Dear Hanna,
You can test the slope in each non-reference group as a linear hypothesis.
You didn’t make the data available for your example, so here’s an example
using the linearHypothesis() function in the car package with the Moore
data set in the same package:
- - - snip - - -
> library(car)
> mod <- lm(conformity ~ fscore*partner.status, data=Moore)
> summary(mod)
Call:
lm(formula = conformity ~ fscore * partner.status, data = Moore)
Residuals:
Min 1Q Median 3Q Max
-7.5296 -2.5984 -0.4473 2.0994 12.4704
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 20.79348 3.26273 6.373 1.27e-07 ***
fscore -0.15110 0.07171 -2.107 0.04127 *
partner.statuslow -15.53408 4.40045 -3.530 0.00104 **
fscore:partner.statuslow 0.26110 0.09700 2.692 0.01024 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.562 on 41 degrees of freedom
Multiple R-squared: 0.2942, Adjusted R-squared: 0.2426
F-statistic: 5.698 on 3 and 41 DF, p-value: 0.002347
> linearHypothesis(mod, "fscore + fscore:partner.statuslow")
Linear hypothesis test
Hypothesis:
fscore + fscore:partner.statuslow = 0
Model 1: restricted model
Model 2: conformity ~ fscore * partner.status
Res.Df RSS Df Sum of Sq F Pr(>F)
1 42 912.45
2 41 853.42 1 59.037 2.8363 0.09976 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
- - - snip - - -
In this case, there are just two levels for partner.status, but for a
multi-level factor you can simply perform more than one test.
I hope this helps,
John
-------------------------------------
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
Web: http://socserv.mcmaster.ca/jfox/
On 2017-03-15, 9:43 PM, "R-help on behalf of li li"
<r-help-boun...@r-project.org on behalf of hannah....@gmail.com> wrote:
>Hi all,
> Consider the data set where there are a continuous response variable, a
>continuous predictor "weeks" and a categorical variable "region" with five
>levels "a", "b", "c",
>"d", "e".
> I fit the ANCOVA model as follows. Here the reference level is region
>"a"
>and there are 4 dummy variables. The interaction terms (in red below)
>represent the slope
>difference between each region and the baseline region "a" and the
>corresponding p-value is for testing whether this slope difference is
>zero.
>Is there a way to directly test whether the slope corresponding to each
>individual factor level is 0 or not, instead of testing the slope
>difference from the baseline level?
> Thanks very much.
> Hanna
>
>
>
>
>
>
>> mod <- lm(response ~ weeks*region,data)> summary(mod)
>Call:
>lm(formula = response ~ weeks * region, data = data)
>
>Residuals:
> Min 1Q Median 3Q Max
>-0.19228 -0.07433 -0.01283 0.04439 0.24544
>
>Coefficients:
> Estimate Std. Error t value Pr(>|t|)
>(Intercept) 1.2105556 0.0954567 12.682 1.2e-14 ***
>weeks -0.0213333 0.0147293 -1.448 0.156
>regionb -0.0257778 0.1349962 -0.191 0.850
>regionc -0.0344444 0.1349962 -0.255 0.800
>regiond -0.0754444 0.1349962 -0.559 0.580
>regione -0.1482222 0.1349962 -1.098 0.280 weeks:regionb
>-0.0007222 0.0208304 -0.035 0.973
>weeks:regionc -0.0017778 0.0208304 -0.085 0.932
>weeks:regiond 0.0030000 0.0208304 0.144 0.886
>weeks:regione 0.0301667 0.0208304 1.448 0.156 ---
>Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
>Residual standard error: 0.1082 on 35 degrees of freedom
>Multiple R-squared: 0.2678, Adjusted R-squared: 0.07946
>F-statistic: 1.422 on 9 and 35 DF, p-value: 0.2165
>
> [[alternative HTML version deleted]]
>
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