A latent class analysis with k classes is performed on the data given by x.

lca(x, k, niter=100, matchdata=FALSE, verbose=FALSE)

Arguments

x

Either a data matrix of binary observations or a list of patterns as created by countpattern

k

Number of classes used for LCA

niter

Number of Iterations

matchdata

If TRUE and x is a data matrix, the class membership of every data point is returned, otherwise the class membership of every pattern is returned.

verbose

If TRUE some output is printed during the computations.

Value

An object of class "lca" is returned, containing

w

Probabilities to belong to each class

p

Probabilities of a `1' for each variable in each class

matching

Depending on matchdata either the class membership of each pattern or of each data point

logl, loglsat

The LogLikelihood of the model and of the saturated model

bic, bicsat

The BIC of the model and of the saturated model

chisq

Pearson's Chisq

lhquot

Likelihood quotient of the model and the saturated model

n

Number of data points.

np

Number of free parameters.

References

Anton K. Formann: ``Die Latent-Class-Analysis'', Beltz Verlag 1984

See also

Examples

## Generate a 4-dim. sample with 2 latent classes of 500 data points each. ## The probabilities for the 2 classes are given by type1 and type2. type1 <- c(0.8,0.8,0.2,0.2) type2 <- c(0.2,0.2,0.8,0.8) x <- matrix(runif(4000),nr=1000) x[1:500,] <- t(t(x[1:500,])<type1)*1 x[501:1000,] <- t(t(x[501:1000,])<type2)*1 l <- lca(x, 2, niter=5) print(l)
#> LCA-Result #> ---------- #> #> Datapoints: 1000 #> Classes: 2 #> Probability of classes #> [1] 0.327 0.673 #> Itemprobabilities #> 1 2 3 4 #> 1 0.87 0.94 0.07 0.13 #> 2 0.31 0.29 0.68 0.68
#> LCA-Result #> ---------- #> #> Datapoints: 1000 #> Classes: 2 #> #> Goodness of fit statistics: #> #> Number of parameters, estimated model: 9 #> Number of parameters, saturated model: 15 #> Log-Likelihood, estimated model: -2499.16 #> Log-Likelihood, saturated model: -2464.486 #> #> Information Criteria: #> #> BIC, estimated model: 5060.489 #> BIC, saturated model: 5032.589 #> #> TestStatistics: #> #> Likelihood ratio: 69.34663 p-val: 5.566658e-13 #> Pearson Chi^2: 74.1275 p-val: 5.806466e-14 #> Degress of freedom: 6
p <- predict(l, x) table(p, c(rep(1,500),rep(2,500)))
#> #> p 1 2 #> 1 366 6 #> 2 134 494