1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confidence interval

  1. Sep 11, 2008 #1
    Let's say we know this:

    \sqrt{n}\left(\widehat{\theta} - \theta\right) \sim \mathcal{N}\left(0, \frac{1}{F(\theta)}\right)

    How do we get from this information to this expression of confidence interval for [itex]\theta[/itex]?

    \left( \widehat{\theta} \pm u_{1-\frac{\alpha}{2}}\frac{1}{\sqrt{nF\left(\widehat{\theta}\right)}}\right)

    Where [itex]u_{1-\frac{\alpha}{2}}[/itex] is appropriate quantil of standard normal distribution.

    Thank you.
  2. jcsd
  3. Sep 11, 2008 #2


    User Avatar
    Homework Helper

    If [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

    \Pr\left(-u < \sqrt{n F(\theta)} \left(\hat \theta - \theta\right) < u)

    because of your stated approximate normality result. That means the event

    -u < \sqrt{n F(\theta)} \left(\hat \theta - \theta\right) < u

    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.)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Confidence interval
  1. Confidence Intervals (Replies: 1)

  2. Confidence Interval (Replies: 0)

  3. Confidence interval (Replies: 0)

  4. Confidence Intervals (Replies: 5)

  5. Confidence interval (Replies: 0)