In earlier blogs, we discussed about one sample and two sample tests but these tests are applicable to calculate whether mean values changed or not.

Here we will talk about tests of variance in which F-test and Chi-Square test are the most important.

Important point to remember- Variance is square of the Standard Deviation.

Test of variance shows whether there is a significant difference between variance / standard deviation or not.

Overview of F-test and Chi-Sqaure test:

F-test: It is used for testing equality of two variances from different population.

Let’s solve one practical problem to get clear understanding of F-test.

Problem: In a factory , we picked 10 samples from Machine A and found the variance 3 and similarly picked 9 samples from Machine B and found the standard deviation 3. Is there any difference of variance in two machines at 90% confidence level?

Solution:
Null Hypothesis: Variance of Machine A=variance of Machine B
Alternate Hypothesis: Variance of Machine A not equal to Machine B

Variance for Machine A=3
Variance for Machine B=9

Put these variance in the F-test formula and rember keep the higher value of variance in the numerator.

Fcalulated=9/3=3

F critical value for 90% confidence level=3.02

Fcalculated< F critical means failed to reject the null hypothesis.

There is no significant difference of variances between these two machines.

Chi-Square Test : It is used for testing the population variance against a specified value.

Here sigma represents variance of specified value ,S is standard deviation of samples and n is the number of samples.

Let’s understand the concept to solve one problem.

Problem: In a factory, we picked 20 samples from Machine A and variance of the samples are 3cc. Has the variance increased from previous established 2cc ? (95% confidence level)

Solution:
Null Hypothesis:Variance <=2cc
Alternate Hypothesis: Variance >2cc

chi-square= (19*3)/2=28.5

chi-square critical value=30.14

chi-square calculated < chi-square critical value ,means failed to reject the null Hypothesis.

There is no changes in the variance, it is still less than 2 cc.

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