# In a replication study, if the sample size decreases does the mean decrease as well?

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.

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

## 1 Answer

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 $$z_2= \frac{z_1}{\sqrt{2}}$$.