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If [$\ a(y)$, $\ b(y)$] is the 99% confidence interval for the parameter $\sqrt\theta$, the confidence interval for $\theta$ is

  1. [$\sqrt a(y)$, $\sqrt b(y)$]
  2. [$\ a(y)$, $\ b(y)$]
  3. [$\ a(y)^2$, $\ b(y)^2$]

I have no idea how to even start solving this. Do I assume that the parameter is from a Normal Distribution or it need a different approach?

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$\frac{99}{100} = \mathbb P(a(y) \leq \sqrt{\theta} \leq b(y)) = \mathbb P(a(y)^2 \leq \theta \leq b(y)^2)$ (assuming that $a(y),b(y) \geq 0$).

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  • $\begingroup$ Thanks, thought it would be harder than that. $\endgroup$ – user120747 Feb 13 '18 at 16:39

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