Definition:Confidence Interval
Definition
Definition 1
Let $\theta$ be a population parameter of some population.
Let $X$ be a random sample from this population.
Let $I = \openint {\map f X} {\map g X}$ for some real-valued functions $f$, $g$.
$I$ is said to be a $100 \gamma \%$ confidence interval for $\theta$ if:
- $\map \Pr {\theta \in I} = \gamma$
where $0 < \gamma < 1$.
Definition 2
Let $X$ be a random variable.
Let $\theta$ be a population parameter of $X$ whose distribution is unknown.
A $100 \paren {1 - \alpha}$ percent confidence interval for $\theta$ is an interval formed by a rule which ensures that, in the long run, $100 \paren {1 - \alpha}$ percent of such intervals will include $\theta$.
This confidence interval is derived from the information obtained from a random sample of observations of $X$.
 | This page needs the help of a knowledgeable authority. In particular: It is unclear at this stage whether these definitions are equivalent, or can be massaged into being equivalent. The first comes from 2011: Morris H. DeGroot and Mark J. Schervish: Probability and Statistics (4th ed.), the second comes from 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) etc. If you are knowledgeable in this area, then you can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Help}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Confidence Level
Consider a $100 \paren {1 - \alpha}$ percent confidence interval for $\theta$.
The confidence level associated with this confidence interval is the coefficient $100 \paren {1 - \alpha}$.
Confidence Limit
The endpoints of a confidence interval are known as confidence limits.
Examples
$95 \%$ Confidence Interval
A $95 \%$ confidence interval is a confidence interval whose $\alpha$ parameter is:
- $\alpha = 0 \cdotp 05$
Let $\bar x$ be the mean of a sample of $n$ observations from a normal distribution with unknown mean $\mu$ and known standard deviation $\sigma$.
Then a $95 \%$ confidence interval for $\mu$ is:
- $\closedint {\bar x - \dfrac {1 \cdotp 96 \sigma} {\sqrt n} } {\bar x + \dfrac {1 \cdotp 96 \sigma} {\sqrt n} }$
Motivation
A $100 \paren {1 - \alpha}$ percent confidence interval for a population parameter $\theta$ derived from a given sample covers all values of $\theta_0$ of that parameter that would be accepted at significance level $\alpha$ in a hypothesis test of:
- the null hypothesis $H_0: \theta = \theta_0$
against:
- the alternative hypothesis $H_1: \theta \ne \theta_0$.
Also see
- Results about confidence intervals can be found here.