R help - Oct 2010 - Variable name as string

Hello,

from Verzani, simpleR (pdf), p. 80, I created the following function to
test the coefficient of lm() against an arbitrary value.

coeff.test <- function(lm.result, var, coeffname, value) {
  # null hypothesis: coeff = value
  # alternative hypothesis: coeff != value
  es <- resid(lm.result)
  coeff <- (coefficients(lm.result))[[coeffname]]
  # degrees of freedom = length(var) - number of coefficients?
  n <- df.residual(lm.result) 
  s <- sqrt( sum( es^2 ) / n )
  SE <- s/sqrt(sum((var - mean(var))^2))
  t <- (coeff - value )/SE
  2 * pt(t,n,lower.tail=FALSE) # times two because problem is two-sided
}

E.g. it needs to be called coeff.test(lm(N ~ D), D, "D", 70) if I want
to test the probability slope of the linear model being 70.

Is there a more elegant way to avoid passing D and "D" both as
parameters?

Also, as a non-professional, I would like to know whether the function
is valid for all coefficients of lm(), e.g. coeff.test(lm(N ~ D + H), H,
"H", 70). I am aware that Verzani gives a different formula for
testing
the intercept.

Thanks!
	Jan Rheinl?nder

You can get the name of var by doing 

coeffname <- deparse(substitute(var))

Is that what you wanted?

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Jan private
> Sent: Saturday, October 16, 2010 9:41 PM
> To: r-help
> Subject: [R] Variable name as string
> 
> Hello,
> 
> from Verzani, simpleR (pdf), p. 80, I created the following function to
> test the coefficient of lm() against an arbitrary value.
> 
> coeff.test <- function(lm.result, var, coeffname, value) {
>   # null hypothesis: coeff = value
>   # alternative hypothesis: coeff != value
>   es <- resid(lm.result)
>   coeff <- (coefficients(lm.result))[[coeffname]]
>   # degrees of freedom = length(var) - number of coefficients?
>   n <- df.residual(lm.result)
>   s <- sqrt( sum( es^2 ) / n )
>   SE <- s/sqrt(sum((var - mean(var))^2))
>   t <- (coeff - value )/SE
>   2 * pt(t,n,lower.tail=FALSE) # times two because problem is two-sided
> }
> 
> E.g. it needs to be called coeff.test(lm(N ~ D), D, "D", 70) if I
want
> to test the probability slope of the linear model being 70.
> 
> Is there a more elegant way to avoid passing D and "D" both as
> parameters?
> 
> Also, as a non-professional, I would like to know whether the function
> is valid for all coefficients of lm(), e.g. coeff.test(lm(N ~ D + H),
> H,
> "H", 70). I am aware that Verzani gives a different formula for
testing
> the intercept.
> 
> Thanks!
> 	Jan Rheinl?nder
> 
> ______________________________________________
> R-help at 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.

> coeffname <- deparse(substitute(var))
> 
> Is that what you wanted?
> 
Yes! That gets rid of the extra parameter.

Thanks

On 10/17/2010 05:41 AM, Jan private wrote:
> Also, as a non-professional, I would like to know whether the function
> is valid for all coefficients of lm(), e.g. coeff.test(lm(N ~ D + H), H,
> "H", 70). I am aware that Verzani gives a different formula for
testing
> the intercept.
In a word, no, it isn't. Your SE formula only holds when "var" is
the
only predictor. It would be more general to extract the SE from
coefficients(summary(lm.result)) (which also avoids having to pass var
as an argument).

-- 
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com

So here is the next version.

Why does the intercept needs lower.tail=TRUE to give the same result as
summary() for value=0?

# See Verzani, simpleR (pdf), p. 80
coeff.test <- function(lm.result, idx, value) {
  # idx = 1 is the intercept, idx>1 the other coefficients
  # null hypothesis: coeff = value
  # alternative hypothesis: coeff != value
  coeff <- coefficients(lm.result)[idx]
  SE <- coefficients(summary(lm.result))[idx,"Std. Error"]
  n <- df.residual(lm.result)
  t <- (coeff - value )/SE
  if (idx == 1) {
    2 * pt(t,n,lower.tail=TRUE) # times two because problem is two-sided
  } else {
    2 * pt(t,n,lower.tail=FALSE)
  }
}