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 $\chi^2_{(1-\alpha/2,n-1)}=\chi^2_{(0.975,26)}=13.844$.

Critical values
Critical Values of of Chi-square

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:

  1. Confidence Interval for Variance Examples

  2. Confidence Interval for Mean Calculator

  3. Confidence Interval for mean examples

  4. Chi Square test calculator for variance

  5. Chi-Square test variance

  6. Normal Approximation Calculator

  7. Mean Median Mode Grouped Data Calculator

  8. Normal Approximation Calculator Examples

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

Leave a Comment