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


Could we use the SD instead since we only have the variance and can't find the SE, like this:

This would result in an interval much too wide, and would in some cases result in non-sense confidence intervals (for example, estimating a binomial parameter which is close to 0.5).

Is this appropriate since we can't find the SE without knowing the n?

No, what you've written can not be considered a confidence interval.
Keep in mind that a confidence interval is intended to reflect uncertainty in the estimate of the parameter, while the standard deviation is (roughly) intended to reflect uncertainty in the data generating process.
