When using a sample to calculate a statistic we are estimating a population parameter. It is just an estimate and the sample due to the nature of drawing a sample may not create a value (statistic) that is close to the actual value (parameter).

We can calculate confidence interval about the statistic to determine where the true and often unknown parameter may exist. This includes the calculation of a variance statistic.

If you were to draw many different samples all the same size from a population and plot the variance statistic the resulting distribution is likely to fit a Ï‡^{2}Â distribution. Plotting the means creates a normal distribution which is symmetrical and produced symmetrical confidence intervals. The Ï‡^{2} distribution is not symmetrical and will produce asymmetrical intervals.

## The Formula

The confidence interval formula is

$$ \large\displaystyleÂ \frac{\left( n-1 \right){{s}^{2}}}{\chi _{\frac{\alpha }{2},\text{ }n-1}^{2}}\le {{\sigma }^{2}}\le \frac{\left( n-1 \right){{s}^{2}}}{\chi _{1-\frac{\alpha }{2},\text{ }n-1}^{2}}$$

Where s^{2} is the sample variance and n is the sample size. The degrees of freedom is n-1. Note you will need to enter the Ï‡^{2} table twice once for each side of the interval.

## Example

Letâ€™s say we have 25 samples and have calculated the sample variance to be 47. What is the 90% confidence interval about the variance? In other words within what range is the true population variance likely to exist?

The degrees of freedom is df = 25 – 1 = 24. Thus using the Ï‡^{2} table we find the lower Ï‡^{2} value is 36.42 and the upper is 13.85. Using the above formula we can then calculate the confidence interval.

$$ \large\displaystyleÂ \begin{array}{l}\frac{\left( 25-1 \right)47}{\chi _{\frac{0.1}{2},\text{ 25}-1}^{2}}\le {{\sigma }^{2}}\le \frac{\left( 25-1 \right)47}{\chi _{1-\frac{0.1}{2},\text{ 25}-1}^{2}}\\\frac{\left( 24 \right)47}{\chi _{0.05,\text{ 24}}^{2}}\le {{\sigma }^{2}}\le \frac{\left( 24 \right)47}{\chi _{0.95,\text{ 24}}^{2}}\\\frac{\left( 24 \right)47}{36.42}\le {{\sigma }^{2}}\le \frac{\left( 24 \right)47}{13.85}\\30.97\le {{\sigma }^{2}}\le 81.44\end{array}$$

Be sure you use the sample variance directly. Sometimes you may be given the sample standard deviation- in this case, you have to square that value and use the sample variance in the formula above.

Related:

Confidence Intervals for MTBFÂ (article)

Tolerance Intervals for Normal Distribution Based Set of DataÂ (article)

Point and Interval EstimatesÂ (article)

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