Question

我正在学习有关创建信息功能的教程，我现在完成了教程，我希望再采取一个步骤并且已经陷入僵局。我正在关注的教程如下。

http://www.gonulates.com/psychometrics/fishers-information-function-for-the-four-parameter-model-4pm/

irt.4PM.Info <- function(theta,item.par,D=1.7) {
# This function calculates the Fisher information for
# Four Parameter Model, given item parameters and ability.
#
# item.par should be a matrix with four columns:
# Each row represents an item,
# first column represents a parameters (item discrimination)
# second column represents b parameters (item difficulty)
# third column represents c parameters(pseudo-guessing parameter)
# fourth column represents d parameters (upper asymptote)
# theta can be a single number or a vector.

return(t(sapply(theta,
   FUN = function (x) (
     (D * item.par[,1])^2 * (item.par[,4] - item.par[,3])^2)/
     ((item.par[,3] + item.par[,4] *
       exp(D*item.par[,1] * (x - item.par[,2]))) *
     (1 - item.par[,3] + (1-item.par[,4]) *
       exp(D*item.par[,1] * (x-item.par[,2]))) *
     (1 + exp(-D * item.par[,1] * (x - item.par[,2])))^2) )))
}
theta.2 <- c(-1,-.5,0,1,2)
itpars <- data.frame(a = c(1,  1,.75),
                 b = c(0,-.5,  1),
                 c = c(0,  0, .2),
                 d = c(1,  1,.98));
itpars

##      a    b   c    d
## 1 1.00  0.0 0.0 1.00
## 2 1.00 -0.5 0.0 1.00
## 3 0.75  1.0 0.2 0.98
theta.3 <- seq(-4,4,by=0.01)
item.1.pars <- data.frame(a=1.2,b=-0.6,c=.1,d=1)
plot(theta.3,irt.4PM.Info(theta.3,item.1.pars),type="l",
 main="Information Functon",
 xlab="Theta", ylab="Information")

我要做的是绘制我现在所具有的功能的反转。也就是说，该函数看起来像一个钟形曲线，我希望得到这个函数的倒数（颠倒的钟形曲线）

我想要的曲线是标准误差，它是信息的倒数SE = 1 / I ^ .5（I =信息）

R，标准误差（IRT）绘制反函数

0 个答案: