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.