[R-sig-eco] Logistic regression with 2 categorical predictors

Andrew Halford andrew.halford at gmail.com
Wed Oct 22 02:21:49 CEST 2014


Hi Thierry,

The multiple comparisons ran just fine but there was a ridiculous amount of
interaction combinations all of which were non-significant even though
there was a highly significant interaction term. I decided to remove test
as a variable to simplify the analysis and run separate single explanatory
variable logistic regressions. I have included a result below which is
still producing an outcome I cant explain. Namely, why am I getting such a
significant result for the ANOVA but when I do the tukey tests nothing is
significant?

> sg_habitat
  Age Prefer Avoid
1   1     17    14
2   2     20    10
3   3     14     9
4   4     13    12
5   5      0    18
6   6      0     5

> model_sg <- glm(cbind(Prefer,Avoid) ~ Age, data=sg_habitat,
family=binomial)

> anova(model_sg, test="Chisq")

Analysis of Deviance Table

Model: binomial, link: logit

Response: cbind(Prefer, Avoid)

Terms added sequentially (first to last)


     Df Deviance Resid. Df Resid. Dev  Pr(>Chi)
NULL                     5     36.588
Age   5   36.588         0      0.000 7.243e-07 ***


> mc_sg <- glht(model_sg, mcp(Age = "Tukey"))

> summary(mc_sg)

         Simultaneous Tests for General Linear Hypotheses

Multiple Comparisons of Means: Tukey Contrasts


Fit: glm(formula = cbind(Prefer, Avoid) ~ Age, family = binomial,
    data = sg_habitat)

Linear Hypotheses:
             Estimate Std. Error z value Pr(>|z|)
2 - 1 == 0     0.4990     0.5294   0.943    0.912
3 - 1 == 0     0.2477     0.5593   0.443    0.997
4 - 1 == 0    -0.1141     0.5390  -0.212    1.000
5 - 1 == 0   -25.8473 53178.5362   0.000    1.000
6 - 1 == 0   -24.7307 57729.9299   0.000    1.000
3 - 2 == 0    -0.2513     0.5767  -0.436    0.997
4 - 2 == 0    -0.6131     0.5570  -1.101    0.844
5 - 2 == 0   -26.3463 53178.5362   0.000    1.000
6 - 2 == 0   -25.2296 57729.9299   0.000    1.000
4 - 3 == 0    -0.3618     0.5855  -0.618    0.985
5 - 3 == 0   -26.0950 53178.5362   0.000    1.000
6 - 3 == 0   -24.9783 57729.9299   0.000    1.000
5 - 4 == 0   -25.7332 53178.5362   0.000    1.000
6 - 4 == 0   -24.6165 57729.9299   0.000    1.000
6 - 5 == 0     1.1167 78490.1364   0.000    1.000
(Adjusted p values reported -- single-step method)


On 21 October 2014 22:53, ONKELINX, Thierry <Thierry.ONKELINX at inbo.be>
wrote:

