F test for two variances
In this tutorial we will discuss some examples on F test for comparing two variances or standard deviations.
F test calculator for two variances
The Ftest calculator for testing two population variances makes it easy to calculate the test statistic, F critical value and the p value given the sample information, level of significance and the type of alternative hypothesis (i.e. lefttailed, righttailed or twotailed.)
F test Calculator for two variances  

Sample 1  Sample 2  
Sample Size  
Standard Deviation  
Level of Significance ($\alpha$)  
Tail  Left tailed Right tailed Two tailed 

Results  
Test Statistics F:  
Degrees of Freedom: (df1)  
Degrees of Freedom: (df2)  
Fcritical value(s):  
pvalue:  
How to use F test calculator for two variances?
Step 1 – Enter the sample size for first sample $n_1$ and second sample $n_2$
Step 2 – Enter the sample standard deviations for first sample $s_1$ and second sample $s_2$
Step 3 – Enter the level of significance $\alpha$
Step 4 – Select the alternative hypothesis (lefttailed / righttailed / twotailed)
Step 5 – Click on "Calculate" button
Step 6 – Calculate p value from f statistics
Step 7 – Calculate F critical value
Example 1 F test Calculator for two variances
A professor from a graduate school claims that there is less variability in the final exam scores of students taking the statistics major than the students taking the mathematics major.
Random samples of 13 mathematics majors and 10 statistics majors are selected from the his large class, and the following results are computed based on the final exam scores:
Mathematics : $n_1= 13$, $s_1^2 = 45$
Statistics: $n_2 = 10$, $s_2^2 = 35.5$
At the 0.05 level of significance, is there evidence to support the professor’s claim?
Solution
Let $X$ denote the final exam scores of students taking the mathematics major and $Y$ denote the final exam scores of students taking the statistics major.
.  Mathematics Major  Statistics Major 

Sample size  $n_1= 13$  $n_2=10$ 
Sample variance  $s_1^2 = 45$  $s_2^2 = 35.5$ 
Step 1 Hypothesis testing problem
The hypothesis testing problem is
$H_0 : \sigma^2_1 = \sigma^2_2$ against $H_1 : \sigma^2_1 >\sigma^2_2$ ($\textit{righttailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} F=\frac{s_1^2}{s_2^2} \end{aligned} $$
The test statistic F follows F distribution with $n_11= 131 = 12$ and $n_21=101= 9$ degrees of freedom.
Step 3 Level of significance
The significance level is $\alpha = 0.05$.
Step 4 Critical value(s)
As the alternative hypothesis is $\textit{righttailed}$, the critical value of F for $12$ and $9$ degrees of
freedom and $\alpha = 0.05$ level of significance $\text{is}$ $\text{3.073}$.
The rejection region (i.e. critical region) is $\text{F > 3.073}$.
Step 5 Computation
The test statistic under the null hypothesis is
$$ \begin{aligned} F_{obs} &=\frac{s_1^2}{s_2^2}\\ &= \frac{45}{35.5}\\ &= 1.2676 \end{aligned} $$
Step 6 Decision (Traditional approach)
The test statistic is $F_{obs} =1.2676$ which falls $\textit{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision (pvalue approach)
This is a $\textit{righttailed}$ test, so the pvalue is the area to the left of the test statistic ($F_{obs}=1.2676$) is pvalue = $0.3674$.
The pvalue is $0.3674$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.
Interpretation
There is no sufficient evidence to support the claim that variability in the final exam scores of students taking the statistics major is less than the students taking the mathematics major.
Example 2 Ftest Calculator for two variances
The same capacity of hard drive is manufactured on two different machines, Machine A and Machine B. Samples are taken from both machines and sample mean manufacturing times and sample variances are recorded.
The table shows the data. Test the belief that the variation in manufacturing times is more with Machine A than with Machine B. Use a 5% level of significance.
Machine A: $n_1=13$, $\overline{x}= 127.4$, $s_1^2= 384.16$
Machine B: $n_2=9$, $\overline{y}= 108.3$, $s_2^2 =106.09$
Solution
Let $X$ denote the manufacturing time for hard drive by machine A and let $Y$ denote the manufacturing time for hard drive by machine B. Given data is as follows:
.  Machine A  Machine B 

