When to use it
Null hypothesis
How the test works
Assumptions
See the Handbook for information on these topics.
Examples
The Mpi example is shown below in the “How to do the test” section.
Graphing the results
Similar tests
See the Handbook for information on these topics.
Logistic regression can be performed in R with the glm (generalized linear model) function. This function uses a link function to determine which kind of model to use, such as logistic, probit, or poisson. These are indicated in the familyand link options. See ?glm and ?family for more information.
Generalized linear models have fewer assumptions than most common parametric tests. Observations still need to be independent, and the correct link function needs to be specified. So, for example you should understand when to use a poisson regression, and when to use a logistic regression. However, the normal distribution of data or residuals is not required.
Logistic regression has a dependent variable with two levels. In R, this can be specified in three ways. 1) The dependent variable can be a factor variable where the first level is interpreted as “failure” and the other levels are interpreted as “success”. (As in the second example in this chapter). 2) The dependent variable can be a vector of proportions of successes, with the caveat that the number of observations for each proportion is indicated in the weights option. 3) The dependent variable can be a matrix with two columns, with the first column being the number of “successes” and the second being the number of “failures”. (As in the first example in this chapter).
Note that in each of these specifications, both the number of successes and the number of failures is known. You should not perform logistic regression on proportion data where you don’t know (or don’t tell R) how many individuals went into those proportions. In statistics, 75% is different if it means 3 out of 4 rather than 150 out of 200. As another example where logistic regression doesn’t apply, the weight people lose in a diet study expressed as a proportion of initial weight cannot be interpreted as a count of “successes” and “failures”. Here, you might be able to use common parametric methods, provided the model assumptions are met; log or arc-sine transformations may be appropriate. Likewise, if you count the number of people in front of you in line, you can’t interpret this as a percentage of people since you don’t know how many people are not in front of you in line. In this case with count data as the dependent variable, you might use poisson regression.
One potential problem to be aware of when using generalized linear models is overdispersion. This occurs when the residual deviance of the model is high relative to the residual degrees of freedom. It is basically an indication that the model doesn’t fit the data well.
It is my understanding, however, that overdispersion is technically not a problem for a simple logistic regression, that is one with a binomial dependent and a single continuous independent variable. Overdispersion is discussed in the chapter on Multiple logistic regression.
R does not produce r-squared values for generalized linear models (glm). My function nagelkerke will calculate the McFadden, Cox and Snell, and Nagelkereke pseudo-R-squared for glm and other model fits. The Cox and Snell is also called the ML, and the Nagelkerke is also called the Cragg and Uhler. These pseudo-R-squared values compare the maximum likelihood of the model to a nested null model fit with the same method. They should not be thought of as the same as the r-squared from an ordinary-least-squares linear (OLS) model, but instead as a relative measure among similar models. The Cox and Snell for an OLS linear model, however, will be equivalent to r-squared for that model. I have seen it mentioned that a McFadden pseudo-R-squared of 0.2–0.4 indicates a good fit. Whereas, I find that the Nagelkerke usually gives a reasonable indication of the goodness of fit for a model on a scale of 0 to 1. That being said, I have found the Cox and Snell and Nagelkerke to sometimes yield values I wouldn’t expect for some glm. The function pR2 in the package pscl will also produce these pseudo-R-squared values.
Note that testing p-values for a logistic or poisson regression uses Chi-square tests. This is achieved through the test=“Wald” option in Anova to test the significance of each coefficient, and the test=“Chisq” option in anova for the significance of the overall model. A likelihood ratio test can also be used to test the significance of the overall model.
