Find the powers of the z-test

Two independent samples ${x_1, x_2, · · · , x_n}$ from population $N(µ_1, σ^2_1)$ and ${y_1, y_2, · · · , y_n}$ from population $N(µ_2, σ^2_2)$ respectively. The hypotheses $H_0 : µ_1 = µ_2$ vs $H_a : µ_1 > µ_2$
Assume $σ_1 = σ_2 = σ = 1.$
Find the powers of the z-test in the following cases: (i) the powers to detect a difference of $µ_1 − µ_2 = 0.1$ and sample sizes are n = 9, 25, 100 respectively

We have a significance level 0.05. Hint: $\bar{x} − \bar{y} ∼ N(µ_1 − µ_2, \frac{2σ^2}{n})$.

This is what i have tried: I used the formula $\frac{\bar{x}-\mu_0}{\sigma/\sqrt{n}} \leq 1.645$; since it is a one-tailed test.
then plug in $\frac{(\bar{x} − \bar{y})-(\mu_1-\mu_2)}{2\sigma^2/n/\sqrt{n}}$
I compute n=9 first
and got $\frac{(\bar{x} − \bar{y})-(0.1)}{2/27} \leq 1.645$
$(\bar{x} − \bar{y})\leq 0.22185$
Look up the value in z-table I found the probability of type II error is .5871
the power of test then is 1-0.5871 = 0.4126
I'm not sure if I have done this right because when I tried n = 25, 100, the power is supposed to get bigger when sample size increase but it doesn't. I'm not sure how I did it wrong.

• Where is the $27$ coming from? – spaceisdarkgreen Oct 3 '17 at 23:59
• It is from n=9. 2$\sigma^2/n/\sqrt{n} = \frac{2}{27}$ – PiCubed Oct 4 '17 at 0:51

You do pick the cutoff so that you reject whenever $$\bar x - \bar y \ge \frac{\sigma}{\sqrt n}(1.645).$$ When the effect size is $m$ the probability of rejection is $$P(\bar x - \bar y \ge \frac{\sqrt{n}}{\sigma}(1.654)) = P\left(\frac{\bar x -\bar y-m}{(\sigma/\sqrt{n})}\ge \frac{\frac{\sigma}{\sqrt n}(1.654)-m}{(\sigma/\sqrt n)}\right) = 1-\Phi\left(1.654 -\frac{m\sqrt n}{\sigma}\right).$$
For $m=0.1,$ $n=9,$ and $\sigma = 1,$ get $1-\Phi(1.354) = 6.25\%.$