Chi-square test for variance 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 Std. Deviation ($\sigma$) | |
Sample Size ($n$) | |
Sample Std. 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 the result
Chi-square test for variance Example 1
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
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 critical value of $\chi^2$ 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 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.
Chi-square test for variance Example 2
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 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.
Chi-square test for variance Example 3
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 critical values of $\chi^2$ 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 $\chi^2$-test for testing population variance. You also learned about the step by step procedure to apply $\chi^2$-test for testing population variance and how to use $\chi^2$-test calculator for testing population variance or population standard deviation to get the value of test statistic, p-value, and z-critical value.
To learn more about other hypothesis testing problems, hypothesis testing calculators and step by step procedure, please refer to the following tutorials:
Let me know in the comments if you have any questions on $\chi^2$-test calculator for variance with examples and your thought on this article.