Some of what you have written appears to be a tangle of confusion. But I think I can help dispel some of the confusion. Let
me start from one of your statements, and use that as a basis for
illustrating a 95% nonparametric bootstrap confidence interval (CI) for
the mean $\mu$ of the population from which available data are sampled:
"I have read some statements about bootstrap method, they say it is effective and does not require any assumptions about the population's distribution. But i have not really understand how to properly use this method."
Data. Suppose you have $n = 50$ observations sampled at random from an unknown
distribution. Suppose these are as follows (the numbers in brackets show
the index of the first observation on each line):
x
# [1] 0.11 0.62 0.61 0.62 0.86 0.64 0.01 0.23 0.67 0.51
#[11] 0.69 0.54 0.28 0.92 0.29 0.84 0.29 0.27 0.19 0.23
#[21] 0.32 0.30 0.16 0.04 0.22 0.81 0.53 0.91 0.83 0.05
#[31] 0.46 0.27 0.30 0.51 0.18 0.76 0.20 0.26 0.99 0.81
#[41] 0.55 0.65 0.31 0.62 0.33 0.50 0.68 0.48 0.24 0.77
Point estimate. The observed sample mean $\bar X = 0.4692.$ In my R code I use a for
this value. It is an estimate of the population mean $\mu.$
a = mean(x); a
## 0.4692
Nonparametric bootstrap CI. In order to make a CI for $\mu,$ I need information about the variability of
$\bar X.$ Specifically, I would like to know the distribution of the
differences $D = \bar X - \mu.$ If I knew this distribution then I could
use it to find $L$ and $U$ such that
$$ 0.95 = P(L < D = \bar X -\mu < U) = P(\bar X - U < \mu < \bar X - L),$$
so that a 95% CI for $\mu$ would be of the form $(\bar X - U, \bar X - L).$
Not knowing the distribution of $D.$ I enter the bootstrap world, where
I repeatedly re-sample from the sample of 50 observations in order to
obtain estimates $L^*$ of $L$ and $U^*$ of $U$.
Specifically, one
re-sample consists of a sample of size $50$ chosen with replacement
from the sample x. I find its mean $\bar X^*.$ And then, temporarily, using the observed $A = 0.4692$ as proxy for the unknown $\mu$, I find
$D^* = \bar X^* - A.$ In this same way, I make a large number $B$
re-samples, obtaining another value $D^*$ from each re-sample.
Back in the real world I find quantiles .025 and .975 of the $B$ values
$D^*,$ which I use as $L^*$ and $U^*,$ respectively. Finally, my 95%
nonparametric bootstrap CI for $\mu$ is of the form
$(\bar X - U^*, \bar X - L^*),$ which I found to be $(0.396, 0.541),$
based on $B = 100,000$ re-samples of size $n=50.$
R code for bootstrap. Because bootstrapping
is a simulation process, you may get slightly different CIs on each
run. (With $B = 10^5,$ often only the last digit of the confidence bounds changes.)
If you use the same data and R with the same seed shown at the head of the
program, you will get exactly the same result I did.
In the R code below, I use suffixes .re instead of $*$ to indicate
re-sampling in the bootstrap world. (I have tried to use simplified R
code so it will be clear what is going on even if you are not familiar
with R.)
set.seed(4321)
B = 10^5; d.re = numeric(B); n = 50
for (i in 1:B) {
a.re = mean(sample(x, n, repl=T))
d.re[i] = a.re - mean(x) }
L.re = quantile(d.re, .025)
U.re = quantile(d.re, .975)
mean(x) - U.re; mean(x) - L.re
97.5%
0.396
2.5%
0.5412
Reality check. By way of confession, I simulated the sample x as you suggested
by taking $n = 50$ values from $\mathsf{Unif}(0,1),$ rounded to two places.
set.seed(1234)
x = round(runif(50),2)
a = mean(x); a; sd(x)
## 0.4692
## 0.2636312
A t confidence interval should come pretty close to finding a valid
95% CI for $\mu = 0.5.$ I obtained the 95% t-interval $(0.394, 0.544),$
which is in excellent agreement with the 95% nonparametric boostrap CI.
(You can generate exactly the same 50 values x, if you use R with the same seed.)
