Editing Confidence interval (section)

== Example ==
[[File:Confidenceinterval.svg|thumb|200px|In this [[bar chart]], the top ends of the brown bars indicate observed means and the red line segments ("[[Error bar|error bars]]") represent the confidence intervals around them. Although the error bars are shown as symmetric around the means, that is not always the case. In most graphs, the error bars do not represent confidence intervals (e.g., they often represent standard errors or standard deviations).]]

Suppose <math>X_1, \ldots, X_n</math> is an [[statistical independence|independent]] sample from a [[normal distribution|normally distributed]] population with unknown parameters [[mean]] <math>\mu</math> and [[variance]] <math>\sigma^2.</math> Define the [[Sample mean and covariance|sample mean]] <math>\bar{X}</math> and [[Variance#Unbiased sample variance|unbiased sample variance]] <math>S^2</math> as

: <math>\bar{X} = \frac{{X_1 + \cdots + X_n}}{{n}},</math>

: <math>S^2 = \frac{{1}}{{n-1}}\sum_{i=1}^n (X_i - \bar{X})^2.</math>

Then the value

: <math>T = \frac{{\bar{X} - \mu}}{{S/\sqrt{n}}}</math>

has a [[Student's t-distribution|Student's ''t'' distribution]] with <math>n - 1</math> degrees of freedom.<ref>Rees, D.G. (2001). ''Essential Statistics'', 4th Edition, Chapman and Hall/CRC. {{isbn|1-58488-007-4}} (Section 9.5)</ref> This value is useful because its distribution does not depend on the values of the unobservable parameters <math>\mu</math> and <math>\sigma^2</math>; i.e., it is a [[pivotal quantity]].

Suppose we wanted to calculate a 95% confidence interval for <math>\mu.</math> First, let <math>c</math> be the 97.5th [[percentile]] of the distribution of <math>T</math>. Then there is a 2.5% chance that <math>T</math> will be less than <math display="inline">-c</math> and a 2.5% chance that it will be larger than <math display="inline">+c</math> (as the ''t'' distribution is symmetric about 0). In other words,

: <math>P_T(-c \leq T \leq c) = 0.95.</math>

Consequently, by replacing <math display="inline">T</math> with <math>\frac{{\bar{X} - \mu}}{{S/\sqrt{n}}}</math> and re-arranging terms,

: <math>P_X {\left(\bar{X} - \frac{{cS}}{{\sqrt{n}}} \leq \mu \leq \bar{X} + \frac{{cS}}{{\sqrt{n}}}\right)} = 0.95</math>

where <math>P_X</math> is the probability measure for the sample <math>X_1, \ldots, X_n</math>.

It means that there is 95% probability with which this condition <math>\bar{X} - \frac{{cS}}{{\sqrt{n}}} \leq \mu \leq \bar{X} + \frac{{cS}}{{\sqrt{n}}}</math> occurs in repeated sampling. After observing a sample, we find values <math>\bar{x}</math> for <math>\bar{X}</math> and <math>s</math> for <math>S,</math> from which we compute the below interval, and we say it is a 95% confidence interval for the mean.

: <math>\left[\bar{x} - \frac{cs}{\sqrt{n}}, \bar{x} + \frac{cs}{\sqrt{n}}\right].</math>