BIOL 458 BIOMETRY Lab 8 - Nested and Repeated Measures ANOVA

BIOL 458 BIOMETRY Lab 8 - Nested and Repeated Measures ANOVA PART 1: NESTED ANOVA Nested designs are used when levels of one factor are not represented within all levels of another factor. Often this is because there is no alternative. For instance, if we were concerned with the effects of acid rain on productivity in British and American lakes, we might select at random 5 lakes in each country and make 10 productivity measurements at the surface. The lakes would constitute a random effect while country would be a fixed effect. However, each lake does not occur in both countries, so lake is, necessarily, nested within country. Such a design confounds the lake by country interaction since to estimate the interaction would require measurements of each lake within both countries, which is impossible. In such a situation, one analyzes the data as if they represent a fully factorial design with all factors completely crossed, but then the interaction term (lake by country in this case) is pooled with the effect of the nested factor (lake in this case), and the country main effect is tested over the effect of the nested factor pooled with the interaction. Source df MS F Country 2-1 SS Country /df Country [ Lake + Lake by Country With Cells (error) 2(5-1) 2 x 5 (10-1) SS Lake + Lake by Country/df Lake + Lake by Country SS error /df error MS Country/ MS Lake + Lake by Country The appropriate F ratio is MS country /MS Lake + Lake by Country. Alternatively, if factor A is the non-nested factor and factor B is the nested factor, the F ratio to test for the effect of the non-nested factor is F = MS A /MS A/B, where A/B connotes the effect of factor B nested within factor A and is equal to the sum of main effect of Factor B and the interaction of Factor A and B that one would obtain from treating the data as being generated by a fully crossed ANOVA design. The results of this analysis should be equivalent to an analysis in which the values in each lake were first averaged and a one-way ANOVA was performed on the averages to test for a country effect. Nested ANOVA in R In R, one can obtain the nested analysis simply using the 'aov' command, or using the 'Anova' command from the 'car' package if you have an unbalanced design and want to use Type II or III sums of squares. However, as for other ANOVA models, F may need to be re-calculated if the model includes a random factor. 1

In a study designed to estimate the number of associated bacterial species as a function of life history stage for a moth, DNA was extracted from 5 field collected individuals of each life stage, and 3 replicate pcr's were run on the bacterial 16s gene. The pcr product was sequenced using next generation sequencing. The resulting DNA sequence reads were groups into clusters at the 98% level of similarity, and the number of bacterial OTU's (organization taxonomic units) were counted for each replicate sample. These data are in the linked file (micro.csv). Note that individuals were selected at random, so we must eventually compute the correct F ratios and p values using the mean squares provided by R. The expected means squares and F-ratios for this specific study design are given in the notes on Repeated Measures and Nested ANOVA, in section III, Table B. First read in the data and examine it. # to perform a nested ANOVA where factor B is nested in A (A/B) # In this case replicate measurements were made on each individual insect and the "individuals" are nested within # life history stage (3rd and 5th instar larvae, and adults) # read in the data file (micro.csv) dat1=read.csv("c:/users/connor/documents/micro.csv",header=true) head(dat1) X individual replicate treat numbact 1 1 1 1 second 4 2 2 1 2 second 6 3 3 1 3 second 1 4 4 2 1 second 4 5 5 2 2 second 3 6 6 2 3 second 3 Perform the ANOVA using the 'aov' command. If factor B is nested within the levels of factor A, then the model syntax will include the term A/B. In our example 'individual' is nested within 'treat" where 'treat' is life history stage. # perform the ANOVA and get summary in a single command summary(aov(dat1$numbact~dat1$treat/factor(dat1$individual))) Df Sum Sq Mean Sq F value Pr(>F) dat1$treat 2 52.04 26.022 7.184 0.00282 ** dat1$treat:factor(dat1$individual) 12 28.53 2.378 0.656 0.77737 Residuals 30 108.67 3.622 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Since individual is a random factor, the correct F ratio MS treat /MS treat/individual = 10.94, with 2 and 12 degrees of freedom. 2

