Confidence interval without sample size, only variance? Suppose we have an estimate ($\bar{y}$) and want to generate a confidence interval. However, we only have the variance of the sampling distribution ($v$) - and not the $\text{n}$.
Instead of:
$$
CI(\bar{Y})=\bar{Y}\pm(1.96)*SE
$$
Could we use the SD instead since we only have the variance and can't find the SE, like this:
$$
CI(\bar{Y})=\bar{Y}\pm(1.96)*\sqrt{v}
$$
Is this appropriate since we can't find the SE without knowing the n? Our interpretation would change due to focusing on variability between samples and not within a single sample, but is this the only approach with that limited information to creating a confidence interval?
Thanks

Your proposal is tantamount to treating the sample as having a single unit (i.e., $n=1$).  This would give you an extremely conservative confidence interval (i.e., too wide unless the true sample size is actually one).  There is really no way to get an accurate confidence interval in this case without knowledge of the sample size.
