The twelveth assignment of Linear Regression. The assignment is written in Rmarkdown, a smart syntax supported by RStudio helping with formula, plot visualization and plugin codes running.

most recommend: click here for html version of assignment, you can see codes as well as plots.

You may also find the PDF Version of this assignment from github. Or if you can cross the fire wall, just see below:

1

a

x <- seq(-5, 18, length=1000)
y1 <- dnorm(x, mean=5.1, sd=2.8)
y2 <- dnorm(x, mean=6.3, sd=2.8)
y3 <- dnorm(x, mean=7.9, sd=2.8)
y4 <- dnorm(x, mean=9.5, sd=2.8)
plot(x, y1, type="l", lwd=1)
lines(x, y2, type="l", lwd=1)
lines(x, y3, type="l", lwd=1)
lines(x, y4, type="l", lwd=1)
abline(v=5.1)
abline(v=6.3)
abline(v=7.9)
abline(v=9.5)

b

$E(MSE) = \sigma^2 = 7.84 \\ E(MSTR) = \sigma^2 + \frac{\sum _i(\mu_i-\mu )^2}{\gamma -1} \\ =7.84+\frac{100[(5.1-7.2)^2+(6.3-7.2)^2+(7.9-7.2)^2+(9.5-7.2)^2]}{3} \approx 374$

It suggests that the different treatments have substantially impact on Y

c

$\text{Use same equation as b, we have: } E(MSTR) = \sigma^2 + \frac{\sum _i(\mu_i-\mu )^2}{\gamma -1} \\ =7.84+\frac{100[(5.1-7.2)^2+(5.6-7.2)^2+(9-7.2)^2+(9.5-7.2)^2]}{3} \approx 523$

It is because the points distribution are more scattered compared to b.

2

1 2	dat<-read.table("CH16PR10.txt") names(dat)<-c("x","age","num")

a

1	plot(x=dat$age, y=dat$x)

The factor level means seem to differ, at least the middle ge group differs from the other two.
The variability within each factor level seems to be constant.

b

code <- rbind(diag(1,3),-1)
xx <- code[dat$age,]
dat$x1 <- xx[,1]
dat$x2 <- xx[,2]
dat$x3 <- xx[,3]

dat$age <- factor(dat$age)
fit1 <- aov(x~age,data=dat)
yhat <- fitted(fit1)
yhat

c

1 2	resid1 <- resid(fit1) resid1

d

1	summary(fit1)

e

$H_0:\mu_1=\mu_2=\mu_3 \\ H_1: \text{otherwise} \\ F^*=\frac{MSTR}{MSE}$

Reject null hypothesis if F*>$F_{0.99,3,33}$
The p value is 4.769e-12
We reject H_0

f

If seems that middle aged people tend to offer more cash for a used car, while young and old people tend to offer less.

3

a

1 2	fit2 <- lm(x~age,data=dat) summary(fit2)

The model is $Yhat=21.5+6.25X_1-0.0833X_2$
The intercept term estimates the average cell sample mean.

b

1	anova(fit2)

$H_0:\tau_1=\tau_2$

$H_1: otherwise$

$F^*=\frac{MSR}{MSE}$

Reject null hypothesis if F*>$F_{0.99,2,33}$
The p value is 4.769e-12, so We reject H_0

4

b

Young=c(23, 25, 21, 22, 21, 22, 20, 23, 19, 22, 19, 21)
Middle=c(28, 27, 27, 29, 26, 29, 27, 30, 28, 27, 26, 29)
Elderly=c(23, 20, 25, 21, 22, 23, 21, 20, 19, 20, 22, 21)
FactorLevels=c(1,2,3)
n1=length(Young)
n2=length(Middle)
n3=length(Elderly)

MyData=data.frame(
Values=c(Young,Middle,Elderly),
Treatment=c(rep(1,n1),rep(2,n2),rep(3,n3)))

y=MyData$Values
x=factor(MyData$Treatment)
means=tapply(y,x,mean)
n=tapply(y,x,length)
df=sum(n)-2

MSE=2.49
alpha=0.01
l1=means[1]-qt(1-alpha/2,df)*sqrt(MSE/n[1])
u1=means[1]+qt(1-alpha/2,df)*sqrt(MSE/n[1])
print(c(l1,u1))

So the confidence level is (20.2572,22.7428)

c

MSE=2.49
alpha=0.01
l31=means[3]-means[1]-qt(1-alpha/2,df)*sqrt(MSE/n[1]+MSE/n[3])
u31=means[3]-means[1]+qt(1-alpha/2,df)*sqrt(MSE/n[1]+MSE/n[3])
print(c(l31,u31))

This confidence interval contains 0, so we cannot reject the null hypothesis that $\mu_1=\mu_3$

d

MSE=2.49;
alpha=0.01;
lcontrast=-means[1]+2*means[2]-means[3]-qt(1-alpha/2,df)*sqrt(MSE/n[1]+4*MSE/n[2]+MSE/n[3])
ucontrast=-means[1]+2*means[2]-means[3]+qt(1-alpha/2,df)*sqrt(MSE/n[1]+4*MSE/n[2]+MSE/n[3])
print(c(lcontrast,ucontrast))

$H_0:\mu_2-\mu_1=\mu_3-\mu_2$
$H_1: otherwise$
Since the confidence interval for the contrast does not contain 0, we do not reject the null hypothesis.

e

1 2	results<-aov(y~x); TukeyHSD(results)

There is significant difference between young and middle aged people, as well as mlderly and middle aged people. But there is no significant difference between the young and elderly.

f

1	pairwise.t.test(y,x,p.adjust="bonferroni")

The bonferroni method gives the same result. But it won’t be more efficient, since it “overstates” the significnce level.

5

a

This has been done in the previous problem, question d.
The confidence interval is (-15.627664,-9.539003)

b

D_1 = 6.2500, D_2 = −6.3333, D_3 = −0.0833,

L_1 =−12.5833, s{D_i} =0.6442 (i = 1, 2, 3), s{L_1} = 1.1158,

F(0.90, 2, 33) = 2.47, S = 2.223

Then we can obtain the family intervals:

(4.818,7.682)

(−7.765,−4.901)

(−1.515,1.349)

(−15.064,−10.103)