Confidence Interval for a function of a parameter

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SUMMARY

This discussion focuses on calculating the confidence interval (CI) for the square of the proportion parameter (p^2) in a binomial model. It establishes that for large sample sizes, the sample proportion (p̂) approximates a normal distribution, which allows for the derivation of a normal distribution for p̂^2. The method for obtaining an approximate CI for p^2 is based on the continuous function f(x) = x^2, utilizing the convergence in distribution principle as outlined in statistical texts such as Hogg and Craig.

PREREQUISITES
  • Understanding of binomial models and sample proportions
  • Familiarity with normal distribution and its properties
  • Knowledge of confidence intervals and their calculation
  • Basic calculus, specifically the concept of derivatives
NEXT STEPS
  • Study the derivation of confidence intervals for binomial proportions
  • Learn about the Central Limit Theorem and its implications for sample distributions
  • Explore the method of delta for approximating distributions of functions of random variables
  • Review statistical texts such as Hogg and Craig for deeper insights into confidence intervals
USEFUL FOR

Statisticians, data analysts, and researchers involved in statistical modeling and inference, particularly those working with binomial data and confidence intervals.

alexhleb366
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I'm not sure of your math-stat background in this problem, so bear with me.

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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