Looking at CTmin
#Checking normality
qqnorm(wild.CTmin$CTmin)
qqline(wild.CTmin$CTmin)
shapiro.test(wild.CTmin$CTmin)
##
## Shapiro-Wilk normality test
##
## data: wild.CTmin$CTmin
## W = 0.91984, p-value = 0.1916
#Checking Relationships
plot(wild.CTmin$CTmin~wild.CTmin$Mass*wild.CTmin$Class)
scatter.smooth(wild.CTmin$Mass,wild.CTmin$CTmin)
Did we start checking CTmin too late in trial 2?
as.factor(wild.CTmin$Trial)
## [1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3
## Levels: 1 2 3
plot(wild.CTmin$CTmin~wild.CTmin$Trial)
summary(aov(wild.CTmin$CTmin~wild.CTmin$Trial))
## Df Sum Sq Mean Sq F value Pr(>F)
## wild.CTmin$Trial 1 24.15 24.149 4.981 0.0439 *
## Residuals 13 63.03 4.848
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
p=0.0439 for anova looking at effect of trial; consider removing trial 2
summer.CTmin<-wild.CTmin[-6:-10,1:5]
as.factor(summer.CTmin$Trial)
## [1] 1 1 1 1 1 3 3 3 3 3
## Levels: 1 3
checking more assumptions
#Equal variance between males and females?
male.CTmin<-(summer.CTmin[c(1,3,5,6:7),5])
female.CTmin<-(summer.CTmin[c(2,4,8:10),5])
var.test(male.CTmin,female.CTmin)
##
## F test to compare two variances
##
## data: male.CTmin and female.CTmin
## F = 1.5659, num df = 4, denom df = 4, p-value = 0.6746
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
## 0.1630348 15.0394666
## sample estimates:
## ratio of variances
## 1.565872
GLM model with mass, class and trial
summer.CTmin.glm1<-glm(summer.CTmin$CTmin~summer.CTmin$Class+summer.CTmin$Trial+summer.CTmin$Mass+summer.CTmin$Class*summer.CTmin$Trial)
summary.glm(summer.CTmin.glm1)
##
## Call:
## glm(formula = summer.CTmin$CTmin ~ summer.CTmin$Class + summer.CTmin$Trial +
## summer.CTmin$Mass + summer.CTmin$Class * summer.CTmin$Trial)
##
## Deviance Residuals:
## 1 2 3 4 5 6 7 8
## -0.5919 0.2157 -0.3370 -0.2157 0.9290 1.3631 -1.3631 3.9614
## 9 10
## -2.9345 -1.0269
##
## Coefficients:
## Estimate Std. Error t value
## (Intercept) 3.9524 3.7864 1.044
## summer.CTmin$ClassM -1.9982 3.7158 -0.538
## summer.CTmin$Trial 1.3354 1.1307 1.181
## summer.CTmin$Mass 0.5392 0.4235 1.273
## summer.CTmin$ClassM:summer.CTmin$Trial 0.4596 1.6286 0.282
## Pr(>|t|)
## (Intercept) 0.344
## summer.CTmin$ClassM 0.614
## summer.CTmin$Trial 0.291
## summer.CTmin$Mass 0.259
## summer.CTmin$ClassM:summer.CTmin$Trial 0.789
##
## (Dispersion parameter for gaussian family taken to be 6.098956)
##
## Null deviance: 69.338 on 9 degrees of freedom
## Residual deviance: 30.495 on 5 degrees of freedom
## AIC: 51.528
##
## Number of Fisher Scoring iterations: 2
“Trial” no longer significant. Consider removing “Trial” as a factor.
Proposed final GLM model for CTmin?
summer.CTmin.glm2<-glm(summer.CTmin$CTmin~summer.CTmin$Class+summer.CTmin$Mass, family=gaussian())
summary.glm(summer.CTmin.glm2)
##
## Call:
## glm(formula = summer.CTmin$CTmin ~ summer.CTmin$Class + summer.CTmin$Mass,
## family = gaussian())
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.967 -1.690 -1.194 0.556 5.130
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.3040 2.9566 2.132 0.0704 .
## summer.CTmin$ClassM -2.1094 2.5522 -0.826 0.4358
## summer.CTmin$Mass 0.6408 0.4648 1.379 0.2104
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 7.735783)
##
## Null deviance: 69.338 on 9 degrees of freedom
## Residual deviance: 54.150 on 7 degrees of freedom
## AIC: 53.271
##
## Number of Fisher Scoring iterations: 2
Looking at CTmax
qqnorm(wild.CTmax$Ctmax)
qqline(wild.CTmax$Ctmax)
shapiro.test((wild.CTmax$Ctmax))
##
## Shapiro-Wilk normality test
##
## data: (wild.CTmax$Ctmax)
## W = 0.86223, p-value = 0.03269
##not normally distributed
scatter.smooth(wild.CTmax$Mass,wild.CTmax$Ctmax)
##size effect?
hist(wild.CTmax$Ctmax)
##left skewed. transform?
