Description

Saket Choudhary (USCID: 2170058637) (skchoudh@usc.edu)
09/21/2015
Chapter 9: 16
Part (a)
library(ggplot2)
data <- read.csv(‘data_ch9_16.csv’, header=T)
ggplot(data, aes(x=DurationOfVisit, y=PollenRemoved, color=BeeType)) + geom_point(shape=1) + scale_colour_hue(l=50) + geom_smooth(method=lm,
se=FALSE)

DurationOfVisit From
the graph we see that the relation between proportion removed and duration visited is not linear Part (b)
data$PollenRemovedlogit = log(data$PollenRemoved/(1-data$PollenRemoved)) ggplot(data, aes(x=DurationOfVisit, y=PollenRemovedlogit, color=BeeType)) +
geom_point(shape=1) + scale_colour_hue(l=50) + geom_smooth(method=lm,
se=FALSE)

From the plot above it does NOT seem that the logit transformed PollenRemoved has a linear relationship with duration of visit
Part (c) Model:
µ{PollenRemovedLogit|DuratioOfV isitlog,BeeType} = β0 + β1DurationOfV isitlog
+ β2BeeType
data$DurationOfVisitlog = log(data$DurationOfVisit)
ggplot(data, aes(x=DurationOfVisitlog, y=PollenRemovedlogit, color=BeeType)) + geom_point(shape=1) + scale_colour_hue(l=50) + geom_smooth(method=lm,
se=FALSE)

From the plot of logit(PollenRemoved) vs log(DurationoOfV isit) it seems that these follow a linear relationship.
Part (d) Model:
µ{PollenRemovedLogit|DuratioOfV isitlog,BeeType} = β0 + β1DurationOfV isitlog
+ β2BeeType + β3BeeType ∗ DurationOfV isitlog
lmfit <- lm(PollenRemovedlogit ~ BeeType + DurationOfVisitlog
+ BeeType*DurationOfVisitlog, data=data)
summary(lmfit)
##
## Call:
## lm(formula = PollenRemovedlogit ~ BeeType + DurationOfVisitlog +
## BeeType * DurationOfVisitlog, data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.3803 -0.3699 0.0307 0.4552 1.1611
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -3.0390 0.5115 -5.941 4.45e-07 ***
## BeeTypeWorker 1.3770 0.8722 1.579 0.122
## DurationOfVisitlog 1.0121 0.1902 5.321 3.52e-06 ***
## BeeTypeWorker:DurationOfVisitlog -0.2709 0.2817 -0.962 0.342
## —
## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ‘ 1
##
## Residual standard error: 0.6525 on 43 degrees of freedom ## Multiple R-squared: 0.6151, Adjusted R-squared: 0.5882
## F-statistic: 22.9 on 3 and 43 DF, p-value: 5.151e-09
The p-value of the interaction term is 0.342 which is not significant at the threshold level of 0.05. Thus we cannot reject the null hypothesis that the interaction term is 0. Thus this tells us that there very little evidence that the proportion of pollen depends on duration of visits differently for queens than for workers Part (e)
µ{PollenRemovedLogit|DuratioOfV isitlog,BeeType} = β0 + β1DurationOfV isitlog
+ β2BeeType + β3BeeType ∗ DurationOfV isitlog
lmfit <- lm(PollenRemovedlogit ~ BeeType + DurationOfVisitlog , data=data)
summary(lmfit)
##
## Call:
## lm(formula = PollenRemovedlogit ~ BeeType + DurationOfVisitlog,
## data = data)
##
## Residuals:
## Min 1Q Median 3Q Max ## -1.40852 -0.49627 0.08815 0.43598 1.15562
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.7146 0.3842 -7.065 9.18e-09 ***
## BeeTypeWorker 0.5697 0.2364 2.409 0.0202 *
## DurationOfVisitlog 0.8886 0.1402 6.339 1.07e-07 *** ## —
## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ‘ 1
##
## Residual standard error: 0.652 on 44 degrees of freedom
## Multiple R-squared: 0.6068, Adjusted R-squared: 0.5889
## F-statistic: 33.95 on 2 and 44 DF, p-value: 1.206e-09
The p-value of the BeeTypeWorker coefficient is 0.02 which is significant for a 0.05 level threshold and the estimate of the slope is 0.5697 so this is the amount by which the mean number of pollen proportions exceeds that with bee type queen. And hence YES, there is enough evidence to say that while adjusting for the time spent on flowers, workers remove a larger proportion than queens(alternatively queen removes a smaller proportion than workers) Rearranging the model equation:
µ{PollenRemovedLogit|DuratioOfV isitlog,BeeType} = β0 + β1DurationOfV isitlog
+ (β2 + β3DurationOfV isitlog)BeeType
With the interaction term included, the effect of the indicator variable is now β2 + β3DurationOFV isitlog and hence the difference between the old(0.122) and new p-value(0.02) can be attributed to this difference in the model.
Chapter 10: 20
N
SS(β0,β1 …βn) = X(Yi − β0 − β1X1i − β2X2i − ··· − βpXpi)2
i=1
N
X
(Yi − β0 − β1X1i − β2X2i − ··· − βpXpi) × −1 = 0
i=1
N
nβ0 + β1XX1i + β2XX2i + ··· + βp XXpi = XYi
i=1
N
X
(Yi − β0 − β1X1i − β2X2i − ··· − βpXpi) × −X1i = 0
i=1
N
β0XX1i + β1XX12i + β2XX2iX1i + ··· + βp XXpiX1i = XX1iYi
i=1
Similarly,
N
X
(Yi − β0 − β1X1i − β2X2i − ··· − βpXpi) × −X1i = 0
∂βp i=1
N
β0XXpi + β1XX1iXpi + β2XX2iXpi + ··· + βp XXpi2 = XXpiYi
i=1
To prove that this is indeed the minimum, we need to show that
∂2SS
is convex:
Similarly for any 1 ≤ j ≤ p: ∂2SS
∂2SS XX12i ≥ 0
i
XXji2 ≥ 0
i
And for
k 6= j
:
2

∂ SS X
= 2 XjiXki ≥ 0
∂βjβk
i

 P X12i Pi X1iX2i i
P X
 i 2iX1i Pi X22i
 … …Pi XniX1i … Pi X1iXni 
… Pi X2iXni 
 P X22i … Pi Xni2 i

each element in the hessian matrix in this case can be written as Hij = xiTx גand hence this is a Gram matrix and positive definite, hence minima at the above point is guaranteed.
Chapter 10: 21
Inverse of XTX always exists, unless p + 1 > n that is to say the number of regressors are more than the number of samples. This can still be solved by adding small values to XTX that makes it invertible.