1. Consider the data set given below

1.1. Consider the data set given by x = c(0.725,0.429,-0.372 ,0.863).What value of \(\mu\) minimizes \(\Sigma((x - \mu)ˆ2)\)?

x <- c(0.725,0.429,-0.372 ,0.863)
x
## [1]  0.725  0.429 -0.372  0.863
mean(x)
## [1] 0.41125
x <- c(0.725,0.429,-0.372 ,0.863)
mean(x)
## [1] 0.41125
p<-c(0.25, 0.25, 0.25, 0.25)
xbar<-sum(x*p)
minimum<-sum(p*(x-mean(x))^2)
minimum
## [1] 0.2290782
minimum1<-sum(p*(x-0.41125)^2)
minimum1
## [1] 0.2290782

Using the median for \(\mu\).

median(x)
## [1] 0.577
minimummedian<-sum(p*(x-median(x))^2)
minimummedian
## [1] 0.2565512

As observe in the result above, its the value of the mean that provides the smallest result of $((x - )ˆ2).

Answer the following questions:

Question 1

Try other value other than the value of the mean and the median, is the result smaller than using the mean in the result of \(\Sigma((x - \mu)ˆ2)\)?

1.2. Reconsider the previous question. Suppose that weights were given, w = c(2, 2, 1, 1) so that we wanted to minimize \(\Sigma(w * (x - mu) ˆ 2)\) for \(\mu\). What value would we obtain?

x <- c(0.725,0.429,-0.372 ,0.863)
w = c(2, 2, 1, 1)
p = c(2/6, 2/6, 1/6, 1/6)
xbar<-sum(x * w) / sum(w)
xbar
## [1] 0.4665
xbarorbs<-c(0.577, 0.577,0.577, 0.577)
minimummean<-sum(p*(x-xbar)^2)
minimummean
## [1] 0.1661253

Using the value of the median

median(x)
## [1] 0.577
minimummedian<-sum(p*(x-median(x))^2)
minimummedian
## [1] 0.1783355

Verify

minimummedian1<-sum(p*(x-xbarorbs)^2)
minimummedian1
## [1] 0.1783355

Using the mean

minimummean<-sum(p*(x-xbar)^2)
minimummean
## [1] 0.1661253

Question 2

Consider the data set given by x = c(0.18, -1.54, 0.42, 0.95).What is the value of \(\Sigma((x - \mu)ˆ2)\)? (Note the value of \(\mu\) is the one that minimizes the result of \(\Sigma((x - \mu)ˆ2)\).

  1. Consider the following data set
x<-c(0.8, 0.47, 0.51, 0.73, 0.36, 0.58, 0.57, 0.85, 0.44, 0.42)
y<-c(1.39, 0.72, 1.55, 0.48, 1.19, -1.59, 1.23, -0.65, 1.49, 0.05)
z<-data.frame(x,y)

# Fit the regression through the origin and get the slope treating y as the outcome and x is the regressor.
fit2 <- lm(z$y~z$x)
fit2
## 
## Call:
## lm(formula = z$y ~ z$x)
## 
## Coefficients:
## (Intercept)          z$x  
##       1.567       -1.713

Answer: The expected slope is the coefficient on x, i.e. -1.713.

Question 3

  1. Do data(mtcars) from the datasets package and fit the regression model with mpg as the outcome and weight as the predictor. Give the slope coefficient.