Question

Please include all code from R that was used to calculate everything.

Problem 2 (0.5 x 3 = 1.5 point ) Simulate a sample of y1, ..., 4100 from a simple linear model Y = 1 + 2x + €, where € ~ N(0,62), and x is an arithmetic sequence from 1 to 100, with a step size of 1. Run set.seed(1) to set the seed of R's random number generator so that the simulation can be reproduced. • Make a scatter plot, and fit the data with a regression line. • Perform a two-sided significance test that Bi = 2 versus the alternative that 81 + 2, following the procedure of seven steps in the lecture note (choose a significance level of 0.05). • Replicate the above simulation for 100 times, each of which is with different seed numbers from 1 to 100. Record the Bo and 31 for each replication, so that you have 100 realizations of Bo and 31, respectively. What can you say about the correlation between 2, and Ŝ,?

Answer 1

set.seed(1)
library(ggplot2)
epsilon=rnorm(100,0,6)
x=c(1:100)
y=1+2*x + epsilon
data=data.frame(x,y)

fit1=lm(y~x,data=data)
fit1
fit1$coefficients
ggplot(data , aes(x = x, y = y)) + geom_point(size = 0.8) + stat_smooth(method = 'lm', formula = y ~ x)
summary(fit1)

# t statistic to check b1=2
#from summary we found, standard error of b1 = 0.02001
se=0.02001
b1=2
n=100
t=(as.numeric(fit1$coefficients[2])-2)/se
p_value =1-pt(abs(t),n-1)
# value comes out to be 0.403835 => We fail to reject null hypothesis that b1=2 and conclude that b1=2

#-------------------------
b00=c()
b11=c()
for(i in 1:100){
set.seed(i)
x1=c(1:100)
e=rnorm(100,0,6)
y1=1+2*x1 + e
data1=data.frame(x1,y1)
fit2=lm(y1~x1,data=data1)
b00=c(b00,as.numeric(fit2$coefficients[1]))
b11=c(b11,as.numeric(fit2$coefficients[2]))
}
cor(b00,b11)
#correlation comes out to be -0.892052

200 150 >100- 50- 0- 100 50 75 X -25