# compute the correct F ratio and determine it statistical significance fcor=26.022/2.378 fcor [1] 10.94281 1-pf(fcor,df1=2,df2=12) [1] 0.0019724 To use the 'car' Anova command, first load the 'car' package. # load the 'car' package library(car) # perform the ANOVA using lm, and the Anova command from 'car' # Note that since we have as balanced design the results are the same dd=lm(dat1$numbact~dat1$treat/factor(dat1$individual)) Anova(dd,Type=2) Anova Table (Type II tests) Response: dat1$numbact Sum Sq Df F value Pr(>F) dat1$treat 52.044 2 7.1840 0.002823 ** dat1$treat:factor(dat1$individual) 28.533 12 0.6564 0.777372 Residuals 108.667 30 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Although the summary table from the 'car' Anova command does not give the Mean squares, dividing the Sums of Squares by their respective degrees of freedom will yield the same values as we obtained using the 'aov' command since we have equal sample sizes. Plotting the data shows that the number of associated bacteria is greatest in the adult stage and lower in the larval stages examined. # Plot the data library(ggplot2) p=ggplot(dat1,aes(x=treat,y=numbact)) + geom_boxplot() p 3

Note that the boxplot supports the conclusion that there are differences in the number of bacterial associates between life stages. We do not perform a test on the effect of individuals nested within life stage effect since it is a random effect. Nested ANOVA by pooling Sums of Squares It is also possible to compute a nested ANOVA by treating the data as if it were not nested and pooling the sums of squares. However, the nested factor level must be coded as if it were not nested. In the example above, the data file mico.csv also has a column labeled "individual2" in which we pretend that the same three individuals were observed under each level of the treat factor (stage). For example, if the study described above we would perform a 2 - factor ANOVA with life stage (treat) and individual2 as our two factors. Since individuals are nested within life stage, we know that we really cannot estimate an interaction between these factors. However, we trick R into computing an effect of treat (A), individual2 (B), and an treat:individual2 interaction (A:B). We would then need to pool (add together) the Sum of Squares for individual2 and the Sum of Squares for treat:individual2 to obtain the Sum of Square for individual nested within treatment (treat/individual). We would pool their degrees of freedom as well. In the example from above, the individual effect would have 5-1 = 4 degrees of freedom, and the interaction would have (5-1) X (3-1) = 8 degrees of freedom, so the total would be 12 as obtained when we explicitly include the nested factor in the ANOVA model. In general, any nested design can be computed using this sort of pooling of Sums of Squares approach although it is a bit more cumbersome. Again, to use the pooling approach with factor B nested within the levels of factor A: SS(A/B) = SS(B) + SS(A:B) df(a/b) = df(b) + df(a:b) 4

