beetle2 wrote:
HI Guys,
I know that this forum is not for homework but I am trying to interpret R
output code.
I was just wondering if someone might be able to help.
Well, if homework, we can only give hints.
I have been given the following.
For (X1,X2) distributed bivariate normal with parameters
mu1 = 5.8
mu2 = 5.3
sd1 = sd2 = 0.2
and p = 0.6
That's Greek letter rho, not p, I expect.
The r-code and inpit/output are as follows
input
m <- 5.3 + 0.6*(6.3 - 5.8)
s <- sqrt(0.2^2*(1-0.6^2))
q <- seq(5.12,6.08,0.16)
print(rbind(q,pnorm(mean=m,sd=sd=s,q=q)))
output
q 5.1200 5.280 5.44 5.6 5.76 5.926.1
0.0013 0.023 0.16 0.5 0.84 0.981
I have been asked to interpret
E[X2|X1 = 6.3] and varE[X2|X1 = 6.3]
I take it that s<- = 0.16 is the standard variation
Standard _deviation_
So I am assuming that varE[X2|X1 = 6.3] = 0.16^2 = .0256
m <- 5.3 + 0.6*(6.3 - 5.8) = 5.6this the Expected value of E[X+Y]
What makes you think that? E[X+Y] is mu1+mu2.
The m calculation could also have been written
x1 <- 6.3
m <- 5.3 + 0.6*(x1 - 5.8)
Now go back to your text book and read up on the formulas that connect
correlation and regression coefficients.
I see from the output that this would be correct because the probability of
5.6 = 0.5
to interpret E[X2|X1 = 6.3] I can't see it in the output. And I'm not sure
how to find the conditional probabilty from the output.
Any help would be greatly appreciated
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
