5Stat NoteIn the fifth of a series of articles about statistics for biologists, Anthony Hilton and Richard Armstrong ask:

is one set of data more variable than another?

www.sfam.org.uk34 June 2006

(Hilton & Armstrong, 2005). Ahypothetical experiment wascarried out to investigate theefficacy of two novel mediasupplements (S1 and S2) inpromoting the development ofcell biomass. Three ten-litrefermentation vessels weresterilised and filled withidentical growth media withthe exception that the mediain two of the vessels wassupplemented with ten ml of

of variation and theassumption of homogeneity ofvariance may need to beexplicitly tested. This Statnotedescribes four such tests, viz.,the variance-ratio (F) test,Bartlett’s test, Levene’s test,and Brown and Forsythe’stest.

The scenario

We return to the scenariofirst described in Statnote 3

important assumption for theuse of the ‘t’ test (Hilton &Armstrong, 2005) or analysisof variance (ANOVA)(Armstrong & Hilton, 2004) isthat the variability of thedifferent groups beingcompared is similar, i.e., thatthey exhibit homogeneity ofvariance. Replicatemeasurements within a controland a treated group, however,often exhibit different degrees

HERE MAY BEoccasions when it isnecessary to testwhether the

variability of two or more setsof data differ.

An investigator, forexample, may wish to testwhether a new treatmentreduces the variability of aparticular microbial responsecompared with an oldertreatment. In addition, an

T

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Features

either medium supplement S1or S2. The vessels wereallowed to equilibrate andwere subject to identicalenvironmental / incubationconditions. The vessels werethen inoculated with a cultureof Bacterium x at an equalculture density and thefermentation allowed toproceed until all the availablenutrients had been exhaustedand bacterial growth hadceased. The entire volume ofculture media in eachfermentation vessel was thenremoved and filtered torecover the bacterial biomass,which was subsequently driedand the dry weight of cellsmeasured. This experimentwas repeated 25 times and thedry weight of biomassproduced in each of the threegroups recorded in Table 1.

The variance-ratio test

If there are only two groupsinvolved, then their variancescan be compared by a two-tailvariance ratio test (F-test)(Snedecor & Cochran, 1980).

How is the test done?

The larger variance isdivided by the smaller and theresulting F ratio comparedwith the value in a table of thevariance ratio to obtain a P-value, entering the table forthe number of degrees offreedom (DF) of thenumerator and denominator.This test uses the two-tailprobabilities of F because weare testing whether or not thetwo variances differ ratherthan whether variance A isgreater than variance B.Hence, this calculation differsfrom that carried out during atypical ANOVA, since in thelatter, it is whether thetreatment variance is largerthan the error variance that isbeing tested (Armstrong &Hilton, 2004). Publishedstatistical tables of the F ratio(Fisher & Yates, 1963;Snedecor & Cochran, 1980)are usually in the form of one-tail tables. Hence, the 2.5%

probability column has to beused to obtain the 5%probability.

Interpretation of theresults

When the unsupplementedand S1 data are compared(Table 1), a value of F = 1.03was obtained. This value isless than the F value in the2.5% column (P > 0.05) andconsequently, there is noevidence that the addition ofthe medium S1 increased ordecreased the variance inreplicate flasks.

Bartlett’s test

If there are three or moregroups, then the differentgroups could be tested inpairs using the F-test

described above, but a betterapproach is to test all thevariances simultaneously usingBartlett’s test (Snedecor &Cochran, 1980).

How is the test done?

If there are equal numbersof observations in each group,calculation of the test statisticis straight-forward and aworked example is shown inTable 2. If the three variancesdo not differ from each other,then the ratio M/C is amember of the chi-square (χ2)distribution with (a – 1)degrees of freedom (DF),where ‘a’ is the number ofgroups being compared. If thegroups have different numbersof observations in each(unequal ‘n’), then the

calculations are slightly morecomplex and are given inSnedecor and Cochran (1980).

Interpretation of theresults

In the worked example inTable 2, the value of χ2 washighly significant (P < 0.001)suggesting real differencesbetween the variances of thethree groups. The previous F-test suggested, however, thatthe variance of theunsupplemented data wassimilar to that of the growthmedium S1. Therefore, it isthe effect of the growthmedium S2 that hassubstantially increased thevariance of bacterial biomass.Hence, if these data were tobe analysed by ANOVA(Armstrong & Hilton, 2004),the assumption ofhomogeneity of variancewould not hold and it may benecessary to transform thedata to logarithms beforeanalysis to stabilize thevariance. Data transformationis described in more detail inStatnote 4 (Hilton &Armstrong, 2006).

