- **Calculus & Beyond Homework**
(*http://www.physicsforums.com/forumdisplay.php?f=156*)

- - **Confidence interval**
(*http://www.physicsforums.com/showthread.php?t=255541*)

Confidence intervalLet's say we know this:
[tex] \sqrt{n}\left(\widehat{\theta} - \theta\right) \sim \mathcal{N}\left(0, \frac{1}{F(\theta)}\right) [/tex] How do we get from this information to this expression of confidence interval for [itex]\theta[/itex]? [tex] \left( \widehat{\theta} \pm u_{1-\frac{\alpha}{2}}\frac{1}{\sqrt{nF\left(\widehat{\theta}\right)}}\right ) [/tex] Where [itex]u_{1-\frac{\alpha}{2}}[/itex] is appropriate quantil of standard normal distribution. Thank you. |

Re: Confidence intervalIf [tex] a [/tex] is the value from [tex] Z [/tex] (standard normal) with area [tex] {\alpha}/2[/tex] to its right, you know the value of
[tex] \Pr\left(-u < \sqrt{n F(\theta)} \left(\hat \theta - \theta\right) < u) [/tex] because of your stated approximate normality result. That means the event [tex] -u < \sqrt{n F(\theta)} \left(\hat \theta - \theta\right) < u [/tex] has a known probability of occurring. What can you do with this inequality? (Try some work and include it with your next question if you are unsure.) |

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