### --------------------------------------------------------------
### Logistic regression, amphipod example, p. 247
### --------------------------------------------------------------
Input = ("
Location Latitude mpi90 mpi100
Port_Townsend,_WA 48.1 47 139
Neskowin,_OR 45.2 177 241
Siuslaw_R.,_OR 44.0 1087 1183
Umpqua_R.,_OR 43.7 187 175
Coos_Bay,_OR 43.5 397 671
San_Francisco,_CA 37.8 40 14
Carmel,_CA 36.6 39 17
Santa_Barbara,_CA 34.3 30 0
")
Data = read.table(textConnection(Input),header=TRUE)
Data$Total = Data$mpi90 + Data$mpi100
Data$Percent = Data$mpi100 / + Data$Total
Trials = cbind(Data$mpi100, Data$mpi90) # Sucesses, Failures
model = glm(Trials ~ Latitude,
data = Data,
family = binomial(link="logit"))
summary(model)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -7.64686 0.92487 -8.268 <2e-16 ***
Latitude 0.17864 0.02104 8.490 <2e-16 ***
confint(model)
2.5 % 97.5 %
(Intercept) -9.5003746 -5.8702453
Latitude 0.1382141 0.2208032
exp(model$coefficients) # exponentiated coefficients
(Intercept) Latitude
0.0004775391 1.1955899446
exp(confint(model)) # 95% CI for exponentiated coefficients
2.5 % 97.5 %
(Intercept) 7.482379e-05 0.002822181
Latitude 1.148221e+00 1.247077992
library(car)
Anova(model, type="II", test="Wald")
Analysis of Deviance Table (Type II tests)
Response: Trials
Df Chisq Pr(>Chisq)
Latitude 1 72.076 < 2.2e-16 ***
library(rcompanion)
nagelkerke(model)
$Models
Model: "glm, Trials ~ Latitude, binomial(link = \"logit\"), Data"
Null: "glm, Trials ~ 1, binomial(link = \"logit\"), Data"
$Pseudo.R.squared.for.model.vs.null
Pseudo.R.squared
McFadden 0.425248
Cox and Snell (ML) 0.999970
Nagelkerke (Cragg and Uhler) 0.999970
anova(model,
update(model, ~1), # update here produces null model for comparison
test="Chisq")
Analysis of Deviance Table
Model 1: Trials ~ Latitude
Model 2: Trials ~ 1
Resid. Df Resid. Dev Df Deviance Pr(>Chi)
1 6 70.333
2 7 153.633 -1 -83.301 < 2.2e-16 ***
library(lmtest)
lrtest(model)
Likelihood ratio test
Model 1: Trials ~ Latitude
Model 2: Trials ~ 1
#Df LogLik Df Chisq Pr(>Chisq)
1 2 -56.293
2 1 -97.944 -1 83.301 < 2.2e-16 ***
plot(fitted(model),
rstandard(model))
A plot of standardized residuals vs. predicted values. The residuals should be unbiased and homoscedastic. For an illustration of these properties, see this diagram by Steve Jost at DePaul University: condor.depaul.edu/sjost/it223/documents/resid-plots.gif.
### additional model checking plots with: plot(model)
plot(Percent ~ Latitude,
data = Data,
xlab="Latitude",
ylab="Percent mpi100",
pch=19)
curve(predict(model,data.frame(Latitude=x),type="response"),
lty=1, lwd=2, col="blue",
add=TRUE)
# # #
### --------------------------------------------------------------
### Logistic regression, favorite insect example, p. 248
### --------------------------------------------------------------
Input = ("
Height Insect
62 beetle
66 other
61 beetle
67 other
62 other
76 other
66 other
70 beetle
67 other
66 other
70 other
70 other
77 beetle
76 other
72 beetle
76 beetle
72 other
70 other
65 other
63 other
63 other
70 other
72 other
70 beetle
74 other
")
Data = read.table(textConnection(Input),header=TRUE)
model = glm(Insect ~ Height,
data=Data,
family = binomial(link="logit"))
summary(model)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 4.41379 6.66190 0.663 0.508
Height -0.05016 0.09577 -0.524 0.600
confint(model)
2.5 % 97.5 %
(Intercept) -8.4723648 18.4667731
Height -0.2498133 0.1374819
exp(model$coefficients) # exponentiated coefficients
(Intercept) Height
82.5821122 0.9510757
exp(confint(model)) # 95% CI for exponentiated coefficients
2.5 % 97.5 %
(Intercept) 0.0002091697 1.047171e+08
