lm1 <- data.frame(metab=c(173,162,176,181,164,169,170,185,
164,177,175,172,168),location=
c(0,0,0,0,0,0,0,1,1,1,1,1,1))
#
library(ggplot2)
ggplot(lm1,aes(x=factor(location),y=metab))+
geom_boxplot()+
labs(x="Location",y="Metabole")+
theme_bw()
metab.fit <- glm(metab~factor(location),data = lm1)drop1(metab.fit,test = "F")## Single term deletions
##
## Model:
## metab ~ factor(location)
## Df Deviance AIC F value Pr(>F)
## <none> 532.93 91.167
## factor(location) 1 558.00 89.765 0.5175 0.4869
| name | SS | df | ms | F | p=value |
|---|---|---|---|---|---|
| Location | 25.1 | 1 | 25.1 | 0.52 | 0.4869 |
| Residual | 532.93 | 11 | 48.4 | ||
| Total | 558 | 12 |
The F-value is 0.52 with a p-value of 0.4869 which is larger then 0.05 so the null-hypothesis is not rejected. This research cannot show that there is a location effect on the metabole.
lm2 <- data.frame(blcalc=c(17.0,18.9,13.2,14.6,13.3,16.5,14.3,
10.9,15.6,8.9,18.6,16.2,12.5,15.1,16.2,17.1,14.7,15.3,14.2,12.8),
treat=gl(2,10,labels=c("No hormone","hormone")),
sex=gl(2,5,length=20,labels=c("Female","male")))The function gl() makes a factor varable and it works as follows: gl(n, k, length = n*k, labels = 1:n, ordered = FALSE) n=an integer giving the number of levels. k=an integer giving the number of replications. length=an integer giving the length of the result. labels an optional vector of labels for the resulting factor levels. ordered a logical indicating whether the result should be ordered or not.
ggplot(lm2,aes(y=blcalc,x=treat,fill=(sex)))+
geom_boxplot()+
labs(y="Blood calcium")+
theme_bw()
blcalc.fit <- glm(blcalc~treat+sex,data = lm2)drop1(blcalc.fit,test="F")## Single term deletions
##
## Model:
## blcalc ~ treat + sex
## Df Deviance AIC F value Pr(>F)
## <none> 97.732 96.488
## treat 1 102.245 95.390 0.7849 0.3880
## sex 1 109.437 96.750 2.0359 0.1717
SS for treat 102.245-97.733=4.512 and for sex 109.437-97.733=11.704
| name | SS | df | ms | F | p=value |
|---|---|---|---|---|---|
| treat | 4.51 | 1 | 4.51 | 0.79 | 0.3880 |
| sex | 11.70 | 1 | 11.70 | 2.04 | 0.1717 |
| residual | 97.73 | 17 | 5.75 | ||
| total | 113.95 |
For sex p-value= 0.17 so one can not show diffrenece between males and females. For treat the p-value=0.3880 so one can not show diffrenece between the hormone groups.
summary(blcalc.fit)##
## Call:
## glm(formula = blcalc ~ treat + sex, data = lm2)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -4.655 -1.725 0.165 1.948 3.815
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 15.0850 0.9286 16.244 8.69e-12 ***
## treathormone 0.9500 1.0723 0.886 0.388
## sexmale -1.5300 1.0723 -1.427 0.172
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 5.748971)
##
## Null deviance: 113.950 on 19 degrees of freedom
## Residual deviance: 97.733 on 17 degrees of freedom
## AIC: 96.488
##
## Number of Fisher Scoring iterations: 2
In the summary the line factor(hormone)2 gives the difference between the mean blood calcium of the hormone group and the the mean blood calcium of the no hormone group. This difference is .95 with a standard error of 1.0723
Data can be directly read from github. First give the adress in a variable and then read the file:
adress <- "https://raw.github.com/janvdbroek/Generalized-Linear-Models/master/aligator.txt"
lm3 <- read.table(adress,header=TRUE)or with read in the data with the rstudio menu: import dataset, from text.
ggplot(lm3,aes(y=Weight,x=Length))+
geom_point()+
geom_smooth(method = lm,se=FALSE)+
theme_bw()## `geom_smooth()` using formula = 'y ~ x'
ggplot(lm3,aes(y=log(Weight),x=log(Length)))+
geom_point()+
geom_smooth(method = lm,se=FALSE)+
theme_bw()## `geom_smooth()` using formula = 'y ~ x'
##
3c
cor(lm3$Length,lm3$Weight)## [1] 0.7812519
cor(log(lm3$Length),log(lm3$Weight))## [1] 0.8680974
The log version is a better.
