Finite population: CI of Normal Distribution same regardless of sample percent? I am trying to wrap my head around the following.  It seems that the confidence interval for a normal distribution does not rely on the sample percent.
For instance, say I am trying to find the average number of dollars spent at a store on a given day.  I do a sample of 100 with mean \$1000 and stdev \$200.  My 90% CI is [\$967.10, \$1032.90].   What I cant understand is why this is the SAME regardless of the percent sampled.  If there was a population of 101 customers or 10000, the CI is the same.  If this is correct, can someone explain this to me in a digestible way?
 A: If you look at it as analyzing data starting with sample 1, I would argue that the percent sampled will be essentially factored into the standard deviation estimator, you get closer and closer to the true population sigma.
Once you get to a lot of data points, the estimator won't really change much because you've very accurately estimated the population standard deviation.
That estimator gets divided by root n, which keeps dropping the range of the CI.
As far as the total population goes, we are really trying to estimate "forever".  So maybe there are 101 customers today, but if all things stayed the same (a huge assumption!), then the population will tend toward infinity.
Another way to look at it is that "with these conditions (season, prices, et cetera) we believe the mean will be in this range."  So on another warm Wednesday, you'd expect sales in that range.
A: I do not dispute @Porter's Answer (+1). However, it seems useful to
focus directly on the key difficulty with CIs based on data from small finite populations. If
you take too large a sample from a small population, t methods
no longer work.
The t statistic becomes much less variable
than a true Student's t distribution. In particular, the
"95%" CI becomes very short and it's coverage probability
approaches 100% as sample size $n$ approaches the
size $N$ of a finite 'population' chosen to be roughly normal.
I have done too many simulations (in R) to show here, but
a few are shown below to illustrate my main point.
Finite population. Begin with a fictitious "population" s of size $N = 110$
with known $\mu = 1000, \sigma = 200,$ as in the Question.
set.seed(123)
z1 = rnorm(110)
z2 = (z1-mean(z1))/sd(z1)
s = z2 * 200 + 1000
mean(s);  sd(s)
[1] 1000
[1] 200

qqnorm(s, col="skyblue2"); qqline(s, lwd=2)


Take samples of size $n=10,$ less than 10% of $N.$
set.seed(2021)
n = 10    # less than 110 
m = 10^5;  a.sm = s.sm = t0.sm = numeric(m)
for (i in 1:m) {
 x.sm = sample(s, n) 
 a.sm[i] = mean(x.sm)
 s.sm[i] = sd(x.sm)
 t0.sm[i] = (mean(x.sm) - 1000)*sqrt(n)/sd(x.sm)
}
mean(abs(t0.sm) < qt(.975, n-1))
[1] 0.95877   # about 96% coverage

For $n = 50,$ coverage probability is about 99%;
for $n = 100,$ nearly 100%. [Change n in code above,
simulations not shown.]  Graph for $n = 100$ below:
hdr = "n=100, N=110; Null Dist'n of T with T(99) Density"
hist(t0.sm, prob=T, xlim=c(-3,3), col="skyblue2", main=hdr)
 curve(dt(x,99), add=T, col="red", lwd=2)


Infinite population. For samples of size $n=100$ from a population of unrestricted size (using rnorm), of
course, coverage probability is 95%.
set.seed(2021); n = 100
m = 10^5;  a = s = t0 = numeric(m)
for (i in 1:m) {
 x = rnorm(n, 1000, 200) 
 a[i] = mean(x)
 s[i] = sd(x)
 t0[i] = (mean(x) - 1000)*sqrt(n)/sd(x)
}
mean(abs(t0)<qt(.975,n-1))
[1] 0.94945   $ aprx 95%

hdr = "n=100: Null Dist'n of T with T(99) Density"
hist(t0, prob=T, xlim=c(-3,3), col="skyblue2", main=hdr)
 curve(dt(x,99), add=T, col="red", lwd=2)


