According to the central limit theorem: $S\approx N(n*\mu, n*\sigma^2)$
Normally, a confidence interval is calculated according to:
$(μ \pm 1.96*\sqrt{\frac{\sigma^2}{n}})$
Do you divide $\sigma^2$ by n in the case of an approximation of the sum?
I have seen cases, where the interval is calculated as:
$(μ \pm 1.96*\sqrt{\sigma^2})$
What is the right way to calculated the confidence interval in this case? Is the approach without n wrong or does it apply to some cases?
Thank you
There seems to be some confusion about the distinction between $\mu$ and its estimate $\bar X,$ between cases where the numerical value of $\sigma$ is and is not known, and between sample size $n$ and $4.$ Perhaps the following step-by-step approach will help.
If data are a random sample from $\mathsf{Norm}(\mu, \sigma),$ where the second argument is the standard deviation, then
$$\bar X_n = \frac 1n\sum_{i=1}^n \frac{X_i}{n} \sim \mathsf{Norm}\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$$
Thus a 95% confidence interval (CI) for $\mu$ (when the numerical value of $\sigma$ is known) is of the form $$\bar X_n \pm 1.96\frac{\sigma}{\sqrt{n}},$$ where 1.96 cuts probability 0.025 from the upper tail of the (symmetrical) standard normal distribution.
Notice that the center of this CI is the sample mean $\bar X_n,$ which estimates the population mean $\mu.$
Furthermore, if $\sigma$ is unknown and estimated by the sample standard deviation $S_n = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i=\bar X_n)^2},$ then a 95% CI for $\mu$ is of the form $$\bar X_n \pm t^*\frac{S_n}{\sqrt{n}},$$ where $t^*$ cuts probability 0.025 from the upper tail of Student's t distribution (also symmetrical) with degrees of freedom $\nu = n-1.$
Finally, if you are dealing with $T_4 = \sum_{i=1}^4 X_i,$ then $T_4\sim\mathsf{Norm}(\mu_4 = 4\mu, \sigma_4 = 4\sigma).$ So if you want a 95% CI for $E(T_4) = \mu_4 = 4\mu,$ then you simply multiply by $4$ the endpoints of the corresponding CI for $\mu.$ In the case where the numerical value of $\sigma$ is known, this would be:
$$4\bar X_n \pm 1.96\frac{4\sigma}{\sqrt{n}}.$$
In particular, if $n = 4,$ then the CI becomes
$$T_4 \pm 1.96\sqrt{4\sigma^2},$$ which agrees with @Henry's first comment,