# Deriving a $(1-\alpha)$100% confidence interval for $\theta$ pivotal quantity

In the question, $u = \frac{\sqrt{n}(\bar{y}-\theta)}{\sqrt{\theta}}$ is given, where $\theta$ is the variable and unlike the other questions, I can't find a way to solve for $\theta$.

• Welcome to stats.SE! Please take a moment to view our tour. Thank you for using the self-study. It would seem that $\sqrt{\theta}u=\sqrt{n}\left(\overline{y}-\theta \right)$ would be a good place to start, but to be completely honest, I find that software such as matematica (professional software that may be availiable at your school), wolfram alpha (free, but limited in scope and ability to direct), or wXmaxima (similar freeware) are worth learning to use for such situations. – Tavrock Feb 28 '17 at 22:38
• @mager You really should show some kind of attempt or explain what steps you took to try to identify a solution. At the very least, did you take some feasible values for $\bar{y}$ and $n$ and draw a picture of the relationship between $\theta$ and $u$? Did that suggest any approach to you? – Glen_b Mar 1 '17 at 0:13

1. If you're finding an interval for $\theta$ you should strictly be dealing with an inequality in $\theta$ (the interval for the pivotal quantity would be all the values for that where a particular inequality is true; you're trying to get the corresponding inequality for $\theta$).
However, you can deal with the equation part of the inequality if you're careful to check that what you think is the interior of the resulting interval actually is (it's no good just assuming that a solution to the equation of $\theta=\theta_L,\theta_U$ means that $\theta_L<\theta<\theta_U$ is true; in some situations it might just turn out that the resulting interval is actually in two parts $\theta<\theta_L$ and $\theta>\theta_U$). In some situations there can be other things you might need to be concerned about -- jump discontinuities, for example, might in some situaitons cause problems if you're trying to find a root.
2. If you multiply through by $\sqrt\theta$ (as Tavrock indicated), it's quite straightforward to solve, since it's quadratic in $\sqrt{\theta}$.