Confidence Interval for Variance with Examples
In this tutorial we will discuss some numerical examples to understand how to construct a confidence interval for population variance or population standard deviation.
Example 1
The mean replacement time for a random sample of 12 microwaves is 8.6 years with a standard deviation of 3.6 years. Construct a 95% confidence interval for the population standard deviation.
Solution
Here the sample size is $n=27$, sample standard deviation is $s=6.8$. We wish to construct a $95$% confidence interval for population standard deviation $\sigma$.
Step 1 Specify the confidence level $(1-\alpha)$
Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.
Step 2 Given information
Given that sample size $n=27$ and sample standard deviation $s =6.8$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval estimate of population standard deviation $\sigma$ is
$$ \begin{aligned} \bigg(\sqrt{\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}}\bigg) \end{aligned} $$
where $\chi^2_{(\alpha/2,n-1)}$ and $\chi^2_{(1-\alpha/2,n-1)}$ are the critical values from $\chi^2$ distribution with $\alpha$ level of significance and $n-1$ degrees of freedom.
Step 4 Determine the critical value
The critical values of $\chi^2$ for $\alpha$ level of significance and $n-1$ degrees of freedom are $\chi^2_{(\alpha/2,n-1)}=\chi^2_{(0.025,26)}=41.923$ and $\chi^2_{(1-\alpha/2,n-1)}=\chi^2_{(0.975,26)}=13.844$.
Step 5 Determine the confidence interval
$95$% confidence interval estimate for population standard deviation is
$$ \begin{aligned} \sqrt{\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}} &\leq \sigma \leq \sqrt{\frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}}\\ \sqrt{\frac{26*46.24}{41.923}} &\leq \sigma \leq \sqrt{\frac{26*46.24}{13.844}}\\ 5.355 &\leq \sigma \leq 9.319. \end{aligned} $$
Thus $95$% confidence interval for population standard deviation is $(5.355,9.319)$.
We can be $95$% confident that the population standard deviation for the replacement time is between $5.355$ and $9.319$.
Example 2
The percentage rates of home ownership for 8 randomly selected states are listed below. Estimate the population variance and standard deviation for the percentage rate of home ownership with 99% confidence.
66.0, 75.8, 70.9, 73.9, 63.4, 68.5, 73.3, 65.9
Solution
Given that the sample size is $n=8$. The sample variance is given by
$$ \begin{aligned} s^2&=\frac{1}{n-1}\bigg(\sum_{i=1}^nx_i^2-\frac{\big(\sum x_i\big)^2}{n}\bigg)\\ &=\frac{1}{8-1}\bigg(39017.17-\frac{\big(557.7\big)^2}{8}\bigg)\\ &=\frac{1}{7}\bigg(39017.17-\frac{311029.29}{8}\bigg)\\ &=\frac{1}{7}\big(138.5087\big)\\ &=19.787 \end{aligned} $$
and sample standard deviation is $s=\sqrt{19.787}=4.4483$.
We wish to construct a $99$% confidence interval for population variance and population standard deviation $\sigma$.
Step 1 Specify the confidence level $(1-\alpha)$
Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.
Step 2 Given information
Given that sample size $n=8$ and sample variance is $19.787$ and standard deviation $s =4.4483$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval estimate of population variance $\sigma^2$ is
$$ \begin{aligned} \bigg(\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}, \frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}\bigg) \end{aligned} $$
and
$100(1-\alpha)$% confidence interval estimate of population standard deviation $\sigma$ is
$$ \begin{aligned} \bigg(\sqrt{\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}}\bigg) \end{aligned} $$
where $\chi^2_{(\alpha/2,n-1)}$ and $\chi^2_{(1-\alpha/2,n-1)}$ are the critical values from $\chi^2$ distribution with $\alpha$ level of significance and $n-1$ degrees of freedom.
Step 4 Determine the critical value
The critical values of $\chi^2$ for $\alpha$ level of significance and $n-1$ degrees of freedom are $\chi^2_{(\alpha/2,n-1)}=\chi^2_{(0.005,7)}=20.278$ and $\chi^2_{(1-\alpha/2,n-1)}=\chi^2_{(0.995,7)}=0.989$.
Step 5 Determine the confidence interval
$99$% confidence interval estimate for population variance is
$$ \begin{aligned} \frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}} &\leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}\\ \frac{7*19.787}{20.278} &\leq \sigma^2 \leq \frac{7*19.787}{0.989}\\ 6.83 &\leq \sigma^2 \leq 140.049. \end{aligned} $$
Thus $99$% confidence interval for population variance is $(6.83,140.049)$.
We can be $99$% confident that the population variance for the percentage rate of home ownership is between $6.8305$ and $140.0495$.
$99$% confidence interval estimate for population standard deviation is
$$ \begin{aligned} \sqrt{6.83} &\leq \sigma \leq \sqrt{140.049}\\ 2.614 &\leq \sigma \leq 11.834. \end{aligned} $$
Thus $99$% confidence interval for population standard deviation is $(2.614,11.834)$.
We can be $99$% confident that the population standard deviation for the percentage rate of home ownership is between $2.614$ and $11.834$.
Reference
You can read my step by step tutorial on Confidence interval for variance,tutorial will help you to understand how to construct a confidence interval for population variance or population standard deviation.
Use calculator to compute the confidence interval for population variance.
Confidence Interval for Variance
Confidence Interval for population variance calculator
Let me know in the comments if you have any questions on confidence interval for variance examples and your thought on this article.
