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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

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3 Answers 3

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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.

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  • $\begingroup$ Ah that makes sense, thank you! accepted. $\endgroup$ Oct 28, 2021 at 2:21
  • $\begingroup$ Would there be a way to construct a CI with only this information? $\endgroup$ Oct 28, 2021 at 2:26
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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)

enter image description here enter image description here

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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.

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