Call:
glm(formula = Y1 ~ 0 + x1 + x2 + x3 + x4 + x5, family = quasibinomial(link = cauchit))
Deviance Residuals:
Min 1Q Median 3Q Max
-2.5415 0.2132 0.3988 0.6614 1.8426
Coefficients:
Estimate Std. Error t value Pr(>|t|)
x1 -0.7280 0.3509 -2.075 0.03884 *
x2 -0.9108 0.3491 -2.609 0.00951 **
x3 0.2377 0.1592 1.494 0.13629
x4 -0.2106 0.1573 -1.339 0.18151
x5 3.6982 0.8658 4.271 2.57e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for quasibinomial family taken to be 0.8782731)
Null deviance: 443.61 on 320 degrees of freedom
Residual deviance: 270.17 on 315 degrees of freedom
AIC: NA
Number of Fisher Scoring iterations: 12
Here is the output from glm in R. Do you know a way to pull out Dispersion parameter which is 0.8782731 in this case, instead of just copy and paste. Thanks.
The dispersion parameter φ is traditionally estimated by the maximum likelihood or method of moments approaches. Notably, when the sample variance exceeds the sample mean these methods give fairly good estimates for φ.
Over dispersion can be detected by dividing the residual deviance by the degrees of freedom. If this quotient is much greater than one, the negative binomial distribution should be used. There is no hard cut off of “much larger than one”, but a rule of thumb is 1.10 or greater is considered large.
Dispersion parameter Dispersion (variability/scatter/spread) simply indicates whether a distribution is wide or narrow. The GLM function can use a dispersion parameter to model the variability. However, for likelihood-based model, the dispersion parameter is always fixed to 1.
By default, however, dispersion will be estimated by the summary() method, presumably leading to the value of 1271 you report. The underlying variance function would then be VAR[y] = 1271 * mu + 12.71 * mu^2 > Fitting a quasipoisson model also yields a large dispersion parameter > (1300).
You can extract it from the output of summary
:
data(iris)
mod <- glm((Petal.Length > 5) ~ Sepal.Width, data=iris)
summary(mod)
#
# Call:
# glm(formula = (Petal.Length > 5) ~ Sepal.Width, data = iris)
#
# Deviance Residuals:
# Min 1Q Median 3Q Max
# -0.3176 -0.2856 -0.2714 0.7073 0.7464
#
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.38887 0.26220 1.483 0.140
# Sepal.Width -0.03561 0.08491 -0.419 0.676
#
# (Dispersion parameter for gaussian family taken to be 0.2040818)
#
# Null deviance: 30.240 on 149 degrees of freedom
# Residual deviance: 30.204 on 148 degrees of freedom
# AIC: 191.28
#
# Number of Fisher Scoring iterations: 2
summary(mod)$dispersion
# [1] 0.2040818
The str
function in R is often helpful to solve these sorts of questions. For instance, I looked at str(summary(mod))
to answer the question.
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