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 D. putrida high density D. putrida with D. tripuncatata What was our solution? U U U

What’s the prob? D. putrida low density D. putrida high density D. putrida with D. tripuncatata Tested each contrast at p = 0.05 Probability of being correct in rejecting each Ho: U U U

What’s the prob? D. putrida low density D. putrida high density D. putrida with D. tripuncatata Tested each contrast at p = 0.05 Probability of being correct in rejecting all Ho: x0.95x0.95 = 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: x0.95x0.95 = 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: x0.95x0.95 = 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: x0.95x0.95 = 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 =

Probability of being correct in rejecting all Ho: x0.95x0.95 = 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 = So, confidence = 0.983

Probability of being correct in rejecting all Ho: x0.983x0.983 = 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 = 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…

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

Sum of squares n - 1 S 2 =

“Sum of squares” = SS n - 1 S 2 = = SS  x 2 ) -  x) 2 n

“Sum of squares” = SS n - 1 S 2 = = SS  x 2 ) -  x) 2 n n - 1 MS =

Conducting ANOVA’s I.Why? II.How? A. Variance Redux B. The ANOVA Table Source of Variation df SS MS F p

Group AGroup BGroup C (Control)(Junk Food)(Health Food) Weight gain in mice fed different diets Group sums  x  x 2  x) 2 /n

Group AGroup BGroup C (Control)(Junk Food)(Health Food) Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n

Group AGroup BGroup C (Control)(Junk Food)(Health Food) Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n Correction term = (  x) 2 /N = (310.7) 2 /30

Group AGroup BGroup C (Control)(Junk Food)(Health Food) Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL 29 SS total = – = = SS  x 2 ) -  x) 2 n

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL SS total = – = = SS  x 2 ) -  x) 2 n

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL GROUP SS group = – = = SS  x 2 ) -  x) 2 n

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL GROUP MS group = /2 =  x 2 ) -  x) 2 n n - 1 MS =

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL GROUP “ERROR” (within)

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL GROUP “ERROR” (within) MS error = /27 = 0.789

GOOD GRIEF !!!

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL GROUP “ERROR” (within) Variance (MS) between groups Variance (MS) within groups F =

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL GROUP “ERROR” (within) F = = 41.81

Weight gain in mice fed different diets Group sumsTotals  x  x  x) 2 /n = CT Source of Variation df SS MS F p TOTAL GROUP < 0.05 “ERROR” (within) F = = 41.81

Conducting ANOVA’s I.Why? II.How? III.Comparing Means “post-hoc mean comparison tests – after ANOVA TUKEY – CV = q MS error n Q from table A.7 = 3.53 n = n per group (10) =

Means: Health Food 8.58 Control10.29 Junk Food12.25 H – C = 1.70 J – C = 1.93 H – J = 3.67 All greater than , so all mean comparisions are significantly different at an experiment-wide error rate of 0.05.

Means: Health Food 8.58 a Control10.29 b Junk Food12.25 c H – C = 1.70 J – C = 1.93 H – J = 3.67 All greater than , so all mean comparisions are significantly different at an experiment-wide error rate of 0.05.