- #1

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the case i'm particularly interested in is an approximate CI for p^2

where p is the proportion in the binomial model.

http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval

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- Thread starter alexhleb366
- Start date

- #1

- 1

- 0

the case i'm particularly interested in is an approximate CI for p^2

where p is the proportion in the binomial model.

http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval

- #2

statdad

Homework Helper

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For a big sample size [tex] \hat p [/tex] has approximately a normal distribution, right? You can approximate the distribution of [tex] \hat{p}^2 [/tex] (it will also turn out to be a normal distribution - look in (say) Hogg/Craig or any introductory math stat book for the idea, or write back and I can put the method here), and then you can get an approximate confidence interval for [tex] p^2 [/tex]

Note - just so I don't have to post it:

If an estimate [tex] X_n [/tex] for some parameter [tex] \theta [/tex] satisfies

[tex]

\sqrt n \left(X_n - \theta \right) \sim n(0, \sigma^2)

[/tex]

(the [tex] \sim [/tex] means "tends to a normal distribution as [tex] n \to \infty [/tex] - i.e., it represents convergence in distribution)

then for a function [tex] f [/tex] that is continuous and has a non-zero derivative at [tex] \theta [/tex] it is true that

[tex]

\sqrt{n} \left(f(X_n) - f(\theta)\right) \sim n(0, \sigma^2 f'(\theta) \right)

[/tex]

Your statistic is the sample proportion, the parameter is [tex] p [/tex], and the function is [tex] f(x) = x^2 [/tex]

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