MS(A/B)= SS(A/B)/df(A/B) In this example, one could have averaged the values for each individual, and then computed a single factor ANOVA to compare differences among life history stages. The results would be identical to fitting the nested model. PART 2: REPEATED MEASURES ANOVA Within subjects designs are used in agricultural and psychological research and have many applications in biology. These designs are called split-plot, repeated measures, or generically within subjects. The primary purpose of these designs is to eliminate uncontrolled variation due to a priori differences or differences in the level of responsiveness in primary sampling units from the estimate of experimental error. In this sense we can see that these designs are a way to remove confounding variation by adding classificatory controls or strata. Repeated Measures ANOVA has been used increasingly in biology for several reasons. The first is that it allows us to better control for inter-subject variability. It allows us to use a subject as its own control. Secondly, it is more economical in use of subjects, which is especially important when subjects (or study sites) are difficult to locate or get to, or are limited in number. Remember that a repeated measures ANOVA is an extension of the paired t - test to more complicated ANOVA designs. As such, one diagnoses the presence of a repeated measures factor in an ANOVA design by the presence of subjects who are observed under all levels of a factor. ANOVA designs can be comprised entirely of repeated measures factors (a full within subjects design), or have a mixture of repeated measures and non-repeated measures factors (a design with both within and between subjects effects). Univariate repeated measures ANOVA requires, in addition to the normal ANOVA assumptions, an assumption that the correlations between observations within a subject are all the same. This is sometimes referred to as the assumption of compound symmetry or sphericity of the variance-covariance matrix. The multivariate approach to repeated measures ANOVA does not require this assumption, but produces multivariate tests of the hypotheses of interest, which may be more difficult for the average reader to comprehend. Using the lm and Anova commands from the 'car' package in R will generate both the univariate and the multivariate tests. When one violates the assumption of sphericity, the TYPE I error rate is inflated. So to measure the degree of deviation from sphericity, and provide an adjustment to the univariate tests, R calculates the Greenhouse-Geisser ε and the Huynh-Feldt ε. To adjust for non-sphericity, multiply the numerator and denominator degrees of freedom by the appropriate ε value, and evaluate the F value reported by R for the adjusted degrees of freedom. In R, ε is calculated and the adjustment of df s is done automatically. See the paper by O'Brien 5

and Kaiser in the supplemental readings to learn more about these two approaches to repeated measures ANOVA. In R, data for repeated measures factors can be on the same line for a single subject. If a subject is observed under all three levels of factor A, then all three response values can be on a single line of data. For a two factor design with repeated measures on one factor (with 3 factor levels for the repeated measures factor and 2 for the between subjects factor) data would look like this: 1 10.5 20.7 34.7 1 8.9 19.3 27.5 1 9.1 17.5 23.8 2 3.4 7.9 12.4 2 2.5 8.3 13.2 2 3.5 6.9 15.3 The integer code represents the level of the between subjects factor in which each individual subject is nested, and the three real numbers in successive columns represent the observation of the random variable of interest in levels 1, 2, and 3 of the within subjects factor, respectively. Entering data on additional repeated measures factors would entail adding the additional observations to each line of the data file since each line represents an individual subject. Repeated Measures ANOVA in R Feldman and Connor (1992) examined the effects of rock type and stream ph on the colonization of cobble habitats by invertebrates. They placed wire mesh baskets full of similar sized cobbles of 3 different rock types into 3 streams with either slightly acidic ph values of approximately 5.8, or in 3 streams with neutral ph values of approximately 7. These data are in the linked data file (rock.csv). The data represent the number of invertebrates detected in each basket in each combination of ph and rock type. The expected mean squares and F ratios for this specific study design are given in the notes on Repeated Measures and Nested ANOVA in section II, Table C. # read in second data file (rock.csv) dat2=read.csv("k:/biometry/biometry-fall-2015/lab8/rock.csv",header=true) head(dat2) ph Rock1 Rock2 Rock3 1 1 248 206 250 2 1 360 216 198 3 1 332 176 234 4 2 390 404 521 5 2 523 594 486 6 2 416 446 433 6

