R Users - UMasspeople.umass.edu/biep640w/pdf/sol_anova_1 of 2 R USERS.pdfBIOSTATS 640 Spring 2019 Unit 7 Introduction to Analysis of Variance (1 of 2) Solutions R Users Sol_anova_1

BIOSTATS 640 Spring 2019 Unit 7 Introduction to Analysis of Variance (1 of 2) Solutions R Users

Sol_anova_1 of 2 R Users.docx Page 1 of 15

Unit 7 – Analysis of Variance Practice Problems

SOLUTIONS – R Users

Before you begin: Download from the course website: R Users anova_infants.Rdata fishgrowth.Rdata

Preliminaries: Set current directory. Attach libraries needed

setwd("/Users/cbigelow/Desktop/") options(scipen=1000) library(DescTools) # command Freq( ): One way frequencies, relative frequencies, etc library(descr) # command CrossTable( ): Two way contingency table library(doBy) # command summaryBy( ): Descriptives by group library(car) # command leveneTest(): Test of equal variances 2 way anova

Practice with one way analysis of variance Exercises #1-6 Data set: anova_infants.Rdata

Zelazo et al. (1972) investigated the variability in age at first walking in infants. Study infants were grouped into four groups, according to reinforcement of walking and placement: (1) active (2) passive (3) no exercise; and (4) 8 week control. Sample sizes were 6 per group, for a total of n=24. For each infant, study data included group assignment and age at first walking, in months. The following are the data and consist of recorded values of age (months) by group:

Active Group Passive Group No-Exercise Group 8 Week Control 9.00 11.00 11.50 13.25 9.50 10.00 12.00 11.50 9.75 10.00 9.00 12.00

10.00 11.75 11.50 13.50 13.00 10.50 13.25 11.50 9.50 15.00 13.00 12.35

Source: Zelazo et al (1972) “Walking” in the newborn. Science 176: 314-315.



Preliminaries for Exercises 1-‐6: Read in data. Clean. Produce descriptives.

load(file="anova_infants.Rdata”) str(dat1)

## 'data.frame': 24 obs. of 7 variables: ## $ group : int 1 1 1 1 1 1 2 2 2 2 ... ## $ age : num 9 9.5 9.75 10 13 ... ## $ passive : num 0 0 0 0 0 0 1 1 1 1 ... ## $ noex : num 0 0 0 0 0 0 0 0 0 0 ... ## $ I_active : num 1 1 1 1 1 1 0 0 0 0 ... ## $ I_passive: num 0 0 0 0 0 0 1 1 1 1 ... ## $ I_noex : num 0 0 0 0 0 0 0 0 0 0 ... ## -‐ attr(*, "datalabel")= chr "" ## -‐ attr(*, "time.stamp")= chr " 9 Apr 2017 16:58" ## -‐ attr(*, "formats")= chr "%8.0g" "%9.0g" "%9.0g" "%9.0g" ... ## -‐ attr(*, "types")= int 251 254 254 254 254 254 254 ## -‐ attr(*, "val.labels")= chr "groupf" "" "" "" ... ## -‐ attr(*, "var.labels")= chr "" "" "" "" ... ## -‐ attr(*, "version")= int 12

dat1$group <-‐ factor(dat1$group, levels=c(1,2,3,4), labels=c("1=active", "2=passive", "3=noex", "4=control")) library(DescTools) DescTools::Freq(dat1$group,digits=2,dnn=c("Group"))

## level freq perc cumfreq cumperc ## 1 1=active 6 25.0% 6 25.0% ## 2 2=passive 6 25.0% 12 50.0% ## 3 3=noex 6 25.0% 18 75.0% ## 4 4=control 6 25.0% 24 100.0%



1. State the analysis of variance model using notation 2iµ and τ and σ as appropriate. Define

all terms and constraints on the parameters

Answer:

∑e

42

ij i ij ij ii=1

i

Y = µ + τ + ε , where ε ~N(0,σ ) and τ =0

i = 1, 2, ... K indexes method of reinforcement group;K = number of groups = 4j=1, 2, ..., n =6 indexes infant within group;µ = population

i

i i

ij

mean age at first walking, over all groupsµ = mean age at first walking for infants in group "i"τ = [ µ - µ ]Y = observed age at first walking for the jth infant in group "i"

O 1 2 3 4

A i

H : =0, =0, =0, and =0H : At least one 0

τ τ τ ττ ≠

2. By any means you like, produce a side by side box plot showing the distribution of age at first walking, separately for each of the 4 groups.