Wght.fit <- glm(log(Weight)~log(Length),data = lm3)
summary(Wght.fit)##
## Call:
## glm(formula = log(Weight) ~ log(Length), data = lm3)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.23474 -0.06012 0.00835 0.13092 0.67242
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -7.3595 1.3889 -5.299 2.23e-05 ***
## log(Length) 2.6348 0.3141 8.387 1.89e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.1105235)
##
## Null deviance: 10.316 on 24 degrees of freedom
## Residual deviance: 2.542 on 23 degrees of freedom
## AIC: 19.799
##
## Number of Fisher Scoring iterations: 2
log(Weight)=-7.36+2.635*log(Length)+residual
drop1(Wght.fit,test="F")## Single term deletions
##
## Model:
## log(Weight) ~ log(Length)
## Df Deviance AIC F value Pr(>F)
## <none> 2.542 19.799
## log(Length) 1 10.316 52.818 70.341 1.889e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
| name | SS | df | ms | F | p=value |
|---|---|---|---|---|---|
| log(length) | 7.77 | 1 | 7.77 | 70.3 | 1.9*10^-8 |
| Residual | 2.54 | 23 | 0.11 | ||
| Total | 10.36 | 24 |
The nul-hypothesis that the regression coefficient in the population is zero is rejected since the p value is very low.
lm4 <- data.frame(bp=c(87,86.5,89,88.5,87.5,88,86.5,87,85,86,85,83),
dose=c(5,6,7,8,9,10,5,6,7,8,9,10),
group=c(0,0,0,0,0,0,1,1,1,1,1,1))
lm4$fgroup <- factor(lm4$group)
lm4$prod <- with(lm4,group*dose)model.i <- glm(bp~group+dose+group:dose,data = lm4)
model.ii <- glm(bp~fgroup+dose+fgroup:dose,data = lm4)
model.iii <- glm(bp~group+group:dose,data = lm4)
model.iv <- glm(bp~group+prod,data = lm4)
model.v <- glm(bp~fgroup+fgroup:dose,data = lm4)summary(model.i)##
## Call:
## glm(formula = bp ~ group + dose + group:dose, data = lm4)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.9286 -0.6131 -0.2500 0.6250 1.3571
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 86.1429 1.6635 51.785 2.14e-11 ***
## group 4.0952 2.3525 1.741 0.1199
## dose 0.2143 0.2163 0.991 0.3508
## group:dose -0.8571 0.3058 -2.803 0.0231 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.8184524)
##
## Null deviance: 30.9167 on 11 degrees of freedom
## Residual deviance: 6.5476 on 8 degrees of freedom
## AIC: 36.785
##
## Number of Fisher Scoring iterations: 2
y=86.1429+0.2143dose and y=90.2381-0.6428dose
summary(model.ii)##
## Call:
## glm(formula = bp ~ fgroup + dose + fgroup:dose, data = lm4)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.9286 -0.6131 -0.2500 0.6250 1.3571
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 86.1429 1.6635 51.785 2.14e-11 ***
## fgroup1 4.0952 2.3525 1.741 0.1199
## dose 0.2143 0.2163 0.991 0.3508
## fgroup1:dose -0.8571 0.3058 -2.803 0.0231 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.8184524)
##
## Null deviance: 30.9167 on 11 degrees of freedom
## Residual deviance: 6.5476 on 8 degrees of freedom
## AIC: 36.785
##
## Number of Fisher Scoring iterations: 2
same as model.i
summary(model.iii)##
## Call:
## glm(formula = bp ~ group + group:dose, data = lm4)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.2500 -0.7411 0.0000 0.6518 1.2500
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 87.7500 0.3690 237.829 <2e-16 ***
## group 2.4881 1.7023 1.462 0.1779
## group:dose -0.6429 0.2160 -2.976 0.0156 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.8167989)
##
## Null deviance: 30.9167 on 11 degrees of freedom
## Residual deviance: 7.3512 on 9 degrees of freedom
## AIC: 36.174
##
## Number of Fisher Scoring iterations: 2
y=87.75 and y=90.2381 - 0.6429*dose so no relation with dose in group 0
summary(model.iv)##
