Skip to content

fixed effects/log transformations question

2 messages · Gus Jespersen, Ben Bolker

#
Dr. Bolker,
Thanks for the response.  One last question with regard to negative t
values when using log10 transformed data.  I am assuming the correct
interpretation of the following output is: if the t value is negative
and you're using log10 data, to get the fixed effect CI, you must add
your own negative sign to 10^(est.+1/96*SE), such that the
backtransformed CI from the output below would be:

([1] "95 % REML Confidence interval"
[1] -0.58261813  0.02578124

becomes

-.295       -1.05


Is this correct,

Thanks again for the help
Gus

[1] "###############NH4 Results Year Two##################"
Data: data.sub
Models:
Mod.NH4.2.2: log10(NH4Nyeartwo) ~ 1 + (1 | pr)
Mod.NH4.2.1: log10(NH4Nyeartwo) ~ 1 + sitett + (1 | pr)
            Df    AIC    BIC   logLik  Chisq Chi Df Pr(>Chisq)
Mod.NH4.2.2  3 26.427 29.700 -10.2136
Mod.NH4.2.1  4 25.243 29.607  -8.6216 3.1841      1    0.07436 .
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
Linear mixed model fit by REML
Formula: log10(NH4Nyeartwo) ~ 1 + sitett + (1 | pr)
   Data: data.sub
   AIC   BIC logLik deviance REMLdev
 30.37 34.73 -11.18    17.24   22.37
Random effects:
 Groups   Name        Variance Std.Dev.
 pr       (Intercept) 0.010942 0.10460
 Residual             0.130473 0.36121
Number of obs: 22, groups: pr, 12

Fixed effects:
                                                    Estimate Std. Error t value
(Intercept)                                    0.7305     0.1086   6.729
sitettToeAdditionsTreatment  -0.2784     0.1552  -1.794

[1] "95 % REML Confidence interval"
[1] -0.58261813  0.02578124

Gus Jespersen <jesper <at> u.washington.edu> writes:
[snip]
It is very likely that the glht function is having trouble
with the spaces in your level names.  I would strongly suggest
that you reformulate them as legal R variable names: something like

     levels(mydata$myfactor) <- make.names(levels(mydata$myfactor))
should work.
Yes, you are.  You need to back-transform the confidence intervals,
not the elements and their standard errors.  The basic math you
are thinking about is

  10^(est+1.96*stderr) !== 10^est+10^(1.96*stderr)

If you back-transform first, the RHS is what you will be doing;
you want the LHS (this is for the upper CI, the obvious parallel
applies for the lower CI)
#
Gus Jespersen <jesper <at> u.washington.edu> writes:
No, but the interpretation is a little bit subtle.  Here you
are working with (as far as I can tell) the back-transformed
confidence intervals on the effect of the treatment.
10^{-0.5826,0.02578} is {0.26,1.06} (where did you get 0.295??); 
this says that the lower CI is that the proportional effect of
the treatment is to multiply by 0.26 (a 74% decrease); the upper
CI is a 6% increase (you can subtract 1 from the CI values if you
want to get it in terms of proportional changes).
  If you were using the natural log (log_e) rather than the log10
scale, then you could interpret *small* (near zero) parameters as
being approximately equivalent to proportional changes (without
back-transforming), because exp(x)-1 is approximately x when
x is small ...

  For what it's worth, this isn't an R question, or a mixed-model
question, any more, it's become a general statistical question -- you
might try asking similar questions on http://stats.stackexchange.com
...