FTest for Equality of Two Variances  Example
Data
The data for this case study were collected by Said Jahanmir of the NIST Ceramics Division in 1996 in connection with a NIST/industry ceramics consortium for strength optimization of ceramic strength.
The motivation for studying this data set is to illustrate the analysis of multiple factors from a designed experiment
This case study will utilize only a subset of a full study that was conducted by Lisa Gill and James Filliben of the NIST Statistical Engineering Division
The response variable is a measure of the strength of the ceramic material.
The goals of this case study is to determine if the nuisance factors (lab and batch) have an effect on the ceramic strength (y).
Source: http://www.itl.nist.gov/div898/handbook/eda/section4/eda42a1.htm
Analysis
 Open the DataBook jahanmi2.vstz
open this data file via the Help / Open Examples menu; it is in the Sample Data  Choose the menu Analyze and the command F Test, under group Basic Statistics
 In Variables 1, select Y
 In Variables 2, select Batch
 Check the box Variable2 is a grouping variable
 Click OK
Output
By default, R code is printed after parsing and before
evaluation.
You can avoid this in File / Option / Advanced
menu.
DataSheet is converted to DataFrame, in the form DataBook.DataSheet
> ## Summary Statistics:
> summary.vst(jahanmi2.Sheet1[,c('Y')],
statistics = c('mean','semean','sd','nobs'), groups =
jahanmi2.Sheet1$Batch)
Statistic
Group nobs mean sd semean
1 240 688.9986 65.54909 4.231176
2 240 611.1560 61.85425 3.992675
> ## FTest for Equality of Two Variances:
> .res < var.test(formula = Y ~ Batch, data = jahanmi2.Sheet1,
alternative = 'two.sided', ratio = 1, conf.level = 0.95)
> .res
F test to compare two variances
data: Y by Batch
F = 1.123, num df = 239, denom df = 239, pvalue = 0.3704
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.8709874 1.4480271
sample estimates:
ratio of variances
1.123038
> ## Find critical values for the F test:
> qf(0.025,.res$parameter[1],.res$parameter[2])
[1] 0.7755639
>
qf(0.975,.res$parameter[1],.res$parameter[2])
[1] 1.289384
Results
The F test indicates that there is not enough evidence to reject the null hypothesis that the two batch variancess are equal at the 0.05 significance level.
(Reject the null hypothesis if F < 0.7756 or F > 1.2894)
Resource

NIST/SEMATECH eHandbook of Statistical Methods,
http://www.itl.nist.gov/div898/handbook/eda/section3/eda353.htm, 10/30/2013  Probability and statistics EBook