CTmax transformations
log.trans.summer.CTmax<-log(wild.CTmax$Ctmax)
hist(log.trans.summer.CTmax)
qqnorm(log.trans.summer.CTmax)
qqline(log.trans.summer.CTmax)
shapiro.test(log.trans.summer.CTmax)
##
## Shapiro-Wilk normality test
##
## data: log.trans.summer.CTmax
## W = 0.85186, p-value = 0.02354
##still not normally distributed
sqr.trans.wild.CTmax<-sqrt(wild.CTmax$Ctmax)
hist(sqr.trans.wild.CTmax)
qqnorm(sqr.trans.wild.CTmax)
qqline(sqr.trans.wild.CTmax)
shapiro.test(sqr.trans.wild.CTmax)
##
## Shapiro-Wilk normality test
##
## data: sqr.trans.wild.CTmax
## W = 0.8571, p-value = 0.02777
##nope
inverse.summer.CTmax<-1/(wild.CTmax$Ctmax)
hist(inverse.summer.CTmax)
qqnorm(inverse.summer.CTmax)
qqline(inverse.summer.CTmax)
shapiro.test(inverse.summer.CTmax)
##
## Shapiro-Wilk normality test
##
## data: inverse.summer.CTmax
## W = 0.84106, p-value = 0.01683
##nope
power.summer.CTmax<-(wild.CTmax$Ctmax)^15
hist(power.summer.CTmax)
qqnorm(power.summer.CTmax)
qqline(power.summer.CTmax)
shapiro.test(power.summer.CTmax)
##
## Shapiro-Wilk normality test
##
## data: power.summer.CTmax
## W = 0.9497, p-value = 0.5561
plot(power.summer.CTmax~wild.CTmax$Ctmax)
##this works but is not intuitive
##consider non-parametric? Though I don't know how to include mass as a covariate in a non-parametric GLM
kruskal.test(wild.CTmax$Ctmax~wild.CTmax$Class)
##
## Kruskal-Wallis rank sum test
##
## data: wild.CTmax$Ctmax by wild.CTmax$Class
## Kruskal-Wallis chi-squared = 0.11111, df = 1, p-value = 0.7389
boxplot(wild.CTmax$Ctmax~wild.CTmax$Class)
##probably un-equal variance
male.CTmax<-(wild.CTmax[c(1,3:7,12:14),5])
female.CTmax<-(wild.CTmax[c(2,8:11),5])
var.test(male.CTmax,female.CTmax)
##
## F test to compare two variances
##
## data: male.CTmax and female.CTmax
## F = 7.3538, num df = 8, denom df = 4, p-value = 0.07122
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
## 0.8189437 37.1558984
## sample estimates:
## ratio of variances
## 7.353771
##p=0.07122. moving forward with assumption of equal variance
##using transformed data (to 15th power)
##model looking at mass and class
summer.CTmax.glm1<-glm(power.summer.CTmax~wild.CTmax$Class+wild.CTmax$Mass)
summary(summer.CTmax.glm1)
##
## Call:
## glm(formula = power.summer.CTmax ~ wild.CTmax$Class + wild.CTmax$Mass)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.936e+24 -6.299e+23 -1.394e+23 9.629e+23 1.796e+24
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.097e+24 8.288e+23 8.563 3.4e-06 ***
## wild.CTmax$ClassM 9.030e+23 7.549e+23 1.196 0.25675
## wild.CTmax$Mass -3.676e+23 9.913e+22 -3.708 0.00345 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 1.4854e+48)
##
## Null deviance: 3.7076e+49 on 13 degrees of freedom
## Residual deviance: 1.6339e+49 on 11 degrees of freedom
## AIC: 1597.2
##
## Number of Fisher Scoring iterations: 2
mass has an effect on CTmax (p=0.00345), class does not (p=.2568).
##Negative correlation between mass and CTmax
scatter.smooth(wild.CTmax$Ctmax~wild.CTmax$Mass)
##largest individuals are clearly males. Was there an outlier?
interaction.plot(wild.CTmax$Mass,wild.CTmax$Class,wild.CTmax$Ctmax,type="b")
## a 40C CTmax doesnt seem normal. Checking excel file...