Confidence interval

Definition taught in Confidence intervals, part of Statistical inference

Confidence interval

Type	Definition
Taught in	Confidence intervals
Part	Statistical inference
Properties	4
Worked examples	4
Questions using it	21

Let L and U be statistics. The random interval (L,U) is a confidence interval for \theta with confidence 1-\alpha when

\mathbb{P}_\theta(L<\theta<U)=1-\alpha

for every \theta. A one-sided interval leaves one endpoint infinite.

Properties

One-sample t pivot

Main article: One-sample t pivot

If X_1,\ldots,X_n are independent and \operatorname{N}(\mu,\sigma^2), then

T=\frac{\overline X-\mu}{s/\sqrt n}

has a Student t distribution with n-1 degrees of freedom. A Student variable with m degrees of freedom is Z/\sqrt{V/m} where Z is standard normal, V is chi-squared with m degrees of freedom, and Z and V are independent.

Two-sample t pivot

Main article: Two-sample t pivot

For two independent normal samples with a common variance,

T=\frac{(\overline X_1-\overline X_2)-(\mu_1-\mu_2)}{s_p\sqrt{1/n_1+1/n_2}}

has a t distribution with n_1+n_2-2 degrees of freedom.

Variance-ratio F pivot

Main article: Variance-ratio F pivot

For two independent normal samples,

\frac{s_1^2/\sigma_1^2}{s_2^2/\sigma_2^2}

has the F distribution with (n_1-1,\,n_2-1) degrees of freedom.

Slope t pivot

Main article: Slope t pivot

Under simple linear regression,

\frac{\widehat\beta_1-\beta_1}{S_1}=\frac{\widehat\beta_1-\beta_1}{S/\sqrt{S_{xx}}}

has the t_{n-2} distribution.

Examples

Household fuel spending

Worked example from Confidence intervals

An oil company samples 51 households at random from a city. They spent on average \overline x=1136 dollars a year on petrol and heating oil, with bias-corrected standard deviation s=178 dollars.

Question. Give a 95% confidence interval for the mean spending \mu in the city, and say what the model assumes.

Solution. Model the spending of the 51 households as a random sample from \operatorname{N}(\mu,\sigma^2), both unknown. Random selection justifies independence and a common distribution. Normality is an assumption; with n=51 the central limit theorem makes it matter little for an interval about the mean. With t_{50;0.975}=2.009, the standard error is 178/\sqrt{51}=24.93, and

1136\pm2.009\times24.93=1136\pm50.1=(1085.9,\ 1186.1).

Aspirin tablet weights

Worked example from Confidence intervals

A label states that tablet weights have standard deviation 1.0 mg. A sample of 25 tablets has s=1.3 mg.

Question. Give 95% confidence intervals for \sigma^2 and for \sigma, assuming normal weights. Is the label believable?

Solution. With 24 degrees of freedom, \chi^2_{24;0.025}=12.40 and \chi^2_{24;0.975}=39.36. Since (n-1)s^2=24\times1.69=40.56,

\left(\frac{40.56}{39.36},\ \frac{40.56}{12.40}\right)=(1.03,\ 3.27)

for \sigma^2, and (1.02,\ 1.81) mg for \sigma. The interval for \sigma excludes the stated 1.0 mg, so the data cast doubt on the label at the 95% level.

Two standard errors

Question page: Two standard errors

Let X_1,\ldots,X_n be independent and \operatorname{N}(\mu,1). Find the exact coverage of the interval \overline X\pm 2/\sqrt n, and the multiplier c for which \overline X\pm c/\sqrt n has coverage exactly 99%.

Worked answer (click to show)

\sqrt n(\overline X-\mu) is standard normal, so

\mathbb{P}\bigl(\overline X-2/\sqrt n<\mu<\overline X+2/\sqrt n\bigr)=\mathbb{P}(-2<Z<2)=2\Phi(2)-1=0.9545.

For exactly 99%, 2\Phi(c)-1=0.99, so \Phi(c)=0.995 and c=z_{0.995}=2.576.

The interval for a normal mean

Question page: The interval for a normal mean

Let X_1,\ldots,X_n be independent and \operatorname{N}(\mu,\sigma^2) with \sigma unknown. Starting from the t pivot (\overline X-\mu)/(s/\sqrt n), derive the confidence interval for \mu with confidence 1-\alpha.