Erin Hodgess wrote:
Dear R People:
Here is a small data frame and two particular formulas:
test.df
y x
1 -0.9261650 1
2 1.5702700 2
3 0.1673920 3
4 0.7893085 4
5 0.3576875 5
6 -1.4620915 6
7 -0.5506215 7
8 -0.3480292 8
9 -1.2344036 9
10 0.8502660 10
summary(lm(exp(y)~x))
Call:
lm(formula = exp(y) ~ x)
Residuals:
Min 1Q Median 3Q Max
-1.6360 -0.6435 -0.4722 0.4215 2.9127
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.1689 0.9782 2.217 0.0574 .
x -0.1368 0.1577 -0.868 0.4108
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.432 on 8 degrees of freedom
Multiple R-squared: 0.08604, Adjusted R-squared: -0.0282
F-statistic: 0.7532 on 1 and 8 DF, p-value: 0.4108
summary(lm(I(y^2)~x))
Call:
lm(formula = I(y^2) ~ x)
Residuals:
Min 1Q Median 3Q Max
-0.9584 -0.6387 -0.2651 0.5754 1.4412
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.10084 0.62428 1.763 0.116
x -0.03813 0.10061 -0.379 0.715
Residual standard error: 0.9138 on 8 degrees of freedom
Multiple R-squared: 0.01764, Adjusted R-squared: -0.1052
F-statistic: 0.1436 on 1 and 8 DF, p-value: 0.7146
These both work just fine.
My question is: when do you know to use I() and just the function of
the variable, please?
I don't think you need I() on the LHS, at least nowadays
> lm(y^2~x)
Call:
lm(formula = y^2 ~ x)
Coefficients:
(Intercept) x
0.8708 -0.7787
> lm(I(y^2)~x)
Call:
lm(formula = I(y^2) ~ x)
Coefficients:
(Intercept) x
0.8708 -0.7787
on the RHS you use I() to prevent special treatment of model formula
operators e.g. (a+b+c)^2 == a:b+b:c+a:c is all 2 factor interactions
between a,b,c whereas I(a+b+c)^2) is a variable which is the squared sum
of a,b, and c.
--
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
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