Download presentation

Presentation is loading. Please wait.

Published byChristian McElroy Modified over 3 years ago

1
Questions: Spatial level How does abundance of small rodents relate with altitude and latitude? Climatic level How does abundance of small rodents relate with climate? Abundance of small mammals along an altitudinal - latitudinal gradient in the Central Andes of Argentina

2
We explore the data in a graphical way xyplot(ABUND~ALT|lat,groups=lat,type="b",lwd=3) ABUNDANCEABUNDANCE ALTITUDE 35°LAT 32°LAT 34°LAT 33°LAT Spatial level

3
ABUNDABUND lat a=with(an,tapply(ABUND,list(ALT,lat),mean)) barplot(a,beside=TRUE) Bar plot of abundance variation per altitude for each latitude Abundance peaks at 2300m. We observed a quadratic relationship. Spatial level

4
We rescaled the latitude and altitude axes: alt=((ALT-1300)/1000) # to standardize the variable lt=(lat-32) We transformed abundance to log, to get a better fit to a normal distribution. ab=log(ABUND) We performed a linear model: Spatial Model =lm(ab~lt+alt+I(lt^2)+I(alt^2)) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) e-05 *** lt * alt ** I(lt^2) * I(alt^2) ** Signif. codes: 0 *** ** 0.01 * Residual standard error: on 15 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 4 and 15 DF, p-value: Spatial level

5
Lollipop3d graph showing the relationship between log-abundance and the spatial predictor variables, altitude and latitude. Log(abund)=latitude+altitude +latitude 2 +altitude 2 Abundance shows a hump shape pattern along altitude (peak= 2300m) and latitude (peak= 33° S). Spatial level

6
Climatic variables: we reduced the variability of all precipitation variables into one PCA (Prec) and all temperature variables into one PCA (Temp). Temperature seasonality accounts for 94% of the variation in PCA (Temp). Annual mean precipitation (53%), Precipitation of the wettest season (20%) and Precipitation of the coldest season (20%) accounts for 93% of the variation in PCA (Prec). PCA Log abundance has a quadratic relationship with the climatic variables. Climatic level

7
We performed a Linear model: Climatic model=lm(log(ABUND)~PCA_Temp+PCA_Prec+I(PCA_Temp^2)+I(PCA_Prec^2)) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 4.805e e e-11 *** PCA_Temp e e PCA_Prec 1.117e e I(PCA_Temp 2 ) e e * I(PCA_Prec 2 ) e e Signif. codes: 0 *** ** 0.01 * Residual standard error: on 15 degrees of freedom Multiple R-squared: 0.512, Adjusted R-squared: F-statistic: on 4 and 15 DF, p-value: Climatic level

8
Lollipop3d graph showing the relationship between log-abundance and the climatic predictor variables, PCA (Temp), PCA (Prec). Climatic level Temperature Seasonality Abundance is maximum at intermediate values of temperature and precipitation. Annual mean precipitation Precipitation of the wettest season Precipitation of the coldest season

9
AIC df dAIC Weight Spatial model Climatic model We compared the spatial and climatic model using the Akaikes Information Criterion The spatial fits better than the climatic model. Apparently altitude variability is composed by more than climatic variables. Final remarks Conclusion: How does abundance of small rodents relate with altitude and latitude? In a quadratic relationship, with a peak at intermediate altitudes and latitudes. How does abundance of small rodents relate with climate? In a quadratic relationship peaking at intermediate Temperature Seasonality and Precipitation.

10
Thank you

11
Richness #con regresiones lineales tengo parametros significativos?? porq? #cuando hago la permutacion me da valores grandes de p, esto es que el modelo no es significativamente diferente del azar (permutado) summary(glm(RICH~Apreci,family=quasipoisson))#hipotesis de temperatura, utiliza quasipoisson ya que para este modelo la varianza esta sobreestimada para una dist poisson names(an) #la devianza de los residuos tiene ser similar a los grados de libertad q=an #an es el nombre original de los datos; q= es el nuevo dataset q$RICH=sample(q$RICH) #extrae la columna response (abundancia) m1=glm(RICH~Apreci,family=quasipoisson,data=q) summary(m1) coef(summary(m1))[,3] results=matrix(ncol=2,nrow=1000) for(i in 1:1000){ q$RICH=sample(q$RICH) m1=glm(RICH~Apreci,family=quasipoisson,data=q) summary(m1) tstats=coef(summary(m1))[,3] results[i,]=tstats cat(i,"\n") } mo2<-glm(RICH~Apreci,family=quasipoisson) obs=coef(summary(mo2))[,3] sum(abs(results[,1])>abs(obs[1]))/1000 #en esta comparacion me da el valor de p de la primera columna (intercept), tengo que hacerlo para las 6 que tenemos en la regresion sum(abs(results[,2])>abs(obs[2]))/1000 names(an) an1=an[,c(5,6,15,16,17,18,19,20,21,22,23)] pairs(an1,panel=panel.smooth)

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google