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.

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#### Best Answer

In addition to @Henry's answer and to @PeterEllis' comment, here are a few comments about the R code itself:

- The third argument of
`mlog1`

is`sdev`

. You therefore need`sdev`

in`out1`

and`out2`

(not`sd`

); - The likelihood ratio test is the logarithm of the ratio between two likelihoods (up to a multiplicative factor). Equivalently, it is a difference between two log-likelihoods (up to a multiplicative factor). It is not the ratio between two log-likelihoods.
`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.- You forgot the multiplicative factor ($-2$).

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