Sample size  $n_1= 13$  $n_2=9$ 
Sample variance  $s_1^2 = 384.16$  $s_2^2 = 106.09$ 
Step 1 Hypothesis testing problem
The hypothesis testing problem is
$H_0 : \sigma^2_1 = \sigma^2_2$ against $H_1 : \sigma^2_1 >\sigma^2_2$ ($\textit{righttailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} F=\frac{s_1^2}{s_2^2} \end{aligned} $$
The test statistic F follows F distribution with $n_11= 131 = 12$ and $n_21=91= 8$ degrees of freedom.
Step 3 Level of significance
The significance level is $\alpha = 0.05$.
Step 4 Critical value(s)
As the alternative hypothesis is $\textit{righttailed}$, the critical value of F for $12$ and $8$ degrees of
freedom and $\alpha = 0.05$ level of significance $\text{is}$ $\text{3.284}$.
The rejection region (i.e. critical region) is $\text{F > 3.284}$.
Step 5 Computation
The test statistic under the null hypothesis is
$$
\begin{aligned}
F_{obs} &=\frac{s_1^2}{s_2^2}\
&= \frac{384.16}{106.09}\
&= 3.6211
\end{aligned}
$$
Step 6 Decision (Traditional approach)
The test statistic is $F_{obs} =3.6211$ which falls $\textit{inside}$ the critical region, we $\textit{reject}$ the null hypothesis.
OR
Step 6 Decision (pvalue approach)
This is a $\textit{righttailed}$ test, so the pvalue is the area to the left of the test statistic ($F_{obs}=3.6211$) is pvalue = $0.0382$.
The pvalue is $0.0382$ which is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.
Interpretation
There is enough evidence to support the claim that variation in manufacturing times is more with Machine A than with Machine B.
Example 3 F test Calculator two sample of variances
Two different brands of batteries are tested and the variations of their voltage outputs are noted. The table gives the data.
At a 5% significance level, test the claim that the populations of the two brands have the same voltage variations.
Brand A: $n_1=10$, $\overline{x}= 9.31$, $s_1=0.37$
Brand B: $n_2=8$, $\overline{y}= 8.82$, $s=0.31$
Give the null and alternative hypotheses, Find the critical value of this Ftest, find the test value for variances, reject null hypothesis or not, conclusion.
Solution
Let $X$ denote the voltage output for brand A batteries and $Y$ denote the voltage output for brand B batteries.
Given data is as follows :
.  Brand A  Brand B 

Sample Size  $n_1= 10$  $n_2=8$ 
Sample variance  $s_1^2 = 0.1369$  $s_2^2 = 0.0961$ 
Step 1 Hypothesis testing problem
The hypothesis testing problem is
$H_0 : \sigma^2_1 = \sigma^2_2$ against $H_1 : \sigma^2_1 \neq\sigma^2_2$ ($\textit{twotailed}$)
Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} F=\frac{s_1^2}{s_2^2} \end{aligned} $$
The test statistic F follows F distribution with $n_11= 101 = 9$ and $n_21=81= 7$ degrees of freedom.
Step 3 Level of significance
The significance level is $\alpha = 0.1$.
Step 4 Critical value(s)
As the alternative hypothesis is $\textit{twotailed}$, the critical value of $F$ for $9$ and $7$ degrees of
freedom and $\alpha = 0.1$ level of significance $\text{are}$ $\text{0.304 and 3.677}$.
The rejection region (i.e. critical region) is $\text{F < 0.304 or F > 3.677}$.
Step 5 Computation
The test statistic under the null hypothesis is
$$
\begin{aligned}
F_{obs} &=\frac{s_1^2}{s_2^2}\
&= \frac{0.1369}{0.0961}\
&= 1.4246
\end{aligned}
$$
Step 6 Decision (Traditional approach)
The test statistic is $F_{obs} =1.4246$ which falls $\textit{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision (pvalue approach)
This is a $\textit{twotailed}$ test, so the pvalue is the area to the left of the test statistic ($F_{obs}=1.4246$) is pvalue = $0.6553$.
The pvalue is $0.6553$ which is $\textit{greater than}$ the significance level of $\alpha = 0.1$, we $\textit{fail to reject}$ the null hypothesis.
Interpretation
There is enough evidence to support the claim that the populations of the two brands have the same voltage variations.
Endnote
In this tutorial, you learned the about how to solve numerical examples on Ftest for testing two population variances. You also learned about the step by step procedure to apply Ftest for testing two population variances and how to use Ftest calculator for testing two population variances to get the value of test statistic, pvalue, and Fcritical 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 F test calculator for two variances with examples and your thought on this article.