Question

We have a random sample of size 10 from a normal distribution. We wish to estimate the population mean. Damjan suggests taking the average of the sample minimum and sample maximum. Allan thinks this will be a poor estimator and says we should use the sample mean instead. Do a simulation in R to compare these two estimators in terms of their bias and variance. Include a side-by-side boxplot that compares their sampling distributions.

Answer 1

We take 100 simulations at n=10

R-code with output-

> n=10
> sim=100
> th=2        ### Mean
> cnt=0
> eps=0.1
> for(i in 1:sim)
+ {
+   x=rnorm(n,th,2.3)
+   Tn=(min(x)+max(x))/2
+   if(abs(Tn-th)<eps)
+   {
+     cnt=cnt+1
+   }
+ }
> cnt
[1] 7
> prob=cnt/sim
> prob
[1] 0.07
>bias= Tn -th
>bias
[1] 0.6669769

> n=10
> sim=100
> th=2
> cnt=0
> eps=0.1
> for(i in 1:sim)
+ {
+   x=rnorm(n,th,2.3)
+   Tn=mean(x)
+   if(abs(Tn-th)<eps)
+   {
+     cnt=cnt+1
+   }
+ }
> cnt
[1] 14
> prob=cnt/sim
> prob
[1] 0.14

>bias= Tn -th
>bias

[1] -1.418102

_________________________________________________________________________________________

In 100 simulations to estimate mean(th), if we use estimator Tn=(min(x)+max(x))/2 then probability of estimating th (mean) is 0.07 and it's bias is 0.6669769 and if we use estimator Tn=mean(x) then probability of estimating th (mean) is 0.14 this means Tn=mean(x) is a good estimator and its bias is -1.418102 which is 0.