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Given two samples $X_1$ and $X_2$ of sizes $n_1$ and $n_2$. I need to test $$ H_0 : EX_1 = EX_2 \\ H_1 : EX_2 < EX_1 $$ with fixed type 1 error $\alpha$ and type 2 error $\beta$. How can I derive the needed $n$ for this?

My thoughts are the following: we can use T test, and calculate $$ {\displaystyle t={\frac {{\overline {X}}_{1}-{\overline {X}}_{2}}{\sqrt {{\frac {s_{1}^{2}}{n_{1}}}+{\frac {s_{2}^{2}}{n_{2}}}}}}.} $$ If $H_0$ is correct then it would be T distribution (which is close to normal distribution). Thus we can from $P(t > c) = \alpha$ calculate threshold $c$ to ensure level of significance $\alpha$ . I suppose I need to derive $n$ (for simplicity assume $n_1 = n_2 = n$) from $\beta$ but how can I do that? Also can I use this test for such (one-sided) task? It looks good when $H_1:EX_2 \neq EX_1$ because in that case because of symmetry of T distribution it would be $P(t>c) = \frac{\alpha}{2}$ , can I use it in general in case of $H_1 : EX_2 < EX_1$?

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Remember the formula for the margin of error=ME for a $2$ sided t confidence interval for $\mu_{1,2}$ with $n_1=n_2=n$:

$$\text{ME}=t^*\ \sqrt{\frac{\text{var}(x_1)+\text{var}(x_2)}{n}}$$

Therefore:

$$n=({\text{var}(x_1)+\text{var}(x_2)})\left(\frac{t^*}{\text{ME}}\right)^2 $$

When you know the margin of error, $3$ percent points for example, the confidence level, $s_1^2$, and $s_2^2$, then you estimate $t^*\approx\text{InvNormCD}\left(\text{area}=\frac{1-C}2\right)$, assuming $n\ge 30$ by the large counts condition:

$$n\ge({\text{var}(x_1)+\text{var}(x_2)})\left(\frac{t^*}{\text{ME}}\right)^2 $$

If $n$ should be less than $30$, then maybe use an estimate

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