The use of the χ2

distribution to test thesignificance of M/C isquestionable if the DF withinthe groups are less than fiveand in such a case, there arespecial tables for calculatingthe significance of the statistic(Pearson & Hartley, 1954).Bartlett’s test is used lesstoday and may not normallybe available as part of astatistics software package.This is because the test isregarded as being too‘sensitive’ resulting in toomany significant resultsespecially with data from long-tailed distributions (Snedecor& Cochran, 1980). Hence useof the test may raiseunjustified concerns aboutwhether the data conform tothe assumption ofhomogeneity of variance. As aconsequence, Levene (1960)developed a more robust testto compare three or more

Variances: US = 463.36. S1 = 447.88. S2 = 18695.24Variance-ratio test comparing US and S1: F = 463.36/447.88 = 1.03(2-tail distribution of F, P > 0.05)

US

461

472

473

481

482

482

494

493

495

S1

562

573

574

581

582

586

591

592

592

S2

354

359

369

403

425

476

511

513

534

US

506

502

501

505

508

500

513

512

511

S1

607

600

603

605

607

609

611

611

615

S2

556

578

604

623

644

668

678

698

703

US

518

527

524

529

537

535

542

S1

617

622

626

628

631

637

645

S2

714

721

722

735

754

759

765

Table 1. Dry weight of bacterial biomass under unsupplemented(US) and two supplemented (S) growth conditions (S1 and S2) ina sample of 25 fermentation vessels.

M = v[a (ln s*2) – Σ ln si2] where s*2 is the mean of the variances, ‘a’ the

number of groups, v = DF of each group, and ln = logarithms to base e.Hence, M = 102.62 C = 1 + (a +1)/(3av) = 1.018χ2 = M/C = 102.62/1.018 = 100.8 (DF = a – 1, P < 0.001)

Group

Unsupplemented

S1

S2

Total

Variance

436.36

447.88

18695.24

19606.48

In (variance)

6.1385

6.1045

9.8360

22.079

Table 2. Comparison of the variances of three groups with equalobservations (v = 25) in each by Bartlett’s test.

www.sfam.org.ukJune 200636

References

■ Armstrong RA & Hilton A(2004) The use of analysis ofvariance (ANOVA) in appliedmicrobiology. Microbiologist, vol5: No.4 18.

■ Brown MB & Forsythe AB(1974) Robust tests for theequality of variances. J Am StatsAssoc 69: 264-267.

■ Fisher RA & Yates F (1963)Statistical tables. Longman,London.

■ Hilton A & Armstrong RA(2005) Statnote 3: Comparingthe difference between twogroups. Microbiologist, vol 6:No.4 30.

■ Hilton A & Armstrong RA(2006) Statnote 4: What if thedata are not normal?Microbiologist, vol 7: No.1 34

■ Levene H (1960) In:Contributions to Probability andStatistics. Stanford UniversityPress, Stanford, California.

■ Pearson ES & Hartley HO(1954) Biometrika Tables forStatisticians, vol1. CambridgeUniversity Press.

■ Snedecor G W & Cochran W G (1980) Statistical Methods,7th Ed. Iowa State UniversityPress, Ames Iowa.

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Dr Anthony* Hilton and Dr Richard Armstrong***Pharmaceutical Sciences and**Vision Sciences, AstonUniversity, Birmingham, UK

variances (Snedecor &Cochran, 1980).

Levene’s test. How isthe test done?

Levene’s test makes use ofthe absolute deviation of theindividual measurements fromtheir group means rather thanthe variance to measure thevariability within a group.Avoiding the squaring ofdeviations as in the calculationof variance results in ameasure of variability that isless sensitive to the presenceof a long-tailed distribution.An ANOVA (Armstrong &Hilton, 2004) is thenperformed on the absolutedeviations and if significant,the hypothesis ofhomogeneous variances isrejected.

Interpretation of thedata

A Levene’s test on the datain Table 1 using STATISTICAsoftware, for example, gave avalue of F = 52.86 (DF 2,72;P < 0.001) confirming theresults of Bartlett’s test.

More recently, Levene’s testhas also been called intoquestion since the absolutedeviations from the groupmeans are likely to be highlyskewed and therefore, violateanother assumption required

for an ANOVA, that ofnormality (Armstrong andHilton, 2004). This problembecomes particularly acute ifthere are unequal numbers ofobservations in the variousgroups being compared. As aconsequence, a modificationof the Levene test has beenproposed by Brown andForsythe (1974).

Brown-Forsythe test.How is the test done?

This differs from Levene’stest in that an ANOVA isperformed not on the absolutedeviations from the groupmeans but on deviations fromthe group medians. This testmay be more accurate thanLevene’s test even when thedata deviate from a normaldistribution. Nevertheless,both Levene’s and the Brown-Forsythe tests suffer from thesame defect in that to assessdifferences in variancerequires an ANOVA, and anANOVA requires theassumption of ‘homogeneity ofvariance,’ which some authorsconsider to be a ‘fatal flaw’ ofthese analyses.

Conclusion

There may becircumstances where it isnecessary for microbiologiststo compare variances rather

than means, e,g., in analysingdata from experiments todetermine whether aparticular treatment alters thedegree of variability or testingthe assumption ofhomogeneity of variance priorto other statistical tests.

All of the tests described inthis Statnote have theirlimitations. Bartlett’s test maybe too sensitive but Levene’sand the Brown-Forsythe testsalso have problems. We wouldrecommend the use of thevariance-ratio test to comparetwo variances and the carefulapplication of Bartlett’s test ifthere are more than twogroups.

Considering that these testsare not particularly robust, itshould be remembered thatthe homogeneity of varianceassumption is usually the leastimportant of those consideredwhen carrying out an ANOVA.

If there is concern aboutthis assumption and especiallyif the other assumptions of theanalysis are also not likely tobe met, e.g., lack of normalityor non additivity of treatmenteffects (Armstrong & Hilton,2004) then it may be bettereither to transform the data orto carry out a non-parametrictest on the data.