Let $X_1, . . . , X_n$ be independent N(θ, $σ^2$) random variables. Determine t such that the test δ = $\chi[\bar{X}< a - t\sqrt{\frac{\sigma^2}{n}}] + \chi[\bar{X}> b + t\sqrt{\frac{\sigma^2}{n}}]$ for testing $H_0$ : θ ∈ [a, b] versus $H_1$ : $θ \not{\in} [a, b]$ has size .05.
I think this test has power function;
$\pi_{\delta}(\theta)=P_{\theta}(Z < \frac{a-\theta}{\sigma}\sqrt{n}-t) + P_{\theta}(Z > \frac{b-\theta}{\sigma}\sqrt{n}+t) $
Now the size is defined to be the sup over all $\theta \in [a,b]$ and here I'm stuck. I think this depends very much on the values of a,b; perhaps something like min{|a|,|b|} but this is just a guess based on doodles and I can't begin to think how to show something like this precisely.
Currently trying to write $\pi(\theta)=\int_{-\infty}^{\alpha(\theta)}\phi(x)dx - \int_{\beta(\theta)}^\infty\phi(x)dx$ with $\alpha(\theta)=\frac{a-\theta}{\sigma}\sqrt{n}-t, \beta(\theta)=\frac{b-\theta}{\sigma}\sqrt{n}+t$ So that $$\frac{d}{d\theta}\pi_{\delta}(\theta)=[\phi(\alpha(\theta))-\phi(\beta(\theta))](-\sqrt{\frac{n}{\sigma^2}})=0$$ iff $$\alpha(\theta)^2=\beta(\theta)^2$$
Is this the right approach? I'm suspicious for 2 reasons;
1) the quadratic that results from this suggests the critical values of $\theta$ depends on t, which intuitively I don't think that it can
2) the calculation is suspiciously long and tedious
$\pi(\theta)$ is a probability, so it is bounded by 1, but in the interval $[a,b]$ you can find another sharper bound (that's what you are looking for.) Note that whenever $\theta\to\infty$ or $\theta\to-\infty$, $\pi(\theta)\to 1$. This suggests that $\pi(\theta)$ has a U shape, with a minimum hopefully in $[a,b]$.
Let say that we prove or we are convinced that the minimum is always in $[a,b]$ and that in fact $\pi(\theta)$ has a U shape (I haven prove it, but used my intuition an R to convince myself.) Now, the supremum has to be on the border, i.e., $\theta=a$ or $\theta=b$. But, interestingly $$ \pi(a)=\pi(b). $$ (Check this! I suppose that $t\geq 0$ and used that $P(Z\geq z)=P(Z\leq -z)$, $z\geq 0$, $Z\sim N(0,1)$.) Therefore, the supremum is $\pi(a)=\pi(b)$.
Finally, to find $t$ you will have that $\pi(a)=P(Z\leq \text{something that depends on t})=0.05$, and you can find the value of $t\geq 0$ that satisfies that.
To see that indeed $\pi(\theta)$ has a U shape centered at the middle $\theta_m=(a+b)/2$, just observe by simple substitution that (I always assume $t\geq 0$) $$ \alpha(\theta_m) = -\beta(\theta_m). $$ Since $\phi(x)=\phi(-x)$, then $\frac{d}{d\theta}\pi(\theta)\Big|_{\theta=\theta_m}=0$.
Now, let $\theta(\epsilon)=\theta_m+\epsilon$, $\epsilon>0$. In this case $$ \frac{d}{d\theta}\pi(\theta)\Big|_{\theta=\theta(\epsilon)}=C\left(\phi\left(\alpha_0-\epsilon\sqrt{n/\sigma^2}\right)-\phi\left(-\alpha_0-\epsilon\sqrt{n/\sigma^2}\right)\right), $$ where $C<0$ and $\alpha_0=\alpha(\theta_m)<0$. Since $-\alpha_0>0$, a patient observation to this expresion confirms that the derivative is positive for any $\epsilon$. On the other hand, redefining $\theta(\epsilon)=\theta_m-\epsilon$ with $\epsilon>0$, one gets $$ \frac{d}{d\theta}\pi(\theta)\Big|_{\theta=\theta(\epsilon)}=C\left(\phi\left(\alpha_0+\epsilon\sqrt{n/\sigma^2}\right)-\phi\left(-\alpha_0+\epsilon\sqrt{n/\sigma^2}\right)\right), $$ and the same conclusion can be drawn, just that the derivative is negative.
In conclusion, in the middle, $\pi(\theta)$ reaches its minimum, and moving to the right is increasing and to the left decreasing, i.e., it has a U shape. So the supremum is reached in either $a$ or $b$.