## Call:
## glm(formula = bp ~ group + prod, data = lm4)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.2500 -0.7411 0.0000 0.6518 1.2500
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 87.7500 0.3690 237.829 <2e-16 ***
## group 2.4881 1.7023 1.462 0.1779
## prod -0.6429 0.2160 -2.976 0.0156 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.8167989)
##
## Null deviance: 30.9167 on 11 degrees of freedom
## Residual deviance: 7.3512 on 9 degrees of freedom
## AIC: 36.174
##
## Number of Fisher Scoring iterations: 2
same as model.iii
summary(model.v)##
## Call:
## glm(formula = bp ~ fgroup + fgroup:dose, data = lm4)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.9286 -0.6131 -0.2500 0.6250 1.3571
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 86.1429 1.6635 51.785 2.14e-11 ***
## fgroup1 4.0952 2.3525 1.741 0.1199
## fgroup0:dose 0.2143 0.2163 0.991 0.3508
## fgroup1:dose -0.6429 0.2163 -2.973 0.0178 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.8184524)
##
## Null deviance: 30.9167 on 11 degrees of freedom
## Residual deviance: 6.5476 on 8 degrees of freedom
## AIC: 36.785
##
## Number of Fisher Scoring iterations: 2
Same as model.i but now with regression coefficients explicit in the output. This is the nested version of the model.Note that the difference fgroup1:dose-fgroup0:dose=-.8571 the interaction from model.i and model.ii. So the interaction term in these models shows wether the difference inslopes is the same in both groups.
Usually first model i or ii is fitted. Then if the interaction is important the nested version (model v) can be fitted to see which of the slopes are important. The slope in group2 is significant so then model.iii or model.iv can be fitted.
So by choosing a parameterization various aspects of the regression equations can be made explicit (e.g. the equation for both groups, or the difference in slope and/or intercept).
adress <- "https://raw.github.com/janvdbroek/Generalized-Linear-Models/master/lowbirth.dat"
lm5 <- read.table(adress,header=TRUE)
#
fit.0 <- glm(bwt~ht+smoke+age+smoke:ht+age:ht,data = lm5)
drop1(fit.0,test="F")## Single term deletions
##
## Model:
## bwt ~ ht + smoke + age + smoke:ht + age:ht
## Df Deviance AIC F value Pr(>F)
## <none> 89847460 3020.9
## ht:smoke 1 90112593 3019.5 0.5400 0.463366
## ht:age 1 93340890 3026.2 7.1154 0.008328 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
So the ht:age interaction is needed in the model. Now fit the model without the ht:smoke interaction: interpretation: for instance ht:age, the age effect depends on the hypertension group. ## 5c
fit.1 <- glm(bwt~ht+smoke+age+age:ht,data = lm5)
drop1(fit.1,test="F")## Single term deletions
##
## Model:
## bwt ~ ht + smoke + age + age:ht
## Df Deviance AIC F value Pr(>F)
## <none> 90112593 3019.5
## smoke 1 93454969 3024.4 6.8248 0.009733 **
## ht:age 1 93655051 3024.8 7.2333 0.007814 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
So no futher reduction is possible.(Age stays in the model since the ht:age interaction is in the model.)
summary(fit.1)##
## Call:
## glm(formula = bwt ~ ht + smoke + age + age:ht, data = lm5)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2174.16 -438.51 48.85 478.79 1556.97
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2699.463 240.220 11.237 < 2e-16 ***
## ht 2568.064 1132.100 2.268 0.02447 *
## smoke -272.746 104.403 -2.612 0.00973 **
## age 16.302 9.847 1.655 0.09955 .
## ht:age -130.504 48.524 -2.689 0.00781 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 489742.4)
##
## Null deviance: 99917053 on 188 degrees of freedom
## Residual deviance: 90112593 on 184 degrees of freedom
## AIC: 3019.5
##
## Number of Fisher Scoring iterations: 2