HW6comp

Newton Raphson function

newtonraphson <- function(ftn, x0, tol = 1e-9, max.iter = 100) {
        x <- x0 # x0: the initial value
        fx <- ftn(x)
        iter <- 0
        while ((abs(fx[1]) > tol) & (iter < max.iter)) {
                x <- x - fx[1]/fx[2]
                fx <- ftn(x)
                iter <- iter + 1
                cat("At iteration", iter, "value of x is:", x, "\n")
        }
        if (abs(fx[1]) > tol) {
                cat("Algorithm failed to converge\n")
                return(NULL)
        } else { # abs(fx[1]) <= tol
                cat("Algorithm converged\n")
                return(x)
        }
}

Ex 17-1:

Consider a class with 100 students, if the probability of passing the final exam is p and we observe 20 students passing the final exam. (1)Please find the 95% exact confidence interval for p. (2)Please find the 95% asymptotic confidence interval for p.

let’s make plota to find the initial value first!

par(mfrow = c(1,2))
p <- seq(0,1,0.01)
fp <- (-0.975)
for(i in 0:19){
        fp <- fp + (choose(100,i)*(p^i)*((1-p)^(100-i)))
}
plot(p,fp, type = "l",lwd = 2)
abline(h = 0, col = "pink", lwd = 4)

p <- seq(0,1,0.01)
fp <- (-0.025)
for(i in 0:20){
        fp <- fp + (choose(100,i)*(p^i)*((1-p)^(100-i)))
}
plot(p,fp, type = "l", lwd = 2)
abline(h = 0, col = "pink", lwd = 4)

decided the intial value for lower bound to be 0.2,and upper bound to be 0.25.

then write the functions.

ftn9 <- function(p){
        fp1 <- (-0.975)
        dfp1 <- 0
        for(i in 0:19){
                fp1 <- fp1 + (choose(100,i)*(p^i)*((1-p)^(100-i)))
                dfp1 <- dfp1 + choose(100,i)*(i*(p^(i-1))*((1-p)^(100-i))-(p^i)*(100-i)*((1-p)^(99-i)))
        }
        return(c(fp1,dfp1))
}

ftn10 <- function(p){
        fp2 <- (-0.025)
        dfp2 <- 0
        for(i in 0:20){
                fp2 <- fp2 + (choose(100,i)*(p^i)*((1-p)^(100-i)))
                dfp2 <- dfp2 + choose(100,i)*(i*(p^(i-1))*((1-p)^(100-i))-(p^i)*(100-i)*((1-p)^(99-i)))
        }
        return(c(fp2,dfp2))
}

the values are same from

newtonraphson(ftn9,0.2,tol = 1e-9,100)

## At iteration 1 value of x is: 0.1481533 
## At iteration 2 value of x is: 0.1338861 
## At iteration 3 value of x is: 0.1278661 
## At iteration 4 value of x is: 0.1266966 
## At iteration 5 value of x is: 0.1266556 
## At iteration 6 value of x is: 0.1266556 
## Algorithm converged

## [1] 0.1266556

newtonraphson(ftn10,0.25,tol = 1e-9,100)

## At iteration 1 value of x is: 0.2735477 
## At iteration 2 value of x is: 0.2867346 
## At iteration 3 value of x is: 0.2913356 
## At iteration 4 value of x is: 0.2918372 
## At iteration 5 value of x is: 0.2918427 
## Algorithm converged

## [1] 0.2918427

library(binom)

## Warning: package 'binom' was built under R version 4.3.2

binom.confint(20,100,0.95,"exact")

use the formula and go on to find out the asymptotic confidence interval.

phat <- 20/100
phat-(qnorm(0.975)*sqrt((phat*(1-phat)/100)))

## [1] 0.1216014

phat+(qnorm(0.975)*sqrt((phat*(1-phat)/100)))

## [1] 0.2783986

library(binom)
binom.confint(20,100,conf.level = 0.95, methods = "asymptotic")