Brown Foorsythe and welch test is used to find out significant difference among the means of more than two samples same as ANOVA when the assumption of homogeneity of variance is viod. 

The Brown and Forsythe Test is a test for equal population variances. It is a robust test based on the absolute differences within each group from the group median. It is a suitable alternative to Bartlett's Test for equal variances, which is sensitive to lack of normality and unequal sample sizes. The Brown and Forsythe Test is better known as The Modified Levene Test. Levene had the idea to modify the data in such a way such that an F test on the new data will actually be a test for equal population variances.

Welch's Test is an alternative test for the one factor analysis of variance F test. It is a test for equal population means to be used when we do not have equal population variances.  It is a parametric test. 

The BF test implemented in the SPSS ONEWAY (ANOVA) procedure is a test of means, not variances.  But as Raid notes, there is also a Brown-Forsythe test for equality of variances. 

In the SPSS documentation (see link below), I found these two citations (emphasis added):

Brown, M. B., and A. B. Forsythe. 1974a. The small sample behavior of some statistics which test the equality of several means. Technometrics, 16, 129-132.

Brown, M. B., and A. B. Forsythe. 1974b. Robust tests for the equality of variances. Journal of the American Statistical Association, 69:, 364-367.

I believe Yiju is asking about the BF test described in the Technometrics article. 

HTH.

http://www-01.ibm.com/support/knowledgecenter/SSLVMB_21.0.0/com.ibm.spss.statistics.help/alg_oneway_forsythe.htm

===========================================

Brown, M. B., and A. B. Forsythe. 1974a. The small sample behavior of some statistics which test the equality of several means. Technometrics, 16, 129-132.

What is the assumption for Brown Forsythe test and Welch test?. Available from: https://www.researchgate.net/post/what_is_the_assumption_for_Brown_Forsythe_test_and_Welch_test#5553bfc56225ffa2348b45d7 [accessed May 13, 2015].

I don't think that any of these 4 tests is by Brown and Forsythe. They compared them.

It is tricky, Bruce. The B&F Test compares the mean absolute differences. It compares "means", but the test in the end compares variability between groups.

If you have access to SPSS, try the following example.  If not, see the attached PDF of the output from this demo.  HTH.

*------------------------------------------------------.

* Unpaired t-test example from

* www.statsdirect.com/help/default.htm#parametric_methods/unpaired_t.htm.

NEW FILE.

DATASET CLOSE all.

DATA LIST list / Group Y (2F5.0).

BEGIN DATA

1 134

1 146

1 104

1 119

1 124

1 161

1 107

1 83

1 113

1 129

1 97

1 123

2 70

2 118

2 101

2 85

2 107

2 132

2 94

END DATA.

T-TEST GROUPS=Group(1 2)

 /MISSING=ANALYSIS

  /VARIABLES=Y

  /CRITERIA=CI(.95).

ONEWAY Y BY Group

  /STATISTICS DESCRIPTIVES HOMOGENEITY BROWNFORSYTHE WELCH

  /MISSING ANALYSIS.

* Notice that for the two-group case, the

* Welch and Brown-Forsythe statistics in

* the "Robust Tests of Equality of Means"

* table are equivalent.

DO IF $CaseNum EQ 1.

- COMPUTE t1Sq = 1.891436**2.

- COMPUTE t2Sq = 1.910700**2.

END IF.

FORMATS t1Sq t2Sq (F8.3).

LIST t1sq t2sq /CASES from 1 to 1.

* t1Sq = Square of "Equal variances assumed t-test result".

* t2Sq = Square of "Equal variances not assumed t-test result".

* As expected, t1Sq = F-ratio from the ANOVA summary table,

* and t2Sq = the Welch and Brown-Forsythe statistics.

Thank you for this information, Bruce. I have to admit that I very rarely use SPSS, and it is very possible that I was simply unaware of a B&F Test for equal means.

So Welch's Test and this B&F Test for equal means give the same results. It is surprising to me that I have not run across this B&F Test before! How is it different from the well-known Welch Test?