After attaching the data object, dat2, combine the 3 levels of the repeated measures factor (within-subjects factor) into a matrix called rocks. Then define the factor 'rocktype'. Finally, using the 'lm' function from R, fit a multivariate linear model where 'rocks' is the dependent variable and ph is the only factor. # attach(dat2) rocks=cbind(rock1,rock2,rock3) rocktype=factor(c("rock1","rock2","rock3")) # Create a multivariate linear model using: mlm1=lm(rocks~factor(ph)) In this last step we are essentially asking if a linear combination of the levels of the rock type factor differ by ph. Now load the 'car' package, and apply the 'Anova' command. Note that we need to define a data.frame for our repeated measures factor, and a model with the within subjects design the "idesign~rocktype" part of the command. # load the package "car" library(car) # Using the Car Anova command perform an ANOVA with Type 2 sums of squares ww=anova(mlm1,idata=data.frame(rocktype), idesign=~rocktype, type="ii") Next we will obtain a summary of this last model. This summary is lengthy, but I will describe its contents. This summary includes results of both a "Multivariate" and a "univariate" approach to repeated measures. The first part of the summary focuses on the Multivariate approach. Here you will find Multivariate F- values, degrees of freedom, and significance levels for all your withinsubjects tests. The reason for this is that one can conceive of a within-subjects factor as either a single response variable examined under a series of different treatment levels (a univariate approach), or a as several different response variables (a multivariate approach). Usually the Univariate and Multivariate approaches give the same answer, although F-values, dfs, and p-values will not be exactly the same. The reason for using the Multivariate approach is that one only has to meet the assumption of Multivariate Normality and homogeneity of the variance/covariance matrix for this test to work well. Tests for multivariate normality are found in the MVN package in R, and Bartlett's test 7

for homogeneity of variance/covariance matrices is available in the base installation of R (bartlett.test). However, in the univariate approach one has to meet the more restrictive assumption of sphericity of the variance/covariance matrix. The second part of the summary focuses on the Univariate approach. It first shows an ANOVA table which includes between and within-subjects factors with no adjustment for non-sphericity of the variance/covariance matrix (assuming sphericity). After the ANOVA table, R reports Mauchly s test for Sphericity of your variance/covariance matrix, and two different estimates of epsilon (a measure of how much your data depart from meeting this assumption). Epsilon is a value between 1 and the lower bound estimate and it is multiplied by the degrees of freedom for both the numerator and denominator mean squares to adjust the test for non-sphericity. Deviations from sphericity lead to too many Type I errors, so by adjusting dfs downward the resulting univariate F-tests have error rates equal to the nominal error rates (if you claim the p value is 0.05 then it will be so if you meet the other assumptions of the test). I recommend using the Greenhouse-Geisser adjustment. # Obtain a summary summary(ww) # This warning indicates that if a value of the Huynh-Feldt epsilon greater than 1 is obtained, the analysis proceeds as if the value were equal to 1. Warning in summary.anova.mlm(ww): HF eps > 1 treated as 1 # This section shows the Multivariate tests. First for the intercept (not int eresting in our case), then ph, rock type, and then the interaction of ph and rocktype. Type II Repeated Measures MANOVA Tests: ------------------------------------------ Term: (Intercept) Response transformation matrix: (Intercept) Rock1 1 Rock2 1 Rock3 1 Sum of squares and products for the hypothesis: (Intercept) (Intercept) 6897248 8

Sum of squares and products for error: (Intercept) (Intercept) 61858.67 Multivariate Tests: (Intercept) Df test stat approx F num Df den Df Pr(>F) Pillai 1 0.99111 446.0004 1 4 2.9718e-05 *** Wilks 1 0.00889 446.0004 1 4 2.9718e-05 *** Hotelling-Lawley 1 111.50011 446.0004 1 4 2.9718e-05 *** Roy 1 111.50011 446.0004 1 4 2.9718e-05 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ------------------------------------------ Term: factor(ph) Response transformation matrix: (Intercept) Rock1 1 Rock2 1 Rock3 1 Sum of squares and products for the hypothesis: (Intercept) (Intercept) 662008.2 Sum of squares and products for error: (Intercept) (Intercept) 61858.67 Multivariate Tests: factor(ph) Df test stat approx F num Df den Df Pr(>F) Pillai 1 0.914544 42.80779 1 4 0.0028205 ** Wilks 1 0.085456 42.80779 1 4 0.0028205 ** Hotelling-Lawley 1 10.701947 42.80779 1 4 0.0028205 ** Roy 1 10.701947 42.80779 1 4 0.0028205 ** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ------------------------------------------ Term: rocktype Response transformation matrix: rocktype1 rocktype2 Rock1 1 0 Rock2 0 1 Rock3-1 -1 9

