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

To complement the existing answer and since I saw your follow up question in the comments, I believe that it is important to stress that the confidence interval formula you mentioned $CI = \bar Y \pm 1.96*SE$ is derived from the sampling distribution of the mean.
So, the standard error in the formula is actually the standard deviation of the sampling distribution of the mean. It provides information about the spread of the sampling distribution of the mean. $s/\sqrt{n}$ is an estimator of the standard error. So, if you have only the mean and the variance you cannot say much.
Below is an example that simulates the sampling distribution of the mean, using bootstrapping, of two samples of different size but with same mean and same variance.
As you can see, the confidence interval gets narrower as the sample size increase.
set.seed(007)

spl1 <- 170+20*scale(rnorm(10))[,1]  #Forcing to have same mean and sd for all samples | n = 10
spl2 <- 170+20*scale(rnorm(100))[,1]  #Forcing to have same mean and sd for all samples | n = 100

library(boot)
myFunc <- function(data, i){
  return(mean(data[i]))
}

bootMean1 <- boot(spl1 , statistic=myFunc, R=10000)
bootMean2 <- boot(spl2 , statistic=myFunc, R=10000)

mean(spl1) + 1.96*c(-1,1)*sd(bootMean1$t)

mean(spl2) + 1.96*c(-1,1)*sd(bootMean2$t)