Here is another example:

http://www.microbiologybytes.com/maths/spss4.html

 Which test does SPSS use for testing the homogeneity of variances when it is named "LEVENE"? Is it the means based original Levene Test or is it the newer test by B&F that is also called Modified Levene Test?  Levene's test lacks power and it is sensitive to non-normality.

Raid, I am now away from the office and don't have SPSS on this machine.  I believe SPSS implements the original version of Levene's test.  But you have the data used to generate the output I posted earlier, so you could perform your preferred modified Levene test and see if you get the same result as SPSS.  You can also find the SPSS algorithms for the ONEWAY procedure at the link shown below.

Cheers,

Bruce

http://www-01.ibm.com/support/knowledgecenter/SSLVMB_20.0.0/com.ibm.spss.statistics.help/alg_oneway_variance.htm

 The B&F test for homogeneity of variances has a p-value=0.905.  SPSS seems to be using the "Modified Levene Test". The Levene Test has p-value=0.83.  There is no need for Welch's test here.

Brown and Forsythe's Test for Homogeneity of resp Variance

ANOVA of Absolute Deviations from Group Medians

Source    DF      Sum of Squares Mean Square F Value Pr > F

group        1           2.4085               2.4085             0.01       0.9051

Error       17            2793.4                164.3

I used SAS.

Raid, here is another example using two larger datasets (from http://www.unm.edu/~marcusj/2Sampletex2.pdf) with SDs that differ quite markedly.  Again, I'll paste the SPSS syntax at the bottom of this post, and attach the output as a PDF. 

Notice that SPSS computes the following for Levene's test on these data:  F = 2.239, p = .140.  At the end of my syntax, I perform a one-way ANOVA using the absolute deviation from the Group Mean as the outcome variable, and get F(1, 63) = 2.239, p = .140.  I.e., it matches exactly the Levene's test result SPSS computes. 

But when I perform the same ANOVA on the absolute deviation from the Group Median, I get F(1, 63) = 1.688, p = .199.  I suspect this matches the Brown-Forsythe test for equality of variances that you will get if you analyze these data with SAS.

Here's the syntax, for those who have SPSS.

*---------------------------------------------.

* Unpaired t-test example from

* http://www.unm.edu/~marcusj/2Sampletex2.pdf.

NEW FILE.

DATASET CLOSE all.

DATA LIST free / Y (F8.4).

BEGIN DATA

19.7146 22.8245 26.3348 25.4338 20.8310

19.3516 29.1662 21.5908 25.0997 18.0220

20.8439 28.8265 23.8161 27.0340 23.5834

18.6316 22.4471 27.8443 25.3329 26.6790

23.7872 28.4952 27.9284 22.2871 13.2098

END DATA.

DATASET NAME G1.

DATA LIST free / Y (F8.4).

BEGIN DATA

140.0790 18.5252 35.8091 26.5560 31.3332

39.6987 25.1476 29.6046 24.2808 23.5064

39.7922 21.4682 13.1078 25.3269 30.2518

39.1803 34.6926 30.9565 29.9376 23.9296

27.6245 37.2205 33.9531 32.0166 37.1757

29.3769 40.7894 39.6987 27.1912 27.3089

36.1267 28.7846 26.5954 19.7374 33.9418

30.6148 26.8967 28.4069 30.6148 33.8551

END DATA.

DATASET NAME G2.

ADD FILES

  FILE = 'G1' / IN = G1 /

  FILE = 'G2' / IN = G2 .

EXECUTE.

DATASET NAME AllData.

DATASET ACTIVATE AllData.

DATASET CLOSE all.

RECODE G1 (1=1) (0=2) INTO Group.

FORMATS Group (F1).

T-TEST GROUPS=Group(1 2)

  /MISSING=ANALYSIS

  /VARIABLES=Y

  /CRITERIA=CI(.95).

ONEWAY Y BY Group

  /STATISTICS DESCRIPTIVES HOMOGENEITY BROWNFORSYTHE WELCH

  /MISSING ANALYSIS.

