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I have a sample of 50 subjects, where every subject completes a task with two possible outcomes (left or right hand use, with 50% probability) 30 times. On an individual level, this leads to a binomial distribution of outcomes for each subject, where a z-score can be calculated for each subject to see how far the hand uses are from the expected 50% (15 left and right hand use) level. Based on the z-scores, we deem those with larger values than 1.96 (›1.96) as right-handed, those with lower values than -1.96 (‹-1.96) as left-handed and those with values in between (-1.96‹‹1.96) as ambilateral.

Question: what is the expected ratio of z-scores (or distribution of z-scores) in my sample of 50 subjects? Is it the same as at the individual level with 95% chance of being ambilateral and 2.5-2.5% as being left or right-handed?

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Let $S_i$, for $i=1,2,\ldots,50$, be the raw score for the $i$th subject, i.e. the number of tasks completed with the right hand. If $S_i \sim \mathrm{Bin}(30, 0.5)$, I'm assuming that you calculate the z-score as $Z_i=(S_i-15)/\sqrt{7.5}$.

The probability an individual is labelled right-handed is $\Pr(Z_i>1.96)=\Pr(S_i>20.37)$, which is around 2.14% (not exactly 2.5% because $S_i$ is discrete), and similarly for left-handed.

The number of people labelled right-handed scores in your sample is $R \sim \mathrm{Bin}(50, 0.0214)$. The expected proportion of right-handers is, thus, 0.214.

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  • $\begingroup$ 2.14% = 0.0214. Did you slip up in your last sentence, or is there a reason you multiplied by 10? $\endgroup$
    – Lukas Lohse
    Commented Jun 12 at 12:05
  • $\begingroup$ Thank you very much for your answer! I guess I made the mistake of assuming that with enough repetitions, that is 30 (on an individual level), the distribution of outcomes will approximate the continuous normal distribution, that's why I assumed 2.5% for a z-score of 1.96. $\endgroup$
    – AnnaB
    Commented Jun 12 at 12:13
  • $\begingroup$ and just to make sure, you used this formula to get to the 2.14% probability, right? Pr(k)=(n! / k!*(n-k)!) * p^k*(1-p)^(n-k), with n=30, k=20.37 (20 here, I guess) and p=0.5? $\endgroup$
    – AnnaB
    Commented Jun 12 at 13:06
  • $\begingroup$ Thanks Lukas! I fixed the typo. $\endgroup$
    – Doctor Milt
    Commented Jun 13 at 12:39
  • $\begingroup$ @AnnaB That's right. I took $\Pr(S_i \geq 21)$. $\endgroup$
    – Doctor Milt
    Commented Jun 13 at 12:40

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