> Hi Andrew,
>
> Please keep the mailing list in cc.
>
> The estimates in mc are the differences of the parameter estimates (betas)
> between both levels. E.g. 5.LR -1.LR = -1.168 or 5.LR = 1.LR - 1.168
>
> summary(mc) should give you the significance of those differences. That
> should work. If it doesn't, please provide more info: at least your code
> and the error message. A small reproducible example is better.
> confint(mc) gives the confidence intervals of the differences.
>
> Best regards,
>
> Thierry
>
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
> Forest
> team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
> Kliniekstraat 25
> 1070 Anderlecht
> Belgium
> + 32 2 525 02 51
> + 32 54 43 61 85
> Thierry.Onkelinx at inbo.be
> www.inbo.be
>
> To call in the statistician after the experiment is done may be no more
> than asking him to perform a post-mortem examination: he may be able to say
> what the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
> Van: Andrew Halford [mailto:andrew.halford at gmail.com]
> Verzonden: dinsdag 21 oktober 2014 16:19
> Aan: ONKELINX, Thierry
> Onderwerp: Re: [R-sig-eco] Logistic regression with 2 categorical
> predictors
>
> Hi Thierry,
> Thanks for the response. I have run your code but it seems you cant call
> the summary function, you just have to call the object on its own i.e. mc.
> The results are I get are below but I am not sure how to interpret these,
> exactly what does the estimate represent? How do I measure significance
> here?
>
>                    Estimate
> 2.LR - 1.LR == 0  1.252e-01
> 3.LR - 1.LR == 0 -5.390e-01
> 4.LR - 1.LR == 0  1.802e-02
> 5.LR - 1.LR == 0 -1.168e+00
> 6.LR - 1.LR == 0 -2.575e+01
> 1.SD - 1.LR == 0  7.411e-02
> 2.SD - 1.LR == 0 -2.408e-01
> 3.SD - 1.LR == 0  2.675e-01
> etc etc
>
> Andy
>
> On 20 October 2014 23:04, ONKELINX, Thierry <Thierry.ONKELINX at inbo.be>
> wrote:
> Dear Andrew,
>
> anova() and summary() test different hypotheses. anova() tests is at least
> one level is different from the others. summary() tests if the coefficient
> is different from zero.
>
> Multiple comparison of different interaction levels is probably the most
> relevant in this case. The easiest way is to make a new variable.
>
> snapper2$inter <- with(snapper2, interaction(age, test))
> model <- glm(cbind(prefer,avoid) ~ 0 + inter, data=snapper2,
> family=binomial)
> library(multcomp)
> mc <- glht(model, mcp(inter = "Tukey))
> summary(mc)
>
> Best regards,
>
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
> Forest
> team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
> Kliniekstraat 25
> 1070 Anderlecht
> Belgium
> + 32 2 525 02 51
> + 32 54 43 61 85
> Thierry.Onkelinx at inbo.be
> www.inbo.be
>
> To call in the statistician after the experiment is done may be no more
> than asking him to perform a post-mortem examination: he may be able to say
> what the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
>
> -----Oorspronkelijk bericht-----
> Van: r-sig-ecology-bounces at r-project.org [mailto:
> r-sig-ecology-bounces at r-project.org] Namens Andrew Halford
> Verzonden: maandag 20 oktober 2014 16:06
> Aan: r-sig-ecology at r-project.org
> Onderwerp: [R-sig-eco] Logistic regression with 2 categorical predictors
>
> Hi Listers,
>
> I am trying to run a logistic regression to look at the effects of
> experiment type and age on the behavior of fish in a choice chamber
> experiment.
>
> I am using the glm approach and would like some advice on how or whether
> to perform contrasts to work out what levels of Factor1 (Age) and Factor 2
> (Test) are significantly different from each other. I have not been able
> to clarify from my reading what is the appropriate approach to take when
> dealing with a significant interaction term. I am also not sure as to how
> one interprets a model when all the coefficients are non-significant but
> the chi-square ANOVA shows a highly significant interaction term.
>
> I have graphed up the data as dot plots and there is definitely evidence
> of changes in proportions in later ages.
>
> I want to provide evidence for when and for which tests there was a
> 'significant' change in behavior.
>
> > snapper2
>    age test prefer avoid
> 1    1   LR     15    14
> 2    1   SD     15    13
