Suppose you have 150 locations altogether, and you decide to base your confidence interval for the mean of the population (for some attribute) from a sample of size 10.

whole = rnorm(150, 50, 7)
x = sample(whole, 10)
summary(x);  length(x);  sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  37.86   43.03   45.24   47.92   52.61   59.93 
[1] 10         # sample size
[1] 7.470816   # sample standard deviation

t.test(x)$conf.int
[1] 42.57347 53.26207
attr(,"conf.level")
[1] 0.95

The mean for the whole company is 50; a 95% confidence interval for the mean is $(42.6, 53.3).$ I used the t.test procedure in R, but the 95% CI can be found from the formula $\bar X \pm t^* S/\sqrt{n},$ where $t^* = 2.262$ cuts probability 2.5% from the upper tail of Student's t distribution with $\nu = n-1 = 9$ degrees of freedom

qt(.975, 9)
[1] 2.262157

mean(x) + qt(c(.025,.975),9)*sd(x)/sqrt(10)
[1] 42.57347 53.26207

If you knew the population standard deviation $\sigma=7,$ then you could use $\bar X \pm 1.96(7/\sqrt{10}),$ which computes to $(42.6,53.3)).$ In general, this method has the potential to be a little more accurate, but there is no difference (to one place accuracy) from the CI above for this example.

mean(x) + qnorm(c(.025,.975))*7/sqrt(10)
[1] 43.57920 52.25633

Notes: (1) You are sampling from a finite population of size 150. As long as the sample size (here $n=10)$ is less than 10% of the population size, these formulas for sampling from essentially infinite populations should give useful results.

(2) These methods assume that the population values are approximately normally distributed. These methods would not work well if you had a few locations that are hugely different from any of the others.

(3) Your idea of doing some sort of stratified sampling so several provinces are represented or that some observations are from urban and some are from rural location might be useful. That would depend on whether there are large differences among provinces or between rural or urban locations. Stratified sampling would make it somewhat more difficult to make a confidence interval.

(4) Here, because I simulated the whole population, we can find the exact population mean and standard deviation and we know that the data are normal. In most actual applications this information would not necessarily be known.

(5) If you have some data for all 100+ scores, you might try the ttest` on a sample of a dozen or so locations to how well it workd in your application.