Food consumption of rats. Two-way vs one-way vs nested ANOVA

Transcription

1 Food consumption of rats Lard Gender Fresh Rancid Male Female Two-way vs one-way vs nested NOV T 1 2 * * M * * * * * * F * * * * T M1 M2 F1 F2 * * * * * * * * * * * * T 1 2 M F M F * * * * * * * * * * * *

2 Two-way versus one-way NOV In the lard example, we could consider the lard by gender groups as four different treatments, and carry out a standard one-way NOV. Let r c n be the number of rows in the two-way NOV, be the number of columns in the two-way NOV, be the number of observations within each of those r c groups. One-way NOV table source sum of squares df between groups within groups SS between = n i j (Ȳij Ȳ ) 2 rc 1 SS within = i j k (Y ijk Ȳij ) 2 rc(n 1) total SS total = i j k (Y ijk Ȳ ) 2 rcn 1

3 Example source SS df MS F p-value between within ut this doesn t tell us anything about the separate effects of freshness and sex. ll sorts of means Fat Gender Fresh Rancid Male Female This table shows the cell, row, and column means, plus the overall mean. (The discussion today is like the analysis of two-dimensional tables, as opposed to one-dimensional tables.)

4 picture Consumption Female Male Female Male Fresh Fresh Rancid Rancid Two-way NOV table source sum of squares df between rows SS rows = c n i (Ȳi Ȳ ) 2 r 1 between columns SS columns = r n j (Ȳ j Ȳ ) 2 c 1 interaction SS interaction (r 1)(c 1) error SS within = i j k (Y ijk Ȳij ) 2 rc(n 1) total SS total = i j k (Y ijk Ȳ ) 2 rcn 1

5 Example source sum of squares df mean squares sex freshness interaction error The NOV model Let Y ijk be the k th item in the subgroup representing the i th group of treatment (r levels) and the j th group of treatment (c levels). We write Y ijk = µ + α i + β j + γ ij + ǫ ijk The corresponding analysis of the data is y ijk = ȳ + (ȳ i ȳ ) + (ȳ j ȳ ) + (ȳ ij ȳ i ȳ j + ȳ ) + (y ijk ȳ ij )

6 Towards hypothesis testing source mean squares expected mean squares between rows c n i (Ȳi Ȳ ) 2 r 1 σ 2 + c n r 1 i α 2 i between columns r n j (Ȳ j Ȳ ) 2 c 1 σ 2 + r n c 1 j β 2 j interaction error n i j (Ȳij Ȳi Ȳ j + Ȳ ) 2 (r 1) (c 1) i j k (Y ijk Ȳij ) 2 r c (n 1) σ 2 + σ 2 n (r 1) (c 1) i j γ 2 ij This is for fixed effects, and equal number of observations per cell! Example (continued) source SS df MS F p-value Sex Freshness interaction error

7 Interaction in a 2-way NOV model Let Y ijk be the k th item in the subgroup representing the i th group of treatment (r levels) and the j th group of treatment (c levels). We write Y ijk = µ + α i + β j + γ ij + ǫ ijk no interaction positive interaction negative interaction N + N + N + Interaction plots The R function interaction.plot() lets you compare the cell means by treatments. no interaction positive interaction negative interaction N Y N Y N Y treatment treatment treatment

8 Interaction plots (2) ssume treatment has four levels and treatment has three levels. The interaction plots could look like one of these: no interaction interaction mean response Example Strain differences and daily differences in blood ph for five (r = 5) inbred strains of mice. Five (n = 5) mice from each strain were tested six times (c = 6) at one-week intervals. Source SS df MS F P-value mouse strains < test days interaction error

9 Two-way versus nested NOV revisited Strains Days * * * * * 2 * * * * * 3 * * * * * 4 * * * * * 5 * * * * * 6 * * * * * Strains Days Mice * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * NOV tables source SS df MS F Correct: two-way anova mouse strains test days interaction error total Incorrect: nested anova mouse strains days within strains error total

10 Unequal number of observations The following data were obtained in a study on energy utilization (in kcal/g) of the pocket mouse during hibernation at different temperatures. Restricted food Unrestricted food 8 o C 18 o C 8 o C 18 o C R is for rescue... The computations for the NOV table get rather complicated if the numbers of observations per cell are not equal. However, you can simply use aov() to get the results. > mouse.aov <- aov(log(rsp) food * temp, data=mouse) > anova(mouse.aov) Df Sum Sq Mean Sq F value Pr(>F) food temp food:temp Residuals

11 Interaction plot temp food UnR R mean of log2(rsp) mean of log2(rsp) R UnR 18 8 food temp Two-way NOV without replicates elow are the development periods (in days) for three strains of houseflies at seven densities. Strain Density OL ELL bwb

12 NOV table source df SS MS fly strains condition interaction total 20 We have 21 observations. estimate an error! That means we have no degrees of freedom left to Interactions strain 12 ELL bwb OL measurement density

13 Result If we assume that there are no interactions, i.e., we assume Y ijk = µ + α i + β j + ǫ ijk we get the following results using aov() in R: > fly.aov <- aov(rsp strain + density, data=fly) > anova(fly.aov) Df Sum Sq Mean Sq F value Pr(>F) strain density Residuals Random within blocks D C C D C C D C C D D D

14 Example Mean weight of 3 genotypes of beetles, reared at a density of 20 beetles per gram of flour. Four series of experiments represent blocks. genotype block ++ +b bb We must assume the effects of the genotypes is the same within each block. Interaction plot 1.05 block mean of rsp b bb genotype

15 NOV table Df Sum Sq Mean Sq F value Pr(>F) genotype block Residuals