# Descriptives summaryBy(age~group,data=dat1, FUN=c(length,mean,var,sd,min,max), fun.names=c("n","Mean", "Var", "SD", "Min", "Max"))

## group age.n age.Mean age.Var age.SD age.Min age.Max ## 1 1=active 6 10.12500 2.093750 1.4469796 9.0 13.00 ## 2 2=passive 6 11.37500 3.593750 1.8957189 10.0 15.00 ## 3 3=noex 6 11.70833 2.310417 1.5200055 9.0 13.25 ## 4 4=control 6 12.35000 0.740000 0.8602325 11.5 13.50



# Side-‐by-‐side box plot boxplot(age~group,data=dat1, main="Age (Month) at First Walking", xlab="Group", ylab="Month")

Exercise #3 - In these data, first walking occurs earlier when infants are reinforced - Distributions differ markedly with respect to variability with greatest seen among infants in the passive group and smallest among infants in the control group - Distributions also differ markedly in their patterns of symmetry with long right tails in the active and passive groups, long left tail in the no-exercise group, and symmetry in controls

- Plot confirms impressions from the descriptive statistics.



3. By any means you like, obtain the entries of the analysis of variance table for this one way analysis of variance. Use your computer output (or excel work or hand calculations or whatever) to complete the following table: Source

df

Sum of Squares SSQ

Mean Square MSQ

F-Statistic

p-value

Between Groups

3

15.74

5.25

2.40

.10

Within Groups

20

43.69

2.18

Total, corrected 23 59.43

q3anova <-‐ aov(age~group, data=dat1) anova(q3anova)

## Analysis of Variance Table ## ## Response: age ## Df Sum Sq Mean Sq F value Pr(>F) ## group 3 15.74 5.2468 2.4018 0.09787 . ## Residuals 20 43.69 2.1845 ## -‐-‐-‐ ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

bartlett.test(age ~ group, data=dat1)

## ## Bartlett test of homogeneity of variances ## ## data: age by group ## Bartlett's K-‐squared = 2.6355, df = 3, p-‐value = 0.4513

4. Write a 2-5 sentence report of your description and hypothesis test findings using language as appropriate for a client who is intelligent but is not knowledgeable about statistics. Include figure and table as you think is appropriate.

In this sample, the data suggest a trend towards earlier age at first walking with increasing reinforcement and placement. The median age at first walking is greatest among controls (12.35 months) and lowest among infants in the “active” group (10.13 months); see also the box plots. Tests of statistical significance were limited to the overall F test for group differences and this did not achieve statistical significance (p-value = .10), possibly due to the small sample sizes (6 in each group). Interestingly, examination of the data also suggests that the variability in age at first walking differed, depending on the intervention received. The variability was greater in the three intervention groups (“active”, “passive”, “no exercise”) compared to in the “control” group; this was not statistically significant however (p-value = .45). Further study, utilizing larger sample sizes and additional hypothesis tests to investigate trend are needed.



5. For the brave Using appropriately defined indicator variables, perform a multivariable linear regression analysis of these same data! Use your computer output to complete the following table: Source df Sum of Squares Mean Square Overall F due model (p) = 3 ( )2

1

ˆn

ii

SSR Y Y=

= −∑ = 15.74 SSR/p = 5.25

2.40

due error (residual)

(n-1-p) = 20 ( )21

ˆn

i ii

SSE Y Y=

= −∑ = 43.69 SSE/(n-1-p) =2.18

Total, corrected (n-1) = 23 ( )21

n

ii

SST Y Y=

= −∑ = 59.43

Some of this has already been done for you: I considered the following parameterization Y = age at first walking I_act = 0/1 indicator of group assignment to “active” I_pass = 0/1 indicator of group assignment to “passive” I_noex = 0/1 indicator of group assignment to “no exercise” Thus, I used a reference cell coding approach with “8 week control” as my reference. I fit the following multivariable linear model of Y Y = β0 + β1 [I_act] + β2 [I_pass] + β3 [I_noex] + error

dat1$I_active <-‐ factor(dat1$I_active) dat1$I_passive <-‐ factor(dat1$I_passive) dat1$I_noex <-‐ factor(dat1$I_noex) q5model <-‐ lm(age ~ I_active + I_passive + I_noex, data=dat1) anova(q5model)

## Analysis of Variance Table ## ## Response: age ## Df Sum Sq Mean Sq F value Pr(>F) ## I_active 1 12.793 12.7934 5.8565 0.02516 * ## I_passive 1 1.712 1.7117 0.7836 0.38656 ## I_noex 1 1.235 1.2352 0.5654 0.46083 ## Residuals 20 43.690 2.1845 ## -‐-‐-‐ ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1



summary(q5model)

## ## Call: ## lm(formula = age ~ I_active + I_passive + I_noex, data = dat1) ## ## Residuals: ## Min 1Q Median 3Q Max ## -‐2.7083 -‐0.8500 -‐0.2792 0.5062 3.6250 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 12.3500 0.6034 20.468 6.94e-‐15 *** ## I_active1 -‐2.2250 0.8533 -‐2.607 0.0169 * ## I_passive1 -‐0.9750 0.8533 -‐1.143 0.2667 ## I_noex1 -‐0.6417 0.8533 -‐0.752 0.4608 ## -‐-‐-‐ ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 1.478 on 20 degrees of freedom ## Multiple R-‐squared: 0.2649, Adjusted R-‐squared: 0.1546 ## F-‐statistic: 2.402 on 3 and 20 DF, p-‐value: 0.09787

The prediction equation is thus:

Y = 12.35 - 2.225*I_active - 0.97*I_passive - 0.64*I_noex

The two analyses match (hooray), thus confirming that a multiple linear regression model utilizing appropriately defined indicator variables is equivalent to an analysis of variance.

6. For the brave: Using your output from your two analyses (1st-analysis of variance, 2nd – regression), obtain the predicted mean of Y =age at first walking twice in two ways.

Prediction Using One Way Analysis of Variance

Prediction Using Multiple Linear Regression

Active 10.125 1 0 1

ˆ ˆµ = (β +β ) = 12.35 - 2.225 10 = .125 Passive 11.375

2 0 2ˆ ˆµ = (β +β ) = 12.35 - 0.97 = 11.38

No-Exercise 11.71 3 0 3

ˆ ˆµ = (β +β ) = 12.35 - 0.64 = 11.71 Control 12.35

4 0ˆµ = β = 12.35



Question #6 For the brave -‐ Obtain Predicted Values: a) anova b) regression

# Predicted Means from ANOVA aactive <-‐ predict(q3anova,data.frame(group="1=active")) apassive <-‐ predict(q3anova,data.frame(group="2=passive")) anoex <-‐ predict(q3anova,data.frame(group="3=noex")) acontrol <-‐ predict(q3anova,data.frame(group="4=control")) anames <-‐ c("Active", "Passive", "NoEx", "Control") ahat <-‐ c(aactive,apassive,anoex,acontrol) means.anova <-‐ data.frame(anames,ahat) means.anova

## anames ahat ## 1 Active 10.12500 ## 2 Passive 11.37500 ## 3 NoEx 11.70833 ## 4 Control 12.35000

# Predicted Means from Regression active <-‐ predict(q5model,data.frame(I_active="1",I_passive="0",I_noex="0")) passive <-‐ predict(q5model,data.frame(I_active="0",I_passive="1",I_noex="0")) noex <-‐ predict(q5model,data.frame(I_active="0",I_passive="0",I_noex="1")) control <-‐ predict(q5model,data.frame(I_active="0",I_passive="0",I_noex="0")) names <-‐ c("Active", "Passive", "NoEx", "Control") yhat <-‐ c(active,passive,noex,control) means.regression <-‐ data.frame(names,yhat) means.regression

## names yhat ## 1 Active 10.12500 ## 2 Passive 11.37500 ## 3 NoEx 11.70833 ## 4 Control 12.35000



Practice with two-way factorial analysis of variance Exercises #7-12 Data set used: fishgrowth.dta Consider again the fish growth data on page 36 of Notes 7. Introduction to Analysis of Variance.

Light (light)

Water Temp (temp)

Fish Growth (growth)

1=low 1=cold 4.55 1=low 1=cold 4.24 1=low 2=lukewarm 4.89 1=low 2=lukewarm 4.88 1=low 3=warm 5.01 1=low 3=warm 5.11 2=high 1=cold 5.55 2=high 1=cold 4.08 2=high 2=lukewarm 6.09 2=high 2=lukewarm 5.01 2=high 3=warm 7.01 2=high 3=warm 6.92

Coding Manual:

Variable Coding growth continuous light 1 = low

2 = high temp 1 = cold

2 = lukewarm 3 = warm



Preliminaries for Exercises 7-‐12: Read in data. Clean. Descriptives

load(file="fishgrowth.Rdata”) str(fishgrowth)

## 'data.frame': 12 obs. of 3 variables: ## $ light : Factor w/ 2 levels "1=low","2=high": 1 1 1 1 1 1 2 2 2 2 ... ## $ temp : Factor w/ 3 levels "1=cold","2=lukewarm",..: 1 1 2 2 3 3 1 1 2 2 ... ## $ growth: num 4.55 4.24 4.89 4.88 5.01 ... ## -‐ attr(*, "datalabel")= chr "" ## -‐ attr(*, "time.stamp")= chr " 9 Apr 2017 16:59" ## -‐ attr(*, "formats")= chr "%8.0g" "%10.0g" "%9.0g" ## -‐ attr(*, "types")= int 251 251 254 ## -‐ attr(*, "val.labels")= chr "lightf" "tempf" "" ## -‐ attr(*, "var.labels")= chr "" "" "" ## -‐ attr(*, "version")= int 12 ## -‐ attr(*, "label.table")=List of 2 ## ..$ lightf: Named int 1 2 ## .. ..-‐ attr(*, "names")= chr "1=low" "2=high" ## ..$ tempf : Named int 1 2 3 ## .. ..-‐ attr(*, "names")= chr "1=cold" "2=lukewarm" "3=warm"

#Crosstab of factors I and II CrossTable(fishgrowth$light,fishgrowth$temp,dnn=c("Light","Temp"), format=c("SAS"),prop.r=FALSE, prop.c=FALSE,prop.t=FALSE, prop.chisq=FALSE)

## Cell Contents ## |-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐| ## | N | ## |-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐| ## ## ============================================== ## Temp ## Light 1=cold 2=lukewarm 3=warm Total ## -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ ## 1=low 2 2 2 6 ## -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ ## 2=high 2 2 2 6 ## -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ ## Total 4 4 4 12 ## ==============================================



# Descriptives on y=growth in all 4 groups summaryBy(growth ~ light + temp, data = fishgrowth, FUN=c(length,mean,var,sd,min,max), fun.names=c("n","Mean", "Var", "SD", "Min", "Max"))

## light temp growth.n growth.Mean growth.Var growth.SD ## 1 1=low 1=cold 2 4.395 4.805013e-‐02 0.219203399 ## 2 1=low 2=lukewarm 2 4.885 4.999752e-‐05 0.007070892 ## 3 1=low 3=warm 2 5.060 4.999990e-‐03 0.070710611 ## 4 2=high 1=cold 2 4.815 1.080450e+00 1.039447157 ## 5 2=high 2=lukewarm 2 5.550 5.831999e-‐01 0.763675270 ## 6 2=high 3=warm 2 6.965 4.050014e-‐03 0.063639718 ## growth.Min growth.Max ## 1 4.24 4.55 ## 2 4.88 4.89 ## 3 5.01 5.11 ## 4 4.08 5.55 ## 5 5.01 6.09 ## 6 6.92 7.01

7. State the analysis of variance model using notation 2

i j ijµ, α , β , (αβ) and σ as appropriate. Define all terms and constraints on the parameters. Answer:

( ) 2ijk i j ijk ijkij

Y = µ + α + β + αβ + ε , where ε ~N(0,σ )

with i = 1, 2 indexing light j = 1, 2, 3 indexing temperature k = 1, 2 indexing individual fish under light "i" at water tempera

( )

∑

∑

2

i ii=1

3

j jj=1

ij

ture "j"µ = population mean fish growth, over all groups

α = effect of light level "i", with α =0

β = effect of water temperature "j", with β =0

αβ = interaction effect of the combination of

∑ ∑2 3

ij iji=1 j=1

"ith" light level and "jth" water temperature

with (αβ) = 0 and (αβ) = 0.



8. Create the following new variables with accompanying definitions

Variable Coding i_high = 1 if (light is 2)

0 otherwise i_luke = 1 if (temp is 2)

0 otherwise i_warm = 1 if (temp is 3)

0 otherwise hi_luke = (i_high) * (i_luke) hi_warm = (i_high)*(i_warm)

fishgrowth$i_high <-‐ ifelse(fishgrowth$light=="2=high",1,0) fishgrowth$i_luke <-‐ ifelse(fishgrowth$temp=="2=lukewarm",1,0) fishgrowth$i_warm <-‐ ifelse(fishgrowth$temp=="3=warm",1,0) fishgrowth$hi_luke <-‐ fishgrowth$i_high*fishgrowth$i_luke fishgrowth$hi_warm <-‐ fishgrowth$i_high*fishgrowth$i_warm

9. Perform a two way analysis of variance so as to reproduce the following table

Source Df SSQ MSQ F p-value Due LIGHT 1 2.98 2.98 10.39 .018 Due TEMP 2 3.984 1.992 6.95 .027 Due Interaction 3 1.268 0.634 2.21 .191 Error 6 1.721 0.287 Total (Corrected) 11 9.953 -

fishgrowth$light <-‐ as.factor(fishgrowth$light) fishgrowth$temp <-‐ as.factor(fishgrowth$temp) q9anova <-‐ aov(growth ~ light + temp + light:temp, data=fishgrowth) anova(q9anova)

## Analysis of Variance Table ## ## Response: growth ## Df Sum Sq Mean Sq F value Pr(>F) ## light 1 2.9800 2.98003 10.3906 0.01806 * ## temp 2 3.9843 1.99216 6.9462 0.02744 * ## light:temp 2 1.2676 0.63381 2.2099 0.19093 ## Residuals 6 1.7208 0.28680 ## -‐-‐-‐ ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1



leveneTest(growth ~ light*temp, data=fishgrowth)

## Warning in anova.lm(lm(resp ~ group)): ANOVA F-‐tests on an essentially ## perfect fit are unreliable

## Levene's Test for Homogeneity of Variance (center = median) ## Df F value Pr(>F) ## group 5 2.7375e+31 < 2.2e-‐16 *** ## 6 ## -‐-‐-‐ ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

10 Perform a regression analysis to obtain the following table and estimates

Source Df SSQ MSQ F p-value Due Model 5 8.2320 1.6464 5.74 .0276 Due Residual 6 1.7208 0.2868 - Total (Corrected) 11 9.953 -

Predictor beta Se(beta) T=beta/se p-value I_high 0.42 0.5355 0.78 0.46 I_luke 0.49 0.5355 0.91 0.395 I_warm 0.665 0.5355 1.24 0.261 Hi_luke 0.245 0.7574 0.32 0.757 Hi_warm 1.485 0.7574 1.96 0.098 Intercept 4.395 0.3787 11.61 0.000

q10model <-‐ lm(growth ~ i_high + i_luke + i_warm + hi_luke + hi_warm, data=fishgrowth) anova(q10model)

## Analysis of Variance Table ## ## Response: growth ## Df Sum Sq Mean Sq F value Pr(>F) ## i_high 1 2.9800 2.9800 10.3906 0.018059 * ## i_luke 1 0.0222 0.0222 0.0774 0.790164 ## i_warm 1 3.9621 3.9621 13.8149 0.009889 ** ## hi_luke 1 0.1650 0.1650 0.5753 0.476877 ## hi_warm 1 1.1026 1.1026 3.8445 0.097594 . ## Residuals 6 1.7208 0.2868 ## -‐-‐-‐ ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1



summary(q10model)

## ## Call: ## lm(formula = growth ~ i_high + i_luke + i_warm + hi_luke + hi_warm, ## data = dat2) ## ## Residuals: ## Min 1Q Median 3Q Max ## -‐0.73500 -‐0.07625 0.00000 0.07625 0.73500 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 4.3950 0.3787 11.606 2.46e-‐05 *** ## i_high 0.4200 0.5355 0.784 0.4627 ## i_luke 0.4900 0.5355 0.915 0.3955 ## i_warm 0.6650 0.5355 1.242 0.2607 ## hi_luke 0.2450 0.7574 0.323 0.7573 ## hi_warm 1.4850 0.7574 1.961 0.0976 . ## -‐-‐-‐ ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.5355 on 6 degrees of freedom ## Multiple R-‐squared: 0.8271, Adjusted R-‐squared: 0.683 ## F-‐statistic: 5.741 on 5 and 6 DF, p-‐value: 0.02755

11. Using the output from each of your anova and regression analyses, complete the following tables and notice that they are the same.

# Predicted Means from ANOVA ameans <-‐ data.frame(fishgrowth$temp,fishgrowth$light, fitted=predict(q9anova)) ameans

## dat2.temp dat2.light fitted ## 1 1=cold 1=low 4.395 ## 2 1=cold 1=low 4.395 ## 3 2=lukewarm 1=low 4.885 ## 4 2=lukewarm 1=low 4.885 ## 5 3=warm 1=low 5.060 ## 6 3=warm 1=low 5.060 ## 7 1=cold 2=high 4.815 ## 8 1=cold 2=high 4.815 ## 9 2=lukewarm 2=high 5.550 ## 10 2=lukewarm 2=high 5.550 ## 11 3=warm 2=high 6.965 ## 12 3=warm 2=high 6.965



Estimated Mean Growth, by Conditions of Light and Temperature – Anova Analysis Cold Lukewarm Warm Low light 4.395 4.885 5.06 High light 4.815 5.55 6.965

# Predicted Means from Regression rmeans <-‐ data.frame(fishgrowth$temp,fishgrowth$light, fitted=predict(q10model)) rmeans

## dat2.temp dat2.light fitted ## 1 1=cold 1=low 4.395 ## 2 1=cold 1=low 4.395 ## 3 2=lukewarm 1=low 4.885 ## 4 2=lukewarm 1=low 4.885 ## 5 3=warm 1=low 5.060 ## 6 3=warm 1=low 5.060 ## 7 1=cold 2=high 4.815 ## 8 1=cold 2=high 4.815 ## 9 2=lukewarm 2=high 5.550 ## 10 2=lukewarm 2=high 5.550 ## 11 3=warm 2=high 6.965 ## 12 3=warm 2=high 6.965

Estimated Mean Growth, by Conditions of Light and Temperature – Regression Analysis

Cold Lukewarm Warm Low light 4.395 4.885 5.06 High light 4.815 5.55 6.965

12. Compare your answer to #11 with the observed means. Observed Mean Growth, by Conditions of Light and Temperature

Cold Lukewarm Warm Low light 4.395 4.885 5.06 High light 4.815 5.55 6.965

They match!

Documents

R Users - UMasspeople.umass.edu/biep640w/pdf/sol_anova_1 of 2 R USERS.pdfBIOSTATS 640 Spring 2019 Unit 7 Introduction to Analysis of Variance (1 of 2) Solutions R Users Sol_anova_1