To all

I have some simple questions pertaining to weights used in regression.
If the variability of the dependent variable (y) is a function of the magnitude of predictor  
variable (x), can the use of weights give an appropriate answer to the regression parameters
and the std errors?

Assume that y at x=1 and 6 has a standard deviation of 0.1 and at x=11 it is 0.4
Then according to a web page on weighted regression for a calibration curve at
http://member.nifty.ne.jp/mniwa/rev006.htm, I should use 1/(std^2) for each weight.

i.e. for x=1 and 6, w = 100 and x=11, w = 6.25

In R the run is:

>y<-c(1,6,11)
>x<-c(6.7,6.7,6.6)
>w<-c(100,100,6.25)
>reg <-lm(x~y, weight=w)
> summary(reg)

Call:
lm(formula = x ~ y, weights = w)

Residuals:
       1        2        3 
-0.04762  0.09524 -0.19048 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)   
(Intercept)  6.707619   0.025431 263.762  0.00241 **
y           -0.002857   0.005471  -0.522  0.69361   
---
Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 

Residual standard error: 0.2182 on 1 degrees of freedom
Multiple R-Squared: 0.2143,     Adjusted R-squared: -0.5714 
F-statistic: 0.2727 on 1 and 1 DF,  p-value: 0.6936 

Am I using the weight method correctly?
And if so does the Estimated Std. Error for the Intercept and slope make sense?

On another note.  How does one do a regression with the origin fixed at 0?

Merry Christmas

REX





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