NTHU STAT 5410, 2019 Solution to Homework 7 Linear Models...

Linear Models Homework 7

1.

(i)

year model: temp~year year

year model: temp~year + (year)2 year year

2

linear trend

NTHU STAT 5410, 2019 Solution to Homework 7

made by 蔡明諺, 邱奕豪, 黃俊閔助教

(ii)

AR(1) year 95% 0 temp

(iii)

year

(

orthogonality )



2020 :

(iv)

1930 1930 year 1930

linear trend



1930 fit model year

reject claim



(v)

R-square cubic spline



Plot the fit in comparison to the previous fits.

cubic spline (indicate some

fluctuation in temperature over time) model

extrapolation



2.

:

data <- read.table('http://www.stat.nthu.edu.tw/~swcheng/Teaching/sta

t5410/data/assess.txt', header = T)

1981 1982 :

plot(x=NULL,y=NULL,xlim=c(31,94),ylim=c(812,2013))

lines(sort(data$P),data$y1982[order(data$P)],type='b',col='green')

lines(sort(data$P),data$y1981[order(data$P)],type='l',col='blue')

points(data$P[1:13],data$y1981[1:13],pch='L')

points(data$P[14:29],data$y1981[14:29],pch='S')

points(data$P[44:60],data$y1981[44:60],pch='e')

points(data$P[30:43],data$y1981[30:43],pch='c')

legend("topleft", legend = c("1981", "1982"),lty=c(1,1),col = c('blue

','green'))

:

1. 1981 1982 1982 1981

2. P Sibley Meeker

3.



:

par(mfrow=c(2,2))

> plot(data$P[1:13],data$y1981[1:13],type='b',main=data$County[1])




.

response weight ,

Sample variance

P score County predictor weighted

least square (

P score County interaction)

y = (data$y1981+data$y1982)/2

var = apply(cbind(data$y1981,data$y1982),1,var)

model1 = lm(y~data$P+data$County+data$P:data$County,weights = 1/var)

summary(model1)



:

= 1143.629 + 4.335 + 782.549 349.483 +

366.654 8.736 + 4.366 1.623

0.05 F p-value 0.05

correlation:

cor(model.matrix(model1)[,-1])

P McLeod McLeod 0.998 P Meeker

Meeker 0.934 P Sibley

Sibley 0.997



fit :

model2 = lm(y~data$P+data$County,weights = 1/var)

anova(model1,model2)

0.05

summary(model2)

:

= 898.438 + 8.018 + 55.592 72.002 + 151.665



plot(data$P,model2$fitted.values,cex=0.05)

points(data$P[1:13],model2$fitted.values[1:13],pch='L')

points(data$P[14:29],model2$fitted.values[14:29],pch='s')

points(data$P[30:43],model2$fitted.values[30:43],pch='c')

points(data$P[44:60],model2$fitted.values[44:60],pch='e')

Sibley McLeod LeSueur

Meeker

> 898.438+(0+55.592-72.002+151.665)/4

932.2518

=



.

1981 1982 1982 1981

:

> round(data$y1982/data$y1981, 2)

1.10 ~ 1.2

P score County predictor

:

0.05 F p-value 0.05

fit :

model5 = lm(y~data$P+data$County)

anova(model5,model4)

0.05



> summary(model5)

:

= 1082.564 + 5.445 + 70.564 83.446 + 187.916

> 1082.564+(0+70.564-83.446+187.916)/4

1126.322

= 1126.322 + 5.445

y

Y response

* Y

Y



3.

> data=read.table("C:/Users/Desktop/E2.7.txt",header=T)

> DAO=diff(data$AO)

> DGNP=diff(data$GNP);DCP=diff(data$CP);DOP=diff(data$OP)

DAO response DGNP, DCP and DOP

> lm3=lm(DAO~DGNP+DCP+DOP)

transformation

> library(MASS)

> b=boxcox(lm3,plotit=T,lambda=seq(-1,1,0.1))

> b$x[which.max(b$y)

maximum -0.2525253 95% [-1 , 0.7]

-0.2525253 transformation response

( ) response log

> logDAO=log(DAO)

> lm3new=lm(logDAO~DGNP+DCP+DOP)



transformation

> summary(lm3)

> summary(lm3new)

coefficients DOP DGNP

QQ-plot

studentized residuals QQ-plot

> lm3r=rstandard(lm3)

> qqnorm(lm3r) # draw the normal probabi


> abline(0,1)

> shapiro.test(lm3r)

> lm3newr=rstandard(lm3new)

> qqnorm(lm3newr) # draw the normal probabi

> abline(0,1)



> shapiro.test(lm3newr)

Transformation response normal

3.

> data=read.table("C:/Users/Desktop/E2.7.txt",header=T)

> DAO=diff(data$AO)

> DGNP=diff(data$GNP);DCP=diff(data$CP);DOP=diff(data$OP)

DAO response DGNP, DCP and DOP

> lm3=lm(DAO~DGNP+DCP+DOP)

transformation

> library(MASS)

> b=boxcox(lm3,plotit=T,lambda=seq(-1,1,0.1))



> b$x[which.max(b$y)

maximum -0.2525253 95% [-1 , 0.7]

-0.2525253 transformation response

( ) response log

> logDAO=log(DAO)

> lm3new=lm(logDAO~DGNP+DCP+DOP)

transformation

> summary(lm3)



> summary(lm3new)

coefficients DOP DGNP

QQ-plot

studentized residuals QQ-plot

> lm3r=rstandard(lm3)



> abline(0,1)

> shapiro.test(lm3r)

> lm3newr=rstandard(lm3new)

> qqnorm(lm3newr) # draw the normal probabi

> abline(0,1)

> shapiro.test(lm3newr)



Transformation response normal



NTHU STAT 5410, 2019 Solution to Homework 7 Linear Models...

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