# 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 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 