ANOVA: Variance Assumption
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 This topic has 10 replies, 10 voices, and was last updated 13 years, 11 months ago by Bulusu.

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June 3, 2008 at 7:31 pm #50202
I’m doing a one way anova. While the data is normal for each of 3 sample sets, and the means are very close, one sample of the 3 exhibits a wider spread, much greater variance (over 3X the others). My pvalue is high .832.
What does this failed equality of variance mean and what should I do given that one of the assumptions of an ANOVA is equal variances?
Dan0June 3, 2008 at 8:37 pm #172532
Ken SchroederParticipant@KenSchroeder Include @KenSchroeder in your post and