* Notice that for the two-group case, the

* Welch and Brown-Forsythe statistics in

* the "Robust Tests of Equality of Means"

* table are equivalent.

DO IF $CaseNum EQ 1.

- COMPUTE t1Sq = -2.447797954**2.

- COMPUTE t2Sq = -3.039526494**2.

END IF.

FORMATS t1Sq t2Sq (F8.3).

LIST t1sq t2sq /CASES from 1 to 1.

* t1Sq = Square of "Equal variances assumed t-test result".

* t2Sq = Square of "Equal variances not assumed t-test result".

* As expected, t1Sq = F-ratio from the ANOVA summary table,

* and t2Sq = the Welch and Brown-Forsythe statistics.

* Now compute Levene's test 'by hand' to see if we

* can match the result computed by SPSS.

* Use AGGREGATE to save the two group means to the file.

AGGREGATE

  /OUTFILE=* MODE=ADDVARIABLES

  /BREAK=Group

  /GroupMean=MEAN(Y)

  /GroupMedian=MEDIAN(Y).

* Compute the absolute deviations from the group mean.

COMPUTE ABSdev1 = ABS(Y - GroupMean).

COMPUTE ABSdev2 = ABS(Y - GroupMedian).

EXECUTE.

VARIABLE LABELS

  ABSdev1 'Absolute deviation from Group MEAN'

  ABSdev2 'Absolute deviation from Group MEDIAN'

.

* Perform a one-way ANOVA on the absolute deviations.

ONEWAY ABSdev1 ABSdev2 BY Group .

* Notice that when I use absolute deviations from

* the Group Mean as the outcome variable, the

* F-test from the ANOVA matches exactly the

* Levene's test result from SPSS.

* When I use absolute deviations from the group

* median as the outcome variable, I get a different

* result, which will likely match Raid's result

* when he performs the Brown-Forsythe test for

* equality of variances on these data.

Hi Bruce,

The Levene Test calculates the absolute deviations from the group mean, while the B&F modified Levene Test uses the absolute deviations from the group median. This is correct.

I have added my results with SAS on the example that you gave us, Bruce. I asked SAS for LEVENE and also for BF.Our results do not seem to agree in some parts. We disagree on Levene.

In SAS, F=0.77 for the Levene Test.

For more information, read this thesis it is small and very organised.

it might help you understanding the differences and assumptions between three different tests

https://scholarscompass.vcu.edu/cgi/viewcontent.cgi?article=5026&context=etd

with great thanks to the author (Hangcheng Liu)

Raid, if you are still following this thread, please note that my current version of SPSS (v25) now includes 4 versions of the homogeneity of variance test in the output from ONEWAY. One of them corresponds to one of the Brown-Forsythe (homogeneity of variance) tests shown in your output from SAS. The SAS BF result based on squared deviations from the mean is not computed automatically by SPSS. But I was able to duplicate it by computing the squared deviations myself and performing one-way ANOVA on them. Syntax and output are attached, the latter in the Excel file.

HTH.

Yes, I am (quietly) following this thread, Bruce. Thanks for the SPSS version of what SAS provides with the option BF (Brown & Forsythe). I recall that the power of the regular Levene Test (based on absolute deviations from the mean) is relatively low. What are you considering using yourself? Why?

Hi Raid. I am not a fan of statistical tests of the assumptions for other tests (either tests of normality or of homogeneity of variance). So when it comes to heterogeneity of variance in the context of ANOVA, I more or less follow the advice found in David Howell's book, Statistical Methods for Psychology. Here is what he says (taken from the 7th edition, 2010, p. 334):

• In general, if the populations can be assumed to be symmetric, or at least similar in shape (e.g., all negatively skewed), and if the largest variance is no more than four times the smallest, the analysis of variance is most likely to be valid. It is important to note, however, that heterogeneity of variance and unequal sample sizes do not mix. If you have reason to anticipate unequal variances, make every effort to keep your sample sizes as equal as possible. This is a serious issue, and people tend to forget that noticeably unequal sample sizes make the test appreciably less robust to heterogeneity of variance.

