Interval for a normal variance
Section of Confidence intervals, part of Statistical inference
Interval for a normal variance
| Type | Section |
| Taught in | Confidence intervals |
| Part | Statistical inference |
The variance uses the chi-squared pivot (n-1)s^2/\sigma^2, which is \chi^2_{n-1} by the normal sample theorem. Writing \chi^2_{m;q} for its quantile of order q,
\mathbb{P}\Bigl(\chi^2_{n-1;\,\alpha/2}<\frac{(n-1)s^2}{\sigma^2}<\chi^2_{n-1;\,1-\alpha/2}\Bigr)=1-\alpha.
Solving for \sigma^2 gives the confidence interval
\left(\frac{(n-1)s^2}{\chi^2_{n-1;\,1-\alpha/2}},\;\frac{(n-1)s^2}{\chi^2_{n-1;\,\alpha/2}}\right).
The upper quantile sits in the lower endpoint, because the pivot has \sigma^2 in the denominator. The chi-squared law is skewed, so the interval is not symmetric about s^2. An interval for \sigma is found by taking square roots of both endpoints; a monotone function carries the coverage with it.
Example 1 (Aspirin tablet weights) 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.
Used in
See also
Confidence interval · Pivot · One-sample t pivot · Pooled variance · Two-sample t pivot · Variance-ratio F pivot · Wald interval · Coverage