At the start of the Tversky's 'Belief in the law of small numbers' is a hypothetical replication study that motivates the essay. A full description is provided by this question. But in essence, a two-tailed study result (sample size $n=20$) has $z = 2.23, p<0.05$. What is the probability of a significant result when $n=10$?

According to the essay, the probability of replicating the significant result is $0.473$ and there is an attempt here which returns the documented probability, but it only works if the mean of the second study is not taken as $2.23$ but as $\frac{2.23}{\sqrt{2}}$ because the second sample size is half the first.

But I don't see that rationale for dividing the mean by $\sqrt{2}$ when the sample size is halved.

The only straws I could clutch were for sampling distributions where standard error $se=\sigma/\sqrt{n}$ and $z=\frac{\overline{x}-\mu}{se}$.

This would mean a z-score would become $\frac{z}{\sqrt{2}}$ when the sample size is halved, ceteris paribus. Maybe this would weight the mean downwards for a sample of z-scores?

But I remain unconvinced. Hopefully someone can provide the correct logic (for dividing mean by $\sqrt{2}$ when sample size halves) or declare it nonsensical.

  • $\begingroup$ Welcome to crossvalidated! Z isn't the mean. It's the test statistic. $\endgroup$ Mar 3, 2021 at 23:59
  • 1
    $\begingroup$ @JeremyMiles, thank you for the correction. I took the language from the original footnote which is confusing at best. $\endgroup$
    – NCT
    Mar 4, 2021 at 8:06

1 Answer 1


As you suggested, the mean is divided by $\sqrt(2)$. If $z_1= \frac{x-\mu}{\sqrt{\frac{\sigma^2}{n}}}$, and $z_2=\frac{x-\mu}{\sqrt{\frac{\sigma^2}{\frac{n}{2}}}}$, noting that you can reexpress this equation as $z_2=\frac{x-\mu}{\sqrt{\frac{\sigma^2}{n}}\sqrt{2}}$ then replacing $z_1$ in $z_2$ yield $z_2= \frac{z_1}{\sqrt{2}}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.