Regarding the robustness of ANOVA to heterogeneity of variance when the sample sizes are equal, I think Howell is actually being quite conservative when he says the ratio of largest to smallest variances should not exceed 4. IIRC, some simulation studies have shown that even with a ratio of 10, the F-test is quite robust.

I would also modify (or add to) Howell's advice a little bit in this way: When the sample sizes become more disparate, and the variances are not homogeneous, I would strongly consider using a form of ANOVA that does not require homogeneity of variance (e.g., the SPSS ONEWAY procedure allows one to select two different "unequal variances" F-tests). Alternatively, if one uses a procedure meant for multilevel modeling, one can typically allow for heterogeneity of variance (e.g., MIXED procedure in SPSS allows this via the REPEATED sub-command, where you can set COVTYPE to DIAGONAL, IIRC).

Cheers,

Bruce

Hi Bruce,

When there are only 2 populations, F=large variance/small variance=4 implies that we look int he F table for 4 as the cut-off. The sample sizes play a role when looking up the critical F values. Most F values > 3.

Thank you very much for this interesting discussion. After reading all the comments I am wondering if there are two Brown-Forsythe tests: one for testing the assumption of variance homogeneity ("the B&F modified Levene Test") and another as a conservative test to compare > 2 means (like the Welch's test) under heteroscedasticity.

Best regards.

Hello Eduardo. Yes, you are right. When Raid talks about the Brown-Forsythe test, he means a test of heterogeneity of variance, similar to Levene's test. But SPSS users who carry out one-way ANOVA via the ONEWAY command can ask to have the Welch and Brown-Forsythe versions of the F-test included in the output. This Brown-Forsythe test is an F-test of the null hypothesis that all k population means are equal, but unlike the ordinary F-test, it does not require homogeneity of variance. The algorithm can be viewed here:

• https://www.ibm.com/support/knowledgecenter/en/SSLVMB_22.0.0/com.ibm.spss.statistics.algorithms/alg_oneway_forsythe.htm

The fact that there are two tests with the same name causes confusion, especially when users of different stats packages that use the names for different things try to have a conversation! HTH. ;-)

Thank you for your replies, can you use the Brown Forsythe test and Welch test also for ordinal level data, such as 1-5 categories?

Hello Christopher Holmberg. If you mean that the dependent variable is an ordinal categorical variable (with no more than about 5 categories), then an ordinal logistic regression model would be a better choice than ANOVA. HTH.

Dear Bruce Weaver , thank you very much for your reply. We are analyzing an established instrument (dependent variable), which uses 12 items on an ordinal scale, ranging from: "no difficulties" (1) to "cannot do" (5). Each two items, make up one dimension. Thus, each dimension (=sum of two items) could have a scale range of 2-10. Our independent variable is three age categories. Do you still think OLR is the best option? I was thinking to use Kruskal-Wallis due to the scale being ordinal and not normally distributed. I was also considering Welch Anova.

I like @Bruce Weaver 's approach to not only (if at all) use statistical tests to inspect homoscedasticity, although they may help to get an overall impression. To make the confusion even bigger, if you use the UNIANOVA command in SPSS, instead of ONEWAY, you can now additionally call the Breusch-Pagan Test, modified Breusch-Pagan Test, White-Test, and F-Test (and there are more, but not in SPSS), to assess homoscedasticity (I am sure you can find one that isn't "significant" ;-) ). An advantage of UNIANOVA is that you can call heteroscedasticity constant standard errors HC0-HC4.

Brown-Forsythe test is used for testing the equality variances (homogeneity of variances) among different groups. It is a robust test when the underlying data does not meet assumptions of normality. Brown-Forsythe test is more robust than Levene’s test when data distributions are skewed or heavy-tailed (Cauchy distribution).

Read more here with example in R https://www.reneshbedre.com/blog/brown-forsythe-test-variance.html