 = 87.274
= SSS(x2) - (Sx)2 n
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29 87.274GROUP 2 65.973
SSgroup = 3283.789 – 3217.816 = 65.973
= SSS(x2) - (Sx)2 n
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29 87.274GROUP 2 65.973 32.986
MSgroup = 65.973/2 = 32.986
S(x2) - (Sx)2 n
n - 1MS =
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29 87.274GROUP 2 65.973 32.986“ERROR” (within) 27 21.301
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29 87.274GROUP 2 65.973 32.986“ERROR” (within) 27 21.301 0.789
MSerror = 21.301/27 = 0.789
GOOD GRIEF !!!
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29 87.274GROUP 2 65.973 32.986“ERROR” (within) 27 21.301 0.789
Variance (MS) between groupsVariance (MS) within groupsF =
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29 87.274GROUP 2 65.973 32.986 41.81“ERROR” (within) 27 21.301 0.789
32.9860.789F = = 41.81
Weight gain in mice fed different diets
Group sums TotalsSx 102.8 122.1 85.8 310.7Sx2 1059.76 1502.41 742.92 3305.09(Sx)2/n 1056.784 1490.841 736.164 3283.789
3217.816 = CT
Source of Variation df SS MS F pTOTAL 29 87.274GROUP 2 65.973 32.986 41.81 < 0.05“ERROR” (within) 27 21.301 0.789
32.9860.789F = = 41.81
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 = 3.53n = n per group (10)
= 0.9915
Means:
Health Food 8.58Control 10.29Junk Food 12.25
H – C = 1.70J – C = 1.93H – 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 8.58 aControl 10.29 bJunk Food 12.25 c
H – C = 1.70J – C = 1.93H – J = 3.67
All greater than 0.9915, so all mean comparisions are significantly different at an experiment-wide error rate of 0.05.
