Cangrejos herradura y regresión logıstica múltiple 20 de enero de

Cangrejos herradura y regresión logı́stica múltiple
20 de enero de 2009
Fijamos el parámetro de la primera categorı́a a 0.
> options(contrasts = c("contr.treatment", "contr.poly"))
Leemos los datos.
> table.4.3 = read.table("..//data/crab.txt", col.names = c("C",
+
"S", "W", "Sa", "Wt"))
Definimos como factor la variable que nos indica el color.
> table.4.3$C.fac <- factor(table.4.3$C, levels = c("5", "4", "3",
+
"2"), labels = c("dark", "med-dark", "med", "med-light"))
Consideramos la variable binaria que nos indica si tiene o no satélites.
> (table.4.3$Sa.bin = ifelse(table.4.3$Sa > 0, 1, 0))
[1]
[38]
[75]
[112]
[149]
1
1
0
1
1
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0
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0
1
1
1
1
1
1
0
1
1
1
0
1
1
0
0
0
0
1
1
1
0
0
1
0
0
0
0
1
0
1
1
0
1
0
1
0
0
1
1
1
1
0
1
0
1
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0
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1
1
0
0
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0
0
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1
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0
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0
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1
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1
Ajustamos el modelo.
> crab.fit.logist <- glm(Sa.bin ~ C.fac + W, family = binomial,
+
data = table.4.3)
> summary(crab.fit.logist, cor = F)
Call:
glm(formula = Sa.bin ~ C.fac + W, family = binomial, data = table.4.3)
Deviance Residuals:
Min
1Q
Median
-2.1124 -0.9848
0.5243
3Q
0.8513
Max
2.1413
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept)
-12.7151
2.7617 -4.604 4.14e-06 ***
C.facmed-dark
1.1061
0.5921
1.868
0.0617 .
C.facmed
1.4023
0.5484
2.557
0.0106 *
C.facmed-light
1.3299
0.8525
1.560
0.1188
W
0.4680
0.1055
4.434 9.26e-06 ***
--Signif. codes: 0 ^
a€~***^
a€™ 0.001 ^
a€~**^
a€™ 0.01 ^
a€~*^
a€™ 0.05 ^
a€~.^
a€™ 0.1 ^
a€~ ^
a€™ 1
1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 225.76
Residual deviance: 187.46
AIC: 197.46
on 172
on 168
degrees of freedom
degrees of freedom
Number of Fisher Scoring iterations: 4
Veamos los intervalos de confianza.
> library(MASS)
> confint(crab.fit.logist)
2.5 %
97.5 %
(Intercept)
-18.45674069 -7.5788795
C.facmed-dark
-0.02792233 2.3138635
C.facmed
0.35269965 2.5260703
C.facmed-light -0.27377584 3.1356611
W
0.27128167 0.6870436
Predicciones.
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>
res1 = predict(crab.fit.logist, type = "response", newdata = data.frame(W = seq(18,
34, 1), C.fac = "med-light"))
res2 = predict(crab.fit.logist, type = "response", newdata = data.frame(W = seq(18,
34, 1), C.fac = "med"))
res3 = predict(crab.fit.logist, type = "response", newdata = data.frame(W = seq(18,
34, 1), C.fac = "med-dark"))
res4 = predict(crab.fit.logist, type = "response", newdata = data.frame(W = seq(18,
34, 1), C.fac = "dark"))
plot(seq(18, 34, 1), res1, type = "l", bty = "L", ylab = "Predicted Probability",
axes = F, xlab = expression(paste("Width, ", italic(x), "(cm)")))
axis(2, at = seq(0, 1, 0.2))
axis(1, at = seq(18, 34, 2))
lines(seq(18, 34, 1), res2)
lines(seq(18, 34, 1), res3)
lines(seq(18, 34, 1), res4)
arrows(x0 = 29, res1[25 - 17], x1 = 25, y1 = res1[25 - 17], length = 0.09)
text(x = 29.1, y = res1[25 - 17], "Color 1", adj = c(0, 0))
arrows(x0 = 23, res2[26 - 17], x1 = 26, y1 = res2[26 - 17], length = 0.09)
text(x = 21.1, y = res2[26 - 17], "Color 2", adj = c(0, 0))
arrows(x0 = 28.9, res3[24 - 17], x1 = 24, y1 = res3[24 - 17],
length = 0.09)
text(x = 29, y = res3[24 - 17], "Color 3", adj = c(0, 0))
arrows(x0 = 25.9, res4[23 - 17], x1 = 23, y1 = res4[23 - 17],
length = 0.09)
text(x = 26, y = res4[23 - 17], "Color 4", adj = c(0, 0))
2
1.0
0.8
0.6
Color 1
0.4
Color 3
0.2
Predicted Probability
Color 2
Color 4
18
20
22
24
26
28
30
32
34
Width, x(cm)
Consideramos un modelo con interacción color y anchura del caparazón.
> crab.fit.logist.ia = update(object = crab.fit.logist, formula = ~. +
+
W:C.fac)
> anova(crab.fit.logist, crab.fit.logist.ia, test = "Chisq")
Analysis of Deviance Table
Model 1: Sa.bin ~ C.fac + W
Model 2: Sa.bin ~ C.fac + W + C.fac:W
Resid. Df Resid. Dev Df Deviance P(>|Chi|)
1
168
187.457
2
165
183.081
3
4.376
0.224
No parece razonable considerar interacción.
3