Reputation: 221
I'm trying to compute Pearson correlation for these two below. And it fits a linear line y=1/3*(x)
. When I calculate covariance, I got 4.5 and standard deviation for x and y are 4.74 and 1.58 respectively, which ends up a positive coefficient. However, the slides I have tells me the covariance is -7.5 and the coefficient -1, which confuses me. Who is actually correct on this?
x<-c(-3,-6,0,3,6)
y<-c(1,-2,0,-1,2)
Upvotes: 0
Views: 77
Reputation: 1075
x<-c(-3,-6,0,3,6)
y<-c(1,-2,0,-1,2)
cor(x,y, method = "pearson") # pearson correlation
[1] 0.6
cov(x,y) # covariance
[1] 4.5
you can also compute the covariance and pearson correlation manually to double check:
N <- length(x)
x_ <- mean(x)
y_ <- mean(y)
cova <- 0
for(i in 1:length(x))
{
cova <- cova + ((x[i]-x_)*(y[i]-y_))/(N-1)
}
cova
[1] 4.5
or:
cova <- as.numeric((x-x_)%*%(y-y_)/(N-1))
cova
[1] 4.5
for the pearson correlation:
cova / (sqrt(var(x)*var(y)))
[1] 0.6
Upvotes: 2