Question

Code in R

Part 1. a) Run the following lines: n<-30 x<-matrix(rnorm(n * 1000), 1000,n) xL,1:3]<-xl,1:3]*10 y<-1+matrix(rnorm (n * 1000), 1000, n) Note: the samples in "x" are contaminated. b) Conduct 1000 two-sample two-sided t-tests for the associated rows of "xand "". (e.g., for the first test, the two samples are "x[1," and "y[1,]", for the second test, the two samples are "x[2," and "y[2,1",etc.). Calculate the total number of rejections of the 1000 tests. (Use significant level a = 0.05) Part 2. Write a code to find all values of x for which the functions below are zero. In other words, find the roots of those function using Newton's method, report the starting values that you used and the number of iterations. Plotting the function might help in selection of the initial values. You must use a while loop to do the computations. Simply using a built-in R function is not acceptable. a) f(x)-2ex2-1 b) f(x) = x2- 251/187x + 82/187

Answer 1

ANSWER :

Please find R code for above

n=30
x=matrix(rnorm(n*1000),1000,n)
x[,1:3]=x[,1:3]*10
y=1+matrix(rnorm(n*1000),1000,n)

#Calculating total nmber of rejections
#rejection if p value is less than 0.05
rejections=0
for(i in 1:1000){
t1=x[i,]
t2=y[i,]
t=t.test(t1,t2)
p=t$p.value
if(p<0.05){
rejections=rejections+1
}
}

The number of rejections are coming out to be 439

Part 2 - Newton's method , iliorations for f(x)=0 are given as, +'&;) (@) f(x) = pets f(0) = teza. o 18771 = tzwaldt tieta fW= 22-2351,+ 345+. f(a) = Dit 29 .: 41 = a; 2a + 18722

In part (A), the function can never be zero, so code will give absurd results

#part 2 a
#let m be the starting point and n be the number of iterations

f1=function(m,n){
it=m
i=1
while(i<(n+1)){
it=it-(2*exp(it^2)-1)/(4*it*exp(it^2))
i=i+1
}
return(it)
}

#part 2 b
f2=function(m,n){
it=m
i=1
while(i<(n+1)){
it=it-(it^2 -(251/187)*(1/it) + 82/187)/(2*it + (251/187)*(1/it^2))
i=i+1
}
return(it)
}

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