Confidence interval for population variance Calculator
Confidence Interval For Population Variance Calculator
Use below Confidence interval for population variance calculator to calculate degees of freedom,chi-square critical values, confidence limits.
Calculator
| Confidence Interval for Variance Calculator | |
|---|---|
| Sample Size ($n$) | |
| Sample Standard Deviation ($s$) | |
| Confidence Level ($1-\alpha$) | |
| Confidence Interval for Variance Results | |
| Degrees of Freedom : (df) | |
| Chi-square critical value 1: | |
| Chi-square critical value 2: | |
| Lower Confidence Limits: | |
| Upper Confidence Limits: | |
How to use Confidence Interval for Variance Calculator?
Step 1 – Enter the Sample Size (n)
Step 2 – Enter the Sample Standard Deviation (s)
Step 3 – Select Confidence level (90%,95%,98% or 99%)
Step 4 – Click on "Calculate" button to calculate Confidence Interval for variance
Step 5 – Calculate Degrees of Freedom (df)
Step 6 – Calculate Chi-Square critical value 1
Step 7 – Calculate Chi-Square critical value 2
Step 8 – Calculate lower confidence limits
Step 9 – Calculate upper confidence limits
Confidence Interval for Variance Theory
Let $X_1, X_2, \cdots , X_n$ be a random sample of size $n$ from $N(\mu, \sigma^2)$.
Let $\overline{X}=\frac{1}{n} \sum X_i$ be the sample mean and $s^2=\dfrac{1}{n-1}\bigg(\sum_{i=1}^nx_i^2-\dfrac{\big(\sum x_i\big)^2}{n}\bigg)$ be the sample variance.
Let $C=1-\alpha$ be the confidence coefficient. We wish to construct a $100(1-\alpha)$% confidence interval of a population variance $\sigma^2$.
$100(1-\alpha)$% confidence interval estimate of population variance $\sigma^2$ is
$$ \begin{aligned} \bigg(\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}, \frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}\bigg) \end{aligned} $$
$100(1-\alpha)$% confidence interval estimate of population standard deviation $\sigma^2$ is
$$ \begin{aligned} \sqrt{\bigg(\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}}\bigg) \end{aligned} $$
Assumptions
a. The sample is a simple random sample.
b. The population has a normal distribution.
Step by Step procedure
Step by step procedure to estimate confidence interval for population variance $\sigma^2$ is as follows:
Step 1 Specify the confidence level $(1-\alpha)$
Step 2 Given information
Specify the given information, sample size $n$, sample mean $\overline{X}$ and sample variance $s^2$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval estimate of population variance $\sigma^2$ is
$$ \begin{aligned} \bigg(\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}, \frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}\bigg) \end{aligned} $$
Step 4 Determine the critical value
Determine the critical values $\chi^2_L = \chi^2_{(\alpha/2,n-1)}$ and $\chi^2_R = \chi^2_{(1-\alpha/2,n-1)}$ from $\chi^2$ statistical table that corresponds to the desired confidence level and the degrees of freedom.
Step 5 Determine the confidence interval
$100(1-\alpha)$% confidence interval estimate for population variance $\sigma^2$ is
$$ \begin{aligned} \bigg(\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}, \frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}\bigg) \end{aligned} $$
Confidence Interval for Variance Examples with steps
In this tutorial we will discuss some numerical examples to understand how to construct a confidence interval for population variance or population standard deviation with steps by steps procedure.
Example 1 – Confidence Interval for Variance Calculator
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 $\sigma$.
Lets calculate confidence interval for variance with steps
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 of Variance
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 – Confidence Interval for Variance Calculator
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 $\sigma$.
Lets calculate confidence interval for variance with steps
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 for variance
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 $11.834$.
Let me know in the comments if you have any questions on confidence interval for population variance calculator and examples
Further Reading
You can read more on Confidence Interval topic here:
