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I am trying to learn F test and on performing the inbuilt var.test() in R, I obtained the following result with var.test(gardenB,gardenC)

F test to compare two variances

data:  gardenB and gardenC
F = 0.09375, num df = 9, denom df = 9, p-value = 0.001624
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.02328617 0.37743695
sample estimates:
ratio of variances 
           0.09375

I understand that based on the p-value, I should reject the Null hypothesis. However, I am unable to understand the meaning conveyed by the 95 percent confidence interval?

I tried reading through the explanation provided for the queries: How to interpret the confidence interval of a variance F-test using R?

But am still able to understand the meaning conveyed in the confidence interval. Any further help would be really appreciated?

Community
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Dilip T Monson
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1 Answers1

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To compare variances, you calculate their ratio. If the variances in the groups are the same, this ratio should be close to 1.0. In your case, it is 0.09375. The 95% confidence interval for this value is 0.02329 - 0.37744. This page lists the definition of a confidence interval, I won't re-type it here. http://www.stats.gla.ac.uk/steps/glossary/confidence_intervals.html
It is an indication of the level of uncertainty around the estimated ratio.

In your example, both the upper and lower bounds are WAY below 1, so I would conclude the variances are very different in the two groups. I would generally ignore the p-value for this type of test--you can get a small p for variances that only slightly differ if your sample size is large enough. Not that this is an issue in your case.

JBeau
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