That is not possible if you do not have a clue about the variation our the mean.. google standard deviation and look at the formula and you will understand. This is very basic descriptive statistics.. Or do you mean approximating sampling variance from a test statistic (that incorporates variance?), for use in e.g. meta-analysis?

That is not possible if you do not have a clue about the variation our the mean.. google standard deviation and look at the formula and you will understand. This is very basic descriptive statistics.. Or do you mean approximating sampling variance from a test statistic (that incorporates variance?), for use in e.g. meta-analysis?

Here is the formula to calculate SD from SEM and sample size.

SDc = SEMc × √nc and SDrx =SEMrx ×√nrx

Where nc and nrx refer to the number of Sample size in the control and treatment group respectively.

Yes sure for SEM, but he says MEAN.. not SE..

Thank you.

Standard deviation is positive square root of the variance. Knowing only mean and sample size is not enough, we need know the values of individual observations. Can you give a specific example so that the situation for this question will be more clear.

It is not possible, one must have access to the full data set in order to obtain information on variation around the mean.

Thank you.

Paul -

Hi.

If you have sample size, and mean of that sample, this will not help to obtain an estimate of the data distribution's standard deviation. However, you may have seen these terms together, because there is an important relationship to be found here. If you did have an estimated standard deviation of the population using a sample, then from that and the sample size, you can obtain an estimated standard error for your estimated mean. As suggested above, you can do an Internet search on these terms, and many statistics books will show you this relationship.

When estimating a mean or total, the variability of that then depends on the sample size and the variability of the data. In regression, something similar holds true, but that involves the use of the term "MSE," used in a different sense than in the context of the above. I won't get into that here, but just caution you that, as in many other fields, terminology can get confusing. I theorize that better terminology would be in use, with less use of the same or similar term for somewhat different purposes, if the scientific community could go back in history now and then, and clean up its notes. :-)

Cheers - Jim

Hi Paul, this is an explanation from Cochrane Handbook. Hope it answer your question.

http://handbook.cochrane.org/chapter_7/7_7_3_data_extraction_for_continuous_outcomes.htm

Check 7.7.3.2 and 7.7.3.3

Good luck

I have the same problem, sadly some authours reported in their studies only figures or as mentioned in the previous question ( sample size and mean). I wonder if you can still do meta-analysis without knowing the standard deviation, especially if your meta-analysis is means based ? anyway, if the p-value is reported in your selected studies like T-test ,, there is an option in the meta-analysis where you can only provide the sample size for the control and experimental groups plus the p-value . in this case you'll be able to meta-analyze what ever you have.

Hi Hamzeh, 

You can still do meta-analysis of means without SD using response ratio to calculate overall effect sizes. The problem you'll have is calculating the variation around the mean. The options you have are: 1) weigth the studies by nonparametric variance:

V= (ne+nc) / (ne*nc)

where ne and nc are experimental and control sample size respectively.

2) unweighted meta-analysis with CI calculated by bootstrapping.

I hope this is useful.

Hi,

Please, see

Wan et al. BMCMedical Research Methodology 2014, 14:135

http://www.biomedcentral.com/1471-2288/14/135

Thank you

 I got you homes, Since you have the mean and the n, this gives the the p value if you divide the n with the P. once you get the p find q which is complimentary. Then multiple n(p)(q) to get the variance. Square root the variance and you find the Standard deviation.

 I have the same problem so, in conclusion, we could not calculate the SD from mean and sample size, exclude the study??

Dear Hamzeh

I have same problem!

http://www.calculator.net/standard-deviation-calculator.html

Standard Deviation Calculator

Standard Deviation Calculator

https://www.easycalculation.com/statistics/standard-deviation.php

Standard Deviation Calculator

https://www.mathsisfun.com/data/standard-deviation-calculator.html

Calculating standard deviation step by step

https://www.khanacademy.org/math/probability/data-distributions-a1/summarizing-spread-distributions/a/calculating-standard-deviation-step-by-step

Standard deviations and standard errors

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1255808/

The mean and standard deviation: what does it all mean?

http://www.sciencedirect.com/science/article/pii/S0022480404000551?via%3Dihub#

How to Calculate a Sample Standard Deviation

https://www.thoughtco.com/calculate-a-sample-standard-deviation-3126345

https://explorable.com/calculate-standard-deviation

How to Interpret Standard Deviation and Standard Error in Survey Research

https://www.greenbook.org/marketing-research/how-to-interpret-standard-deviation-and-standard-error-in-survey-research-03377

Assuming that this is a binomial experiment (e.g. pass/fail, yes/no), a standard deviation can be determined.

sigmaPhat = sqrt((pi(1-pi)/n) where,

sigmaPhat = the sample size standard deviation

pi = the mean of the sample size

n = the number of samples

So, if you only have the mean and the number of samples, you can determine the standard deviation for binomial experiments.

Hi Hamzeh, 

You can still do meta-analysis of means without SD using response ratio to calculate overall effect sizes. The problem you'll have is calculating the variation around the mean. The options you have are: 1) weigth the studies by nonparametric variance:

V= (ne+nc) / (ne*nc)

where ne and nc are experimental and control sample size respectively.

2) unweighted meta-analysis with CI calculated by bootstrapping.

I hope this is useful.

Answer

If the outcome is binomial you don't need a measure of variance. Therefore I presume you are talking about a continuous outcome. Meta-analysis of continuous outcomes usually weight the trials by 1/variance providing more weight to trials with smaller variance. Thus a measure of variance is essential. Potential sources include SD, SE, p-value, or CI. They can also be plucked (less reliably) from graphs, using pixel counting techniques - just be sure you know whether the error bar is for the SE or CI.

Dear Paul Jebaraj P:

For continuous measures you can calculate the standard deviations from either one of these values:

1. the standard error of each arm.

2. the confidence intervals of each arm

3. the confidence intervals of the pool mean difference between arms.

You can check out how to do it either by using the math formulas from the Cochrane Handbook or by using the RevMan Calculator.

Here the URL respectively:

A. "7.7.3.2 Obtaining standard deviations from standard errors and confidence intervals for group means"

https://handbook-5-1.cochrane.org/chapter_7/7_7_3_2_obtaining_standard_deviations_from_standard_errors_and.htm

B. "Part 3: Calculating missing standard deviations"

https://training.cochrane.org/resource/entering-data-revman-calculator

If you only have the median and quartiles you could find a guide from

Wan X et al. BMC Medical Research Methodology201414:135

Hope it still will help.

Mario.

i have the same problem but i still didn't get how to do it