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

Question

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.

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