we know that in order to calculate the confidence interval at 95% we can do mean +/- Z95*SD/sqrt(N) using the mean of the sample data and StdDev of the data with N equals to the data size

But if the mean is known and the standard deviation is known, and we want to check if our data sample is valid using 100 simulations for instance is it correct to say that if we run a simulation 100 times, we should expect 95 of those times for the mean of the sample to be between:

knownMean - 1.96knownStdDev/sqrt(N) < mean of sample < knownMean + 1.96knownStdDev/sqrt(N)

is this equivalent to cheking the following?

sampleMean - 1.96sampleSD/sqrt(N) < known Mean < sampleMean + 1.96sampleSD/sqrt(N)

Are both equivalent ways of testing the validity of the simulated data? And if not, why not?

The distribution is known to be normal with a known mean and std dev

  • $\begingroup$ An unstated assumption here is that data are normal. $\endgroup$
    – BruceET
    Commented Jan 27, 2022 at 8:21

1 Answer 1


Simply subtracting/adding the 1.96... term on both sides will show you that both equations are identical to the condition $$|\mu - \overline{x}| < z_{1-\alpha/2}\sigma/\sqrt{n}$$ So yes, both conditions are equivalent.


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