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I apologize if my question is extremely simple, but you know my name.

I have only the variance of a sample and its 95% confidence intervals, nothing more, nothing less.

Please correct me if I am wrong: I know that sd = sqrt(variance), can I compute the standard deviation and its 95% confidence intervals squaring the estimate and confidence intervals of the variance of the sample?

I read How can the confidence interval for standard deviation not include the sample standard deviation?, but I do not know if the sample originated from a normal distribution and I do not know N.

(I use R)

Edit, at the request of whuber:

I apologize and I hope that it will be clearer than my previous question. I am talking about calculating the standard deviation and its 95% confidence intervals from the variance and its 95% confidence intervals, without any other info. I do not know how they were calculated, I do not know if the distribution was normal. If sd=sqrt(var), my question is similar to asking how you can backtransform a variable and its 95% confidence intervals (unfortunately I did not find relevant Q&A here).

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    $\begingroup$ Could you please edit this question to clarify what you are talking about? In particular, what parameters are these confidence intervals for, how were they calculated, and what parameter do you want to compute CIs for? $\endgroup$ – whuber Jan 1 '18 at 21:47
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    $\begingroup$ I think your approach is valid because: stats.stackexchange.com/questions/187154/… $\endgroup$ – Moss Murderer Jan 2 '18 at 1:11
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  • If your data is following normal distribution (can be checked using QQ plot in R),

    you can just square the std deviation and square it's CI to get the variance and CI for variance.

For more understanding, look at the end in : http://www.kean.edu/~fosborne/bstat/06evar.html

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    $\begingroup$ Your reference depends very strongly on the assumed Normality, whereas this question explicitly states that assumption cannot be made. $\endgroup$ – whuber Jan 1 '18 at 21:48

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