If $X_i$ and $Y_j$ are normal distributions where $i = 1,...,n$ and $j= 1,...,n$ with different $\mu$ but same $\sigma^2$, and $\mu_x$ - $\mu_y$ = $\sigma/3$ what is common sample size $n$ needed to be 95% certain of correctly selecting group with higher mean?

What means common sample size? What test we use (t or Z?) to get 95% confidence?

  • $\begingroup$ I think you mean X(i) and Y(j) are normally distributed? (not themselves normal distributions?). In theory, you could pick n members of X and m members of Y, compute the means, and be 95% confident they are different. In this case, however, you're restricted to choosing an equal number of members from both, so you'd choose the higher of m and n. $\endgroup$ – barrycarter Feb 22 '16 at 0:57
  • $\begingroup$ The uncertainty in a mean depends only on the standard deviation, not on the value of the mean itself. $\endgroup$ – barrycarter Feb 22 '16 at 15:37

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