# bootstrapping for a parameter estimation

At the moment I'm studying bootstrap and I want to do some examples to get a feeling for it. Suppose I generate the following data:

set.seed(10)
data <- rnorm(1000,1,3)
number <- min(data[data <= 0.9])
number


so number.min will contain the minimum number of the data in the interval $(-\infty,0.9]$ which is equal to $-8.036491$. If we know there were again 1000 generated numbers using a normal distribution with standard deviation 3. Suppose the number.min is this time $a:=-12.2385$, is it still plausible to assume that the expected value is $1$?

We denote number.min as a random variable $X$. My idea was to perform a hypothesis test. So $H_0: \mu = 1$. The problem here is we do not know the exact distribution of $X$ and I don't want to assume normal distribution. My idea was to use bootstrap. Under the assumption $H_0$, I can generate as many outcomes of $X$ as I want. So if I'm right, I'm interested in

$$P(X\in [x_l,x_u]|\mu =1)=1-\alpha$$

when choosing $\alpha=0.01$ I want to find $x_u,x_l$. If $a\in[x_l,x_u]$ I would argue that $\mu=1$ is plausible, right?

1. Question: This is theoretical nature: Is my idea correct, i.e. finding $x_l,x_u$ is this whole procedure from a statistical viewpoint ok?

Assuming this is right, we can go over and implement this. First we generate enough outcomes of $X$

 f <- function(n){
data <- rnorm(n,1,3)
number <- min(data[data <= 0.9])
return(number)
}
data <- replicate(500,f(1000))


This leads to 500 realisations of $X$. Now we can generate our boostrap samples

bst.smpl <- boot(data,?,1000)


2. Question: the boot function needs an argument "statistics", but I don't understand which it is in my case?

Last I would just compute the confidence intervals

ci <- boot.ci(bst.smpl,conf=0.99,type="basic",index=1)


3.Question: Is there any easier way to decide if we are still using $\mu=1$?

• I'm not entirely clear on what you are asking. Why should the minimum of your sample carry much information about the expected value? Why do you cherry-pick data <0.9? Apr 7, 2014 at 14:42
• @StephanKolassa about your first question: Well that's just a question. I defined $X$ in this way and I would like to deduce, given an outcome if we are using a specific $\mu$. About the second one. This is basically because then $X$ has a unknown distribution
– math
Apr 7, 2014 at 14:46
• @StephanKolassa I designed this problem by myself to learn bootstrap. The definition of $X$ is arbitrary. My goal was to get outcomes of a unknown distribution to use the bootstrap method.
– math
Apr 7, 2014 at 14:54
• I'll wait for the statistics gurus to chime in, but my suspicion is that the answer to your first question is No, as in here: en.wikipedia.org/wiki/Not_even_wrong Sorry. I'm open to being convinced otherwise. Apr 7, 2014 at 14:59
• @StephanKolassa thanks for your thoughts. Can you explain why do you think it is not true? Suppose you have a normal distributed r.v. $X$ (everything known) and you observe a value $a$. If $a$ is outside $[x_l,x_u]$, you would argue that this outcome is extremely rare. So as a decision point, it seems to be plausible
– math
Apr 7, 2014 at 15:15