Height 0.7789461738 1.147381e+0
library(car)
Anova(model, type="II", test="Wald")
Analysis of Deviance Table (Type II tests)
Response: Insect
Df Chisq Pr(>Chisq)
Height 1 0.2743 0.6004
Residuals 23
library(rcompanion)
nagelkerke(model)
$Pseudo.R.squared.for.model.vs.null
Pseudo.R.squared
McFadden 0.00936978
Cox and Snell (ML) 0.01105020
Nagelkerke (Cragg and Uhler) 0.01591030
anova(model,
update(model, ~1), # update here produces null model for comparison
test="Chisq")
Analysis of Deviance Table
Model 1: Insect ~ Height
Model 2: Insect ~ 1
Resid. Df Resid. Dev Df Deviance Pr(>Chi)
1 23 29.370
2 24 29.648 -1 -0.27779 0.5982
library(lmtest)
lrtest(model)
Likelihood ratio test
Model 1: Insect ~ Height
Model 2: Insect ~ 1
#Df LogLik Df Chisq Pr(>Chisq)
1 2 -14.685
2 1 -14.824 -1 0.2778 0.5982
plot(fitted(model),
rstandard(model))
### Convert Insect to a numeric variable, levels 0 and 1
Data$Insect.num=as.numeric(Data$Insect)-1
library(FSA)
headtail(Data)
Height Insect Insect.num
1 62 beetle 0
2 66 other 1
3 61 beetle 0
23 72 other 1
24 70 beetle 0
25 74 other 1
### Plot
plot(Insect.num ~ Height,
data = Data,
xlab="Height",
ylab="Insect",
pch=19)
curve(predict(model,data.frame(Height=x),type="response"),
lty=1, lwd=2, col="blue",
add=TRUE)
### Convert Insect to a logical variable, levels TRUE and FALSE
Data$Insect.log=(Data$Insect=="other")
library(FSA)
headtail(Data)
Height Insect Insect.num Insect.log
1 62 beetle 0 FALSE
2 66 other 1 TRUE
3 61 beetle 0 FALSE
23 72 other 1 TRUE
24 70 beetle 0 FALSE
25 74 other 1 TRUE
library(popbio)
logi.hist.plot(Data$Height,
Data$Insect.log,
boxp=FALSE,
type="hist",
col="gray",
xlabel="Height")
# # #
### --------------------------------------------------------------
### Logistic regression, hypothetical example
### Abbreviated code and description
### --------------------------------------------------------------
Input = ("
Continuous Factor
62 A
63 A
64 A
65 A
66 A
67 A
68 A
69 A
70 A
71 A
72 A
73 A
74 A
75 A
72.5 B
73.5 B
74.5 B
75 B
76 B
77 B
78 B
79 B
80 B
81 B
82 B
83 B
84 B
85 B
86 B
")
Data = read.table(textConnection(Input),header=TRUE)
model = glm(Factor ~ Continuous,
data=Data,
family = binomial(link="logit"))
summary(model)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -66.4981 32.3787 -2.054 0.0400 *
Continuous 0.9027 0.4389 2.056 0.0397 *
library(car)
Anova(model, type="II", test="Wald")
Analysis of Deviance Table (Type II tests)
Response: Factor
Df Chisq Pr(>Chisq)
Continuous 1 4.229 0.03974 *
Residuals 27
library(rcompanion)
nagelkerke(model)
Pseudo.R.squared
McFadden 0.697579
Cox and Snell (ML) 0.619482
Nagelkerke (Cragg and Uhler) 0.826303
anova(model,
update(model, ~1),
test="Chisq")
Resid. Df Resid. Dev Df Deviance Pr(>Chi)
1 27 12.148
2 28 40.168 -1 -28.02 1.2e-07 ***
plot(fitted(model),
rstandard(model))
### Convert Factor to a numeric variable, levels 0 and 1
Data$Factor.num=as.numeric(Data$Factor)-1
library(FSA)
headtail(Data)
Continuous Factor Factor.num
1 62 A 0
2 63 A 0
3 64 A 0
27 84 B 1
28 85 B 1
29 86 B 1
plot(Factor.num ~ Continuous,
data = Data,
xlab="Continuous",
ylab="Factor",
pch=19)
curve(predict(model,data.frame(Continuous=x),type="response"),
lty=1, lwd=2, col="blue",
add=TRUE)
### Convert Factor to a logical variable, levels TRUE and FALSE
Data$Factor.log=(Data$Factor=="B")
library(FSA)
headtail(Data)
Continuous Factor Factor.num Factor.log
1 62 A 0 FALSE
2 63 A 0 FALSE
3 64 A 0 FALSE
27 84 B 1 TRUE
28 85 B 1 TRUE
29 86 B 1 TRUE
library(popbio)
logi.hist.plot(Data$Continuous,
Data$Factor.log,
boxp=FALSE,
type="hist",
col="gray",
xlabel="Height")
# # #
Power analysis
原文:https://www.cnblogs.com/tecdat/p/12056619.html