# [R] Test individual slope for each factor level in ANCOVA

Fox, John jfox at mcmaster.ca
Thu Mar 16 13:02:16 CET 2017

```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
Web: http://socserv.mcmaster.ca/jfox/

On 2017-03-15, 9:43 PM, "R-help on behalf of li li"
<r-help-bounces at r-project.org on behalf of hannah.hlx at 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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