My solution doesn't match the one given in my course, however I can't quite see what I've done wrong. Can someone give me a heads-up?
Problem
Given the following:
- $y: N(2,3)$
- $z: \chi^2(7 d.f.)$
$$P((y-2)^2>a^2z)=0.9$$ Find the critical value for $a$. Rearranging the equation: $$P(3\frac{(y-2)}{3}>a\sqrt{\frac{z}{7}}\sqrt{7})=0.9$$
Now y got standarised, so i substitute it to $y_s=\frac{(y-2)}{3}$ $$P(\frac{y_s}{\sqrt{\frac{z}{7}}}>a\frac{\sqrt{7}}{3})=0.9$$
My book states that $\frac{y_s}{\sqrt{\frac{z}{7}}}$ now is a $t$-distribution with ($7 d.f.$) Using an online $t$-distribution calculator i get the value $1.415$ for $t_{0.9}(7 d.f.)$ Because of the symmetry, we can invert the sign $$P(-1.415\frac{3}{\sqrt{7}}<a)=0.1$$ $$\leftrightarrow P(-1.6045<a)=0.1$$ So I'd conclude that for the critical value $a=-1.6045$
However the given solution says: $a= 0.1477$
HINT: You made a mistake in this step:
$$P((y-2)^2>a^2z)=0.9 \implies P\left(3\frac{(y-2)}{3}>a\sqrt{\frac{z}{7}}\sqrt{7}\right)=0.9$$
What you should have found was:
$$P((y-2)^2>a^2z)=0.9 \implies P\left(3\left|\frac{(y-2)}{3}\right|>\left|a\right|\sqrt{\frac{z}{7}}\sqrt{7}\right)=0.9$$
If we simplify this and assume $a\geq 0$, then we get:
$$P\left(\left|3\frac{(y-2)}{3}\right|>a\sqrt{\frac{z}{7}}\sqrt{7}\right) =0.9 \implies P\left(-a\sqrt{\frac{z}{7}}\sqrt{7} > 3\frac{(y-2)}{3}\;\textrm{or}\;3\frac{(y-2)}{3}>a\sqrt{\frac{z}{7}}\sqrt{7}\right) =0.9 $$
$$\implies P\left(-a\frac{\sqrt{7}}{3} > \frac{y_s}{\sqrt{\frac{z}{7}}}\;\textrm{or}\;\frac{y_s}{\sqrt{\frac{z}{7}}}>a\frac{\sqrt{7}}{3}\right) = 0.9$$
As your book says $\frac{y_s}{\sqrt{\frac{z}{7}}} \sim t_7$, so by symmetry, this is equivalent to finding $a$ such that:
$$2P\left(\frac{y_s}{\sqrt{\frac{z}{7}}}<-a\frac{\sqrt{7}}{3}\right) = 0.9$$
Can you take it from there?