# Chi Square Test Calculator for Variance with examples

## Chi square test calculator for variance with Examples

In this tutorial we will discuss a method for testing a claim made about the population variance $\sigma^2$ or population standard deviation $\sigma$. To test the claim about the population variance or population standard deviation we use chi-square test.

We will discuss some numerical examples using six steps approach used in hypothesis testing to test hypothesis about the population variance or population standard deviation.

## Chi square test calculator for variance

The chi square test calculator for testing population variance makes it easy to calculate the test statistic, chi-square critical value and the p-value given the sample information, level of significance and the type of alternative hypothesis (i.e. left-tailed, right-tailed or two-tailed.)

Chi Square test Calculator for variance
Population Standard Deviation ($\sigma$)
Sample Size ($n$)
Sample Standard Deviation ($s$)
Level of Significance ($\alpha$)
Tail : Left tailed
Right tailed
Two tailed
Results
Test Statistics $\chi^2$:
Degrees of Freedom:
$\chi^2$-critical value(s):
p-value:

## How to use chi square test calculator for testing population variance

Step 1 - Enter the population standard deviation $\sigma$.

Step 2 - Enter the sample size $n$

Step 3 - Enter the sample standard deviation $s$

Step 4 - Enter the level of significance $\alpha$

Step 5 - Select the alternative hypothesis (left-tailed / right-tailed / two-tailed)

Step 6 - Click on "Calculate" button to get chi square critical value, test statistics,p-value

## Example -1 Chi square test calculator for variance

A statistician wishes to test the claim that the standard deviation of the weights of firemen is less than 25 pounds. She selected a random sample of 20 firemen and found $s = 23.2$ pounds. Assuming that the weights of firemen are normally distributed, test the claim of the statistician at the 0.05 level of significance.

#### Solution

Given that the sample size $n = 20$ and sample standard deviation $s = 23.2$.

#### Step 1 Hypothesis Problem

The hypothesis testing problem is
$H_0 : \sigma = 25$ against $H_1 : \sigma < 25$ ($\text{left-tailed}$)

#### Step 2 Test Statistic calculation

The test statistic for testing above hypothesis testing problem is

$$\chi^2 =\frac{(n-1)s^2}{\sigma^2}$$
The test statistic follows a chi-square distribution with $n-1$ degrees of freedom.

#### Step 3 Level of Significance

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value

As the alternative hypothesis is $\text{left-tailed}$, the chi square critical value for $\alpha=0.05$ level of significance and $n-1 = 19$ degrees of freedom $\text{ is }$ $\text{10.117}$ (from $\chi^2$ statistical table).

The rejection region (i.e. critical region) is $\chi^2 < 10.117$.

#### Step 5 Test Statistic

The test statistic under the null hypothesis is

 \begin{aligned} \chi^2 &=\frac{(n-1)s^2}{\sigma^2_0}\\ &= \frac{(20-1)*(23.2)^2}{(25)^2}\\ &= 16.362 \end{aligned}

#### Step 6 Decision (Traditional approach)

The chi square test statistic is $\chi^2 =16.362$ which falls $outside$ the critical region, we $\text{fail to reject}$ the null hypothesis.

OR

#### Step 6 Decision (p-value Approach)

This is a $\text{left-tailed}$ test, so the p-value is the area to the left of the test statistic ($\chi^2=16.362$) is p-value = $0.367$.

The p-value is $0.367$ which is $\text{greater than}$ the significance level of $\alpha = 0.05$, we $\text{fail to reject}$ the null hypothesis.

## Example -2 Chi square test Calculation for variance

An engineer is investing the amount of standard deviation in the time it takes a 3D printer to make a particular part. The engineer believes that the standard deviation in the time it takes to make the part is more than 2. Test this at $\alpha = 0.01$ level of significance, using 11 sample times taken while the printer was making these parts. The sample standard deviation is 2.3.