Sum of squares and products for the hypothesis: rocktype1 rocktype2 rocktype1 3601.5-1960.000 rocktype2-1960.0 1066.667 Sum of squares and products for error: rocktype1 rocktype2 rocktype1 28376 23794.00 rocktype2 23794 28788.67 Multivariate Tests: rocktype Df test stat approx F num Df den Df Pr(>F) Pillai 1 0.4753856 1.359243 2 3 0.37998 Wilks 1 0.5246144 1.359243 2 3 0.37998 Hotelling-Lawley 1 0.9061619 1.359243 2 3 0.37998 Roy 1 0.9061619 1.359243 2 3 0.37998 ------------------------------------------ Term: factor(ph):rocktype Response transformation matrix: rocktype1 rocktype2 Rock1 1 0 Rock2 0 1 Rock3-1 -1 Sum of squares and products for the hypothesis: rocktype1 rocktype2 rocktype1 22693.5-5412.000 rocktype2-5412.0 1290.667 Sum of squares and products for error: rocktype1 rocktype2 rocktype1 28376 23794.00 rocktype2 23794 28788.67 Multivariate Tests: factor(ph):rocktype Df test stat approx F num Df den Df Pr(>F) Pillai 1 0.790732 5.667849 2 3 0.095731. Wilks 1 0.209268 5.667849 2 3 0.095731. Hotelling-Lawley 1 3.778566 5.667849 2 3 0.095731. Roy 1 3.778566 5.667849 2 3 0.095731. --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 # this section gives the results of the univariate approach to repeated-measu red ANOVA. 10

Univariate Type II Repeated-Measures ANOVA Assuming Sphericity SS num Df Error SS den Df F Pr(>F) (Intercept) 2299083 1 20620 4 446.0004 2.972e-05 *** factor(ph) 220669 1 20620 4 42.8078 0.00282 ** rocktype 4419 2 22247 8 0.7945 0.48447 factor(ph):rocktype 19597 2 22247 8 3.5236 0.07990. --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 # This section shows the estimates of tests for Sphericity of the variance/ covariance matrix. R gives both the Greenhouse-Geisser epsilon (preferred) and the Huynh-Feldt epsilon which measure departure from sphericity. A value of 1 indicates that the variance/covariance matrix is spherical. Adjusted p- values for the within-subjects factors are also given (highlighted in red). Mauchly Tests for Sphericity Test statistic p-value rocktype 0.67552 0.55521 factor(ph):rocktype 0.67552 0.55521 Greenhouse-Geisser and Huynh-Feldt Corrections for Departure from Sphericity GG eps Pr(>F[GG]) rocktype 0.75501 0.4597 factor(ph):rocktype 0.75501 0.1026 HF eps Pr(>F[HF]) rocktype 1.114494 0.48447373 factor(ph):rocktype 1.114494 0.07989851 Finally, it is always good to plot the data to get a better feel for the patterns revealed. # plot the data # first put the data into 3 vectors of equal length numinverts=c(rock1,rock2,rock3) numinverts [1] 248 360 332 390 523 416 206 216 176 404 594 446 250 198 234 521 486 [18] 433 11

rocks=factor(c(rep(1,6),rep(2,6),rep(3,6))) rocks [1] 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 Levels: 1 2 3 phh=factor(rep(ph,3)) phh [1] 1 1 1 2 2 2 1 1 1 2 2 2 1 1 1 2 2 2 Levels: 1 2 # obtain a boxplot p2=ggplot(dat2,aes(x=phh,y=numinverts,fill=rocks)) + geom_boxplot() p2 The graph supports the conclusion that more invertebrates were detected in streams with neutral ph than in low ph streams, and show some suggestion of an interaction between ph and rock type although it was not statistically significant. Lab 8 - Assignment PART 1: Nested ANOVA Data on insect damage on oak trees was collected in order to answer questions concerning the effects of shading on damage levels. Six oak trees were selected at random, 3 in the shade and 3 in the sun. The data are included in the linked data file nest.csv. The file has the following form: 12

