quiz1

http://rpubs.com/kiichi/27763

Q1

x <- c(0.18, -1.54, 0.42, 0.95)
w <- c(2, 1, 3, 1)
plot(x,w)

#sum((wi - xi)^2) ... from zero to n
mu<-c(0.0025,1.077,0.1471,0.300)
mu

## [1] 0.0025 1.0770 0.1471 0.3000

sapply(mu,function(muval){
  sum(w*(x-muval)^2)  
})

## [1] 3.863 9.769 3.717 3.880

Q2

y is outcome x is predictor

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)
coef(lm(y~x-1))

##      x 
## 0.8263

Fit line through the origin. Use -1 in the lm function

plot(x,y,xlim=c(-2,2),ylim=c(-2,4))
abline(lm(y~x-1),col='red')
abline(h=0,col="gray")
abline(v=0,col="gray")

mpg as the outcome and weight as the predictor

data(mtcars)
coef(lm(mtcars$mpg~mtcars$wt))

## (Intercept)   mtcars$wt 
##      37.285      -5.344

plot(mtcars$wt,mtcars$mpg)
abline(lm(mtcars$mpg~mtcars$wt),col="red")

Q4

Consider data with an outcome (Y) and a predictor (X). The standard deviation of the predictor is one half that of the outcome. The correlation between the two variables is .5. What value would the slope coefficient for the regression model with Y as the outcome and X as the predictor?

B1=Cor(Y,X)Sd(Y)/Sd(X) Sd(X) is one half of Sd(Y) Let Sd(Y) = 1 Sd(X) = 0.5 Cor(Y,X) = 0.5 B1=0.51/0.5

Q5

Students were given two hard tests and scores were normalized to have empirical mean 0 and variance 1. The correlation between the scores on the two tests was 0.4. What would be the expected score on Quiz 2 for a student who had a normalized score of 1.5 on Quiz 1?

mean 0 and variance 1 will indicate the line goes through the origin (0,0)

Corr(Y,X)=0.4 (correlationship of coefficient)

if x is 1.5 , then y?

1.5*0.4

## [1] 0.6

Q6

Normalize this.

(actual_val_X - mean(x)) / StdDev(x)

x <- c(8.58, 10.46, 9.01, 9.64, 8.86)
xc<-(x-mean(x))/sd(x)
xc

## [1] -0.9719  1.5310 -0.3994  0.4393 -0.5991

Q7

• x - predictor
• y - 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)
coef(lm(y~x))

## (Intercept)           x 
##       1.567      -1.713

plot(x,y,xlim=c(-2,2),ylim=c(-2,2))
abline(lm(y~x),col="red")
abline(h=0,col="gray")
abline(v=0,col="gray")

#centerilize and normalize
xc<-(x-mean(x))/sd(x)
yc<-(y-mean(y))/sd(y)
plot(xc,yc,ylim=c(-2,2),xlim=c(-2,2))
abline(lm(yc~xc),col="blue")
abline(h=0,col="gray")
abline(v=0,col="gray")

Q8

x<-c(-4,-2,2,4)
mean(x)

## [1] 0

y<-c(-6,-1,3,4)
mean(y)

## [1] 0

coef(lm(y~x))

## (Intercept)           x 
##         0.0         1.2

Q9

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)
mu<-c(0.573,0.8,0.36,0.44)
mu

## [1] 0.573 0.800 0.360 0.440

sapply(mu,function(mu){
  sum( (x-mu)^2 )
})

## [1] 0.2540 0.7693 0.7077 0.4309

Q10

Consider taking the slope having fit Y as the outcome and X as the predictor, β1 and the slope from fitting X as the outcome and Y as the predictor, γ1, and dividing the two as β1/γ1. What is this ratio always equal to?

x<-mtcars$wt
y<-mtcars$mpg
coef(lm(y~x))[2]/coef(lm(x~y))[2]

##     x 
## 37.94

var(y)/var(x)

## [1] 37.94