#### Solution

Given that the sample size $n = 11$ and sample standard deviation $s = 2.3$.

#### Step 1 Hypothesis Problem

The hypothesis testing problem is
$H_0 : \sigma = 2$ against $H_1 : \sigma > 2$ ($\text{right-tailed}$)

#### Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

$$\chi^2 =\frac{(n-1)s^2}{\sigma^2}$$
The test statistic follows a chi square distribution with $n-1$ degrees of freedom.

#### Step 3 Level of Significance

The significance level is $\alpha = 0.01$.

#### Step 4 Critical Value

As the alternative hypothesis is $\text{right-tailed}$, the chi square critical value of $\chi^2$ $\text{ is }$ $\text{23.209}$ (from $\chi^2$ statistical table).

The rejection region (i.e. critical region) is $\chi^2 > 23.209$.

#### Step 5 Test Statistic

The test statistic under the null hypothesis is

 \begin{aligned} \chi^2& =\frac{(n-1)s^2}{\sigma^2_0}\\ &= \frac{(11-1)*(2.3)^2}{(2)^2}\\ &= 13.225 \end{aligned}

#### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =13.225$ which falls $outside$ the critical region, we $\text{fail to reject}$ the null hypothesis.

OR

#### Step 6 Decision (p-value Approach)

This is a $\text{right-tailed}$ test, so the p-value is the area to the left of the test statistic ($\chi^2=13.225$) is p-value = $0.2114$.

The p-value is $0.2114$ which is $\text{greater than}$ the significance level of $\alpha = 0.01$, we $\text{fail to reject}$ the null hypothesis.

## Example -3 Chi-square test Calculator for variance

A cigarette manufacturer wishes to test the claim that the variance of nicotine content of its cigarettes is 0.644. Nicotine content is measured in milligrams and is assumed normally distributed. A sample of 20 cigarettes has a standard deviation of 1.00 milligram. At $\alpha = 0.01$, is there enough evidence to reject the manufacturer's claim?

#### Solution

Given that the sample size $n = 20$ and sample standard deviation $s = 1$.

#### Step 1 Hypothesis Problem

The hypothesis testing problem is
$H_0 : \sigma^2 = 0.644$ against $H_1 : \sigma^2 \neq 0.644$ ($\text{two-tailed}$)

#### Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

$$\chi^2 =\frac{(n-1)s^2}{\sigma^2}$$

#### Step 3 Level of Significance

The significance level is $\alpha = 0.01$.

#### Step 4 Critical Value

As the alternative hypothesis is $\text{two-tailed}$, the chi square critical values for $\alpha=0.01$ level of significance and $n-1 = 19$ degrees of freedom $\text{ are }$ $\text{6.844 and 38.582}$ (from $\chi^2$ statistical table).

The rejection region (i.e. critical region) is $\chi^2 < 6.844 or \chi^2 > 38.582$.

#### Step 5 Test Statistic

The test statistic under the null hypothesis is

 \begin{aligned} \chi^2 &=\frac{(n-1)s^2}{\sigma^2_0}\\ &= \frac{(20-1)*(1)^2}{(0.8024961)^2}\\ &= 29.503 \end{aligned}

#### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =29.503$ which falls $outside$ the critical region, we $\text{fail to reject}$ the null hypothesis.

#### Step 6 Decision (p-value Approach)

This is a $\text{two-tailed}$ test, so the p-value is the area to the left of the test statistic ($\chi^2=29.503$) is p-value = $0.0585$.

The p-value is $0.0585$ which is $\text{greater than}$ the significance level of $\alpha = 0.01$, we $\text{fail to reject}$ the null hypothesis.

## Endnote

In this tutorial, you learned the about how to solve numerical examples on chi square statistics test for testing population variance. You also learned about the step by step procedure to apply chi square test for testing population variance and how to use chi square test calculator for testing population variance or population standard deviation to get the value of test statistic, p-value, and z-critical value.