Column 1: Light level: 1 = shade; 2 = sun Column 2: Tree. Trees 1 3 (six different trees 3 in sun and 3 in shade). Column 3: Damage in percent. Column 4: Tree2. Trees numbered 1-6 Reiterating, this file contains data which describe the response of trees in terms of damage; thus damage is the response variable. The data come from observations taken from six randomly selected oak trees. Three are located in the shade; and three are located in the sun. Thus this is a nested design, with trees nested within the level of light (shade or sun) and leaves nested within trees. A sample of 15 leaves was taken at random from each tree. To use the R syntax to perform a nested ANOVA, use the tree coding variable (tree2). To perform the analysis for pooling the sums of squares use the tree coding variable (tree). 1.1) What kind of design is this? Draw a graphical representation of the experimental design, where A = light level, B = trees, and G = group of subjects. When I ask for a graphical representation of the experimental design, I am looking for a diagram like I have shown in class that shows all the factors in the experiment, their levels, and represents each group of subjects with the letter G appropriately subscripted under each treatment combination. I also ask that you accurately describe the pattern of crossing and nesting of factor levels and subjects. 1.2) Conduct an appropriate analysis of variance on these data. a) specify the null hypotheses being tested and turn in the edited output b) pool the appropriate sums - of - squares and perform the appropriate F test c) interpret the results of the tests (at α = 0.05) d) How could the data have been treated differently so a nested ANOVA could have been avoided? In our particular example, our nested factor (trees) is a random effects factor since we chose them at random from a large number of possible trees. Therefore, the final test of the effect of factor A will involve computing an F ratio with the MS(B(A)) as the denominator F test for factor A, F = MS(A)/MS(A/B). 13

Given that our nested factor is a random effects factor, it also makes little sense to compute an hypothesis test for the B(A) effect, but if one insisted on doing so then the within cell or error mean square would serve as the denominator of the F- ratio. Finally, for our particular problem, an alternative way to analyze the data would be to average the values among the 15 leaves within each tree, and use the tree means to compute a independent groups t-test for differences in leaf damage between trees in the sun versus the shade. The results of this test would be identical to the test of the light effect in the Nested ANOVA (except the t-value would equal the square root of the F value). Hence, it is sometimes possible to turn a nested ANOVA into a simpler problem, particularly if the nesting of factor levels arises because multiple observations are made on the same subject. Here, by using tree averages, we turn a nested random factor (the tree factor) into our subjects, hence removing the nested factor levels and simplifying the analysis. PART 2: Repeated Measures ANOVA In this experiment, ten small-mammal trapping grids were established in order to study the effect of food addition on population densities. Five of the grids received food supplements while the others were not manipulated. The population levels on each grid were monitored 3 times at monthly intervals following the food addition. The data are contained in the linked data file rep.csv. It is in the following form: Column 1: Food. Where: 1 = no addition; 2 = food added Column 2: Grid. Grids 1-10 Column 3-5: Population density in June, July, and August. Reiterating, here we have 10 grids. 5 have food added; 5 do not. The response variable, population density, was measured on each grid (subject) 3 times (a repeated measures factor). 2.1) What kind of design is this? Draw a graphical representation of the experimental design, where A = food treatment, B = time or month, and G = group of subjects. 2.2) Conduct an appropriate analysis of variance on these data and report the results for tests of the main effects of food treatment, time, and their interactions (no pooling of sums - of - squares is necessary. 14

a) specify the null hypotheses being tested and interpret the results of the tests (at α = 0.05). Don't forget to include your syntax and results in your R.markdown file. b) How could the study have been designed differently so that a repeated measures ANOVA could have been avoided? 15