Question 1

Consider the data set given below

x <- c(0.18, -1.54, 0.42, 0.95)

and weights given by

w <- c(2, 1, 3, 1)

Give the value of ?? that minimizes the least squares equation ???ni=1wi(xi?????)2.

x <- c(0.18, -1.54, 0.42, 0.95)
w <- c(2, 1, 3, 1)
mu <- sum(w * x) / sum(w)
mu
## [1] 0.1471429

Question 2

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)

Fit the regression through the origin and get the slope treating y as the outcome and x as the regressor. (Hint, do not center the data since we want regression through the origin, not through the means of the data.)

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)
slope <- lm(y ~ x - 1)
slope
## 
## Call:
## lm(formula = y ~ x - 1)
## 
## Coefficients:
##      x  
## 0.8263

Question 3

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.

data(mtcars)
slope <- lm(mtcars$mpg ~ mtcars$wt)
slope
## 
## Call:
## lm(formula = mtcars$mpg ~ mtcars$wt)
## 
## Coefficients:
## (Intercept)    mtcars$wt  
##      37.285       -5.344

Question 6

Consider the data given by the following

x <- c(8.58, 10.46, 9.01, 9.64, 8.86)

What is the value of the first measurement if x were normalized (to have mean 0 and variance 1)?

x <- c(8.58, 10.46, 9.01, 9.64, 8.86)
norm <- (x - mean(x)) / sd(x)
norm[1]
## [1] -0.9718658

Question 7

Consider the following data set (used above as well). What is the intercept for fitting the model with x as the predictor and y as the outcome?

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)

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)
lm(y ~ x)
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##       1.567       -1.713

Question 8

You know that both the predictor and response have mean 0. What can be said about the intercept when you fit a linear regression?

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)
y <- (y - mean(y)); x <- (x - mean(x))
lm(y ~ x)
## 
## Call:
## lm(formula = y ~ x)
## 
## Coefficients:
## (Intercept)            x  
##   1.003e-16   -1.713e+00

Question 9

Consider the data given by

x <- c(0.8, 0.47, 0.51, 0.73, 0.36, 0.58, 0.57, 0.85, 0.44, 0.42)

What value minimizes the sum of the squared distances between these points and itself?

x <- c(0.8, 0.47, 0.51, 0.73, 0.36, 0.58, 0.57, 0.85, 0.44, 0.42)
mean(x)
## [1] 0.573