# Re: [R] Asking, are simple effects different from 0

From: Chuck Cleland <ccleland_at_optonline.net>
Date: Wed, 05 Mar 2008 09:01:36 -0500

On 3/4/2008 2:45 PM, Jarrett Byrnes wrote:
> Hello, R-i-zens. I'm working on an data set with a factorial ANOVA
> that has a significant interaction. I'm interested in seeing whether
> the simple effects are different from 0, and I'm pondering how to do
> this. So, I have
>
> my.anova<-lm(response ~ trtA*trtB)
>
> The output for which gives me a number of coefficients and whether
> they are different from 0. However, I want the simple effects only,
> incorporating their intercepts, with their error estimates. Can I
> somehow manipulate this object to get that? Or, would a good shortcut
> be
>
> my.simple.anova<-lm(response ~ trtA:trtB + 0)
>
> and then use those coefficients and their error estimates?

fm <- lm(breaks ~ tension*wool, data=warpbreaks)

# names(coef(fm))
# (Intercept) tensionM tensionH woolB tensionM:woolB tensionH:woolB

cm <- rbind(

```"A vs. B at L" = c(0, 0, 0,-1, 0, 0),
"A vs. B at M" = c(0, 0, 0,-1,-1, 0),
"A vs. B at H" = c(0, 0, 0,-1, 0,-1),
"M vs. L at A" = c(0, 1, 0, 0, 0, 0),
"M vs. H at A" = c(0, 1,-1, 0, 0, 0),
"L vs. H at A" = c(0, 0,-1, 0, 0, 0),
"M vs. L at B" = c(0, 1, 0, 0, 1, 0),
"M vs. H at B" = c(0, 1,-1, 0, 1,-1),
"L vs. H at B" = c(0, 0,-1, 0, 0,-1))

```

library(multcomp)

summary(glht(fm, linfct = cm), test = adjusted(type="none"))

Simultaneous Tests for General Linear Hypotheses

Fit: lm(formula = breaks ~ tension * wool, data = warpbreaks)

Linear Hypotheses:

```                   Estimate Std. Error t value  p value
A vs. B at L == 0  16.3333     5.1573   3.167 0.002677 **
A vs. B at M == 0  -4.7778     5.1573  -0.926 0.358867
A vs. B at H == 0   5.7778     5.1573   1.120 0.268156
M vs. L at A == 0 -20.5556     5.1573  -3.986 0.000228 ***
M vs. H at A == 0  -0.5556     5.1573  -0.108 0.914665
L vs. H at A == 0  20.0000     5.1573   3.878 0.000320 ***
M vs. L at B == 0   0.5556     5.1573   0.108 0.914665
M vs. H at B == 0  10.0000     5.1573   1.939 0.058392 .
L vs. H at B == 0   9.4444     5.1573   1.831 0.073270 .
```
```---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Adjusted p values reported -- none method)

> If so, I realize that R gives me t tests for each coefficient.  Now,

> for those I know I'm using the residual degrees of freedom.  Would it
> then be more appropriate to use those, all with the same residual DF
> and apply a bonferroni correction, or, use the mean and SE estimate
> with the sample size for that particular treatment and perform an
> uncorrected one sample t-test to see if the value is different from 0?

I won't comment on whether to adjust and if so how, but you can
implement various adjustments when summarizing.  For example:

summary(glht(fm, linfct = cm), test = adjusted(type="bonferroni"))

Simultaneous Tests for General Linear Hypotheses

Fit: lm(formula = breaks ~ tension * wool, data = warpbreaks)

Linear Hypotheses:
Estimate Std. Error t value p value
A vs. B at L == 0  16.3333     5.1573   3.167 0.02409 *
A vs. B at M == 0  -4.7778     5.1573  -0.926 1.00000
A vs. B at H == 0   5.7778     5.1573   1.120 1.00000
M vs. L at A == 0 -20.5556     5.1573  -3.986 0.00205 **
M vs. H at A == 0  -0.5556     5.1573  -0.108 1.00000
L vs. H at A == 0  20.0000     5.1573   3.878 0.00288 **
M vs. L at B == 0   0.5556     5.1573   0.108 1.00000
M vs. H at B == 0  10.0000     5.1573   1.939 0.52553
L vs. H at B == 0   9.4444     5.1573   1.831 0.65943
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Adjusted p values reported -- bonferroni method)

> Sorry for the bonehead question, but it's a situation I haven't seen

> before.
>
> -Jarrett
>
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--
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
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Received on Wed 05 Mar 2008 - 14:07:09 GMT

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