Hello readers! Today, we will discuss the hypothesis test for the equality of two variances. The 2-variance test is used to assess whether the variances of two independent samples are equal.
Let us understand with an example.
A new filtering device is installed in a chemical unit. Before and after its installation, a random sample yielded the following information about the percentage of impurity.
Steps to conduct a 2-variance test
Step 1- Formulate Hypotheses – Identify null hypothesis and alternative hypothesis
As we know, standard deviation is expressed in terms of sigma (σ), and the square of sigma (σ)2 is called variance.
Now open the Minitab.
Path
Stat>Basic statistics > 2 variance Test then Then select the sample variance as shown in the picture (1.0). Then go to option tab then select sample 1 variance/sample 2 variance. Confidence level = 95.0 , Hypothesized =1 and Alternative hypothesis : Ratio ≠ hypothesized ratio as mentioned in picture (2.0).
Picture (1.0)
Picture (2.0)
Then click Ok.
Now, observing that the p-value is 0.918, which is greater than 0.05, we do not have sufficient evidence to reject the null hypothesis (H0). Therefore, we accept H0, indicating that the two variances are equal.
2) Has the filtering device reduced the percentage of impurity significantly?
To address the second question, we have established from the previous example that the variances are equal. Consequently, we proceed to examine the means of the two samples, employing a 2-sample T-Test in this case.
Hypothesis formulation.
Now open the Minitab.
Path
Stat> Basic statistic > 2 sample T test then choose summarized data.
Put the value in sample size, sample mean, standard deviation for sample 1 & sample 2 as shown in picture (3.0).
Picture (3.0)
Proceed to the ‘Options’ tab and check the box that says ‘Assume Equal Variance.’ (picture 4.0) Afterward, click ‘OK
Picture (4.0)
Conclusion – Now, observing that the p-value is 0.639, which exceeds 0.05, we cannot reject the null hypothesis (H0). Therefore, we accept H0. This suggests that, at the chosen significance level, there is insufficient evidence to conclude a significant reduction in the percentage of impurity after using the filtering device. There is no observed impact.