> 3    1   SG     17    14
> 4    1   SW     14    14
> 5    2   LR     17    14
> 6    2   SD     16    19
> 7    2   SG     20    10
> 8    2   SW     15    21
> 9    3   LR     10    16
> 10   3   SD     14    10
> 11   3   SG     14     9
> 12   3   SW     13    15
> 13   4   LR     12    11
> 14   4   SD     14    11
> 15   4   SG     13    12
> 16   4   SW     11    14
> 17   5   LR      4    12
> 18   5   SD      8     8
> 19   5   SG      0    18
> 20   5   SW     10     6
> 21   6   LR      0     6
> 22   6   SD      3     4
> 23   6   SG      0     5
> 24   6   SW      5     3
>
>  >
>
> dotplot(age~prefer/avoid,data=snapper2,group=snapper2$test,cex=1.5,pch=19,ylab="age",auto.key=list(space="right",title="Tests"))
>
>
> > out2 <- glm(cbind(prefer,avoid) ~ age*test, data=snapper2,
> family=binomial)
>
> > summary(out2)
>
> Call:
> glm(formula = cbind(prefer, avoid) ~ age * test, family = binomial,
>     data = snapper2)
>
> Deviance Residuals:
>  [1]  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
> 0
>
> Coefficients:
>               Estimate Std. Error z value Pr(>|z|)
> (Intercept)  6.899e-02  3.716e-01   0.186   0.8527
> age2         1.252e-01  5.180e-01   0.242   0.8091
> age3        -5.390e-01  5.483e-01  -0.983   0.3256
> age4         1.802e-02  5.589e-01   0.032   0.9743
> age5        -1.168e+00  6.866e-01  -1.701   0.0890 .
> age6        -2.575e+01  9.348e+04   0.000   0.9998
> testSD       7.411e-02  5.307e-01   0.140   0.8890
> testSG       1.252e-01  5.180e-01   0.242   0.8091
> testSW      -6.899e-02  5.301e-01  -0.130   0.8964
> age2:testSD -4.401e-01  7.260e-01  -0.606   0.5444
> age3:testSD  7.324e-01  7.846e-01   0.933   0.3506
> age4:testSD  8.004e-02  7.863e-01   0.102   0.9189
> age5:testSD  1.024e+00  9.301e-01   1.102   0.2707
> age6:testSD  2.532e+01  9.348e+04   0.000   0.9998
> age2:testSG  3.738e-01  7.407e-01   0.505   0.6138
> age3:testSG  7.867e-01  7.832e-01   1.004   0.3152
> age4:testSG -1.321e-01  7.764e-01  -0.170   0.8649
> age5:testSG -2.568e+01  8.768e+04   0.000   0.9998
> age6:testSG  2.121e-02  1.334e+05   0.000   1.0000
> age2:testSW -4.616e-01  7.249e-01  -0.637   0.5242
> age3:testSW  3.959e-01  7.662e-01   0.517   0.6054
> age4:testSW -2.592e-01  7.858e-01  -0.330   0.7415
> age5:testSW  1.678e+00  9.386e-01   1.788   0.0737 .
> age6:testSW  2.626e+01  9.348e+04   0.000   0.9998
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> (Dispersion parameter for binomial family taken to be 1)
>
>     Null deviance: 5.4908e+01  on 23  degrees of freedom Residual
> deviance: 2.6113e-10  on  0  degrees of freedom
> AIC: 122.73
>
> Number of Fisher Scoring iterations: 23
>
>
> > anova(out2, test="Chisq")
>
> Analysis of Deviance Table
>
> Model: binomial, link: logit
>
> Response: cbind(prefer, avoid)
>
> Terms added sequentially (first to last)
>
>
>          Df Deviance Resid. Df Resid. Dev  Pr(>Chi)
> NULL                        23     54.908
> age       5   11.235        18     43.673 0.0469115 *
> test      3    1.593        15     42.079 0.6608887
> age:test 15   42.079         0      0.000 0.0002185 ***
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> cheers
>
> Andy
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>
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>
>
> --
> Andrew Halford Ph.D
> Research Scientist (Kimberley Marine Parks)|  Adjunct Research Scientist
> (Curtin University)
> Dept. Parks and Wildlife
> Western Australia
>
> Ph: +61 8 9219 9795
> Mobile: +61 (0) 468 419 473
> * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * *
> Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver
> weer en binden het INBO onder geen enkel beding, zolang dit bericht niet
> bevestigd is door een geldig ondertekend document.
> The views expressed in this message and any annex are purely those of the
> writer and may not be regarded as stating an official position of INBO, as
> long as the message is not confirmed by a duly signed document.
>



-- 
Andrew Halford Ph.D
Research Scientist (Kimberley Marine Parks)|  Adjunct Research Scientist
(Curtin University)
Dept. Parks and Wildlife
Western Australia

Ph: +61 8 9219 9795
Mobile: +61 (0) 468 419 473

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