# Solved – Likelihood ratio test in R

I am in desperate need for help. I am trying to calculate the likehood ratio test in R, but I don't have allot of experience using R.

For example, to calculate the following

Suppose \$X_1, X_2,ldots, X_n\$ is a random sample from a normal population with mean \$mu\$ and variance \$16\$. Find the test with the best critical region, that is, find the most powerful test, with a sample size of \$n = 16\$ and a significance level \$alpha = 0.05\$ to test the simple null hypothesis \$H_0: mu = 10\$ against the simple alternative hypothesis \$H_A: mu = 15\$.

This is the code I have up until now

``#X values #Added the mean=10 thanks to the comments here X=rnorm(16,10,sd=4)  mlog1=function(media,x,sdev){ sum(-dnorm(x,mean=media, sd=sdev, log=T)) }  #L(10) #also added sdev out1=nlm(mlog1,10,x=X,sdev=4) #L(15) #also added sdev out2=nlm(mlog1,15,x=X,sdev=4)  k1=out1\$estimate     k2=out2\$estimate  #L(10)/L(15)     print( k1/k2) ``

This is the very basic axample and the solution, done by hand, can be found here.

I know you also have the LRT function, but I don't know how to apply this function.

Please help, the code looks correct but it always returns 1 or 0.9999999 for any value in the parameters.

Contents

• The third argument of `mlog1` is `sdev`. You therefore need `sdev` in `out1` and `out2` (not `sd`);
• `k1=out1\$estimate` returns an estimate of \$mu\$. `k2=out2\$estimate` returns another estimate of \$mu\$ (based on another starting value). Both aim to estimate the same thing; that's why you always get something very close to \$1\$. Neither of them returns minus a log-likelihood.