Damage
) within explanatory variables (Trial
, Mite.Strain
).library(ggplot2) # for general plotting
library(car) # for ANOVA (Type II used, better than Type I when there is an unbalanced design)
Damage from TU and TU-A (100 mites) on Moneymaker tomato leaflets, 24 hpi.
damage.data <- read.csv("~/Lab Stuff/Adapted mites/Tomato/Damage assay/TU and TA on Moneymaker/Damage R data.csv", header = TRUE)
# trial as a factor
damage.data$Trial <- factor(damage.data$Trial)
str(damage.data)
## 'data.frame': 24 obs. of 3 variables:
## $ Trial : Factor w/ 3 levels "1","2","3": 1 1 1 1 2 2 2 2 3 3 ...
## $ Mite.Strain: Factor w/ 2 levels "TU","TU-A": 1 1 1 1 1 1 1 1 1 1 ...
## $ Damage : num 0.312 0.125 0.188 0.125 0.375 ...
H0: There will be no difference in damage produced by the mite strains.
HA: TU-A will produce more damage on Moneymaker leaflets than TU mites.
Damage
) within explanatory variables (Trial
, Mite.Strain
).ggplot(damage.data, aes(x = Trial, y = Damage)) + geom_boxplot() + theme_classic()
ggplot(damage.data, aes(x = Mite.Strain, y = Damage)) + geom_boxplot() + theme_classic()
Outlier left in, probably represents real variability.
Des not apply, all explanatory variables are categorical/factorial.
No, I am treating Trial
as a main effect to check for reproducibility (not a random effect/blocking factor).
Interaction betweenTrial
and Mite.Strain
will be performed to test for reproducibility.
No
summary(damage.data)
## Trial Mite.Strain Damage
## 1:8 TU :12 Min. : 0.125
## 2:8 TU-A:12 1st Qu.: 0.375
## 3:8 Median : 17.312
## Mean : 36.885
## 3rd Qu.: 60.469
## Max. :142.000
Yes
# fit linear model and display model fit information and ANOVA table
m <- lm(Damage ~ Mite.Strain + Trial + Mite.Strain:Trial, data = damage.data)
summary(m)
##
## Call:
## lm(formula = Damage ~ Mite.Strain + Trial + Mite.Strain:Trial,
## data = damage.data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -40.484 -3.188 -0.063 0.168 48.922
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.1875 10.5359 0.018 0.9860
## Mite.StrainTU-A 93.2969 14.9000 6.262 6.63e-06 ***
## Trial2 0.8281 14.9000 0.056 0.9563
## Trial3 0.2500 14.9000 0.017 0.9868
## Mite.StrainTU-A:Trial2 -16.7969 21.0717 -0.797 0.4358
## Mite.StrainTU-A:Trial3 -45.0625 21.0717 -2.139 0.0464 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 21.07 on 18 degrees of freedom
## Multiple R-squared: 0.8176, Adjusted R-squared: 0.7669
## F-statistic: 16.13 on 5 and 18 DF, p-value: 4.217e-06
Anova(m)
## Anova Table (Type II tests)
##
## Response: Damage
## Sum Sq Df F value Pr(>F)
## Mite.Strain 31692 1 71.3750 1.118e-07 ***
## Trial 2054 2 2.3127 0.1277
## Mite.Strain:Trial 2074 2 2.3360 0.1253
## Residuals 7992 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Calculate effect size and display
result.anova<-Anova(m)
ss<-result.anova$"Sum Sq" ##ss = sum of squares
pes<-ss/(ss+ss[length(ss)]) ##pes = partial e squared
pes[length(pes)]<-""
result.anova$"Part E Sq"<-pes
result.anova
## Anova Table (Type II tests)
##
## Response: Damage
## Sum Sq Df F value Pr(>F) Part E Sq
## Mite.Strain 31692 1 71.3750 0.00000 0.79860
## Trial 2054 2 2.3127 0.12766 0.20444
## Mite.Strain:Trial 2074 2 2.3360 0.12532 0.20607
## Residuals 7992 18
# plot interactions
interaction.plot(damage.data$Mite.Strain, damage.data$Trial, damage.data$Damage, type="l", leg.bty="o", leg.bg="grey95", lwd=2, ylab="Damage", xlab="Mite Strain", main="Mite.Strain:Trial")
damage.data$m.fit <- fitted(m) # fitted values
damage.data$m.res <- rstandard(m) # Pearson residuals
We assumed normal residuals. This is the least important regression assumption but its can be tested with a qq plot.
ggplot(damage.data, aes(sample = m.res)) + geom_qq() +
geom_abline(intercept = 0, slope = 1) + theme_classic()
Some curvage here, not too bad.
Testing for:
Linearity - there should be no curvilinear pattern in the residuals.
Equal variance - the vertical spread of the residuals should be constant across all fitted values.
ggplot(damage.data, aes(x = m.fit, y = m.res)) +
geom_point() + geom_hline(yintercept = 0) + geom_smooth() + theme_classic()
## `geom_smooth()` using method = 'loess' and formula 'y ~ x'
Linearity - decent - confidence interval includes 0 for all values.
Equal variance - no obvious problems
Should be centered around 0, if not then model requires another explanatory variable(s), to account for observed variation.
ggplot(damage.data, aes(x = Mite.Strain, y = m.res)) +
geom_boxplot() + geom_hline(yintercept = 0) + theme_classic()
ggplot(damage.data, aes(x = Trial, y = m.res)) +
geom_boxplot() + geom_hline(yintercept = 0) + theme_classic()
ggplot(damage.data, aes(x = Mite.Strain:Trial, y = m.res)) +
geom_boxplot() + geom_hline(yintercept = 0) + theme_classic()
Looks good.
Model is valid and interpretation of ANOVA is good.
sessionInfo()
## R version 3.6.0 (2019-04-26)
## Platform: x86_64-w64-mingw32/x64 (64-bit)
## Running under: Windows 10 x64 (build 17763)
##
## Matrix products: default
##
## locale:
## [1] LC_COLLATE=English_Canada.1252 LC_CTYPE=English_Canada.1252
## [3] LC_MONETARY=English_Canada.1252 LC_NUMERIC=C
## [5] LC_TIME=English_Canada.1252
##
## attached base packages:
## [1] stats graphics grDevices utils datasets methods base
##
## other attached packages:
## [1] car_3.0-3 carData_3.0-2 ggplot2_3.1.1
##
## loaded via a namespace (and not attached):
## [1] zip_2.0.2 Rcpp_1.0.1 cellranger_1.1.0
## [4] pillar_1.4.1 compiler_3.6.0 plyr_1.8.4
## [7] forcats_0.4.0 tools_3.6.0 digest_0.6.19
## [10] evaluate_0.14 tibble_2.1.1 gtable_0.3.0
## [13] pkgconfig_2.0.2 rlang_0.3.4 openxlsx_4.1.0
## [16] curl_3.3 yaml_2.2.0 haven_2.1.0
## [19] xfun_0.7 rio_0.5.16 withr_2.1.2
## [22] stringr_1.4.0 knitr_1.23 hms_0.4.2
## [25] grid_3.6.0 data.table_1.12.2 readxl_1.3.1
## [28] foreign_0.8-71 rmarkdown_1.13 magrittr_1.5
## [31] scales_1.0.0 htmltools_0.3.6 abind_1.4-5
## [34] colorspace_1.4-1 labeling_0.3 stringi_1.4.3
## [37] lazyeval_0.2.2 munsell_0.5.0 crayon_1.3.4