if we have $1250$ values(sample), are we then able to calculate $99$% confidence interval less than $0.1$ ? how do i check if its possible ?
my first thought was to try going backwards in the process of finding confidence interval. which is impossible as many parameters are missing
i am a bit confused here so any hints are appreciated !
Thanks in advance!
Your question is how to check whether this is possible.
My guess is that you are supposed to assume that the sample mean $\bar X$ is very nearly normal because of the large sample size $n = 1250.$ And then to assume that an approximate $99\%$ confidence interval is given by $\bar X \pm 2.58 S/\sqrt{n}$, where $S$ is the sample standard deviation. Thus the length of the CI would be about $5.16 S/\sqrt{1250},$ which you could compare with the desired length of the CI (which I take to be) $0.1.$
However, even in applied situations, it is possible to encounter samples from distributions for which the assumptions of the previous paragraph are not valid. In particular, as remarked by @AndreNicholas, the parent population must have a finite standard deviation. Otherwise a CI based on a sample standard deviation $S$ cannot be the basis for a large-sample confidence interval.
If the original data pass a test for normality, then the assumptions are likely OK. Less rigorously, if a histogram of the data seems roughly symmetrical and has no far outliers, then the assumptions may be OK.
The figure below shows histograms of two samples of size $n=1250$. The first (at left) is from a normal populaion, has $\bar X = 99.8,$ $S_X = 14.4$ and a 99% CI $(98.76, 100.86)$ for $\mu$, which is valid (even if somewhat longer than you have in mind). The second shows a sample from a badly-behaved distribution for which the CI formula above is not valid.