I apologize in advance for my rookie question.
I have been looking in the internet how to get SD from CI values, and some sources tell me to apply the formula =SQRT(N)*(CI_upper.limit - CI_lower.limit)/(T.INV.2T(0,05;N-1)*2) in exel, while others give me this one: = SQRT(N)*(CI_upper.limit - CI_lower.limit)/(tinv(0,05;N-1), which gives me results that are twice as much as the SD I obtain with the first formula.
Which one should I use?
Thanks!
https://stats.stackexchange.com/questions/321119/calculate-standard-deviation-from-sample-size-mean-and-confidence-interval
When you are talking about CI, you need to specify the associated confidence level (CL). I guess the CL associated with your question is 90% so that alpha/2 = (1-CL)/2 = 0.05. Is this correct ?
To solve your problem, you need to know how the CI was calculated, because there are many different possible ways that the CI was calculated. For instance, if the distribution is of normal distribution, and the CI was calculated using t-test, then the SD can be estimated with following equation:
SD = sqrt(n) * (ci_upper - ci_lower)/(2 * tinv (alpha/2; n-1)),
where 'tinv()' is the inverse of Student's T cdf.
On the other hand, if still it is of normal distribution, but a known SD was used in calculating CI, then the SD can be estimated with following equation:
SD = sqrt(n) * (ci_upper - ci_lower)/(sqrt(8) * erfinv (CL)),
where 'erfinv()' is the inverse error function.
in addition, if you know what distribution it is, it may greatly simplify your problem. For instance, if it is of Poisson distribution, then the variance equals its mean; otherwise if it is of Bernoulli distribution, then its variance equals u * (1-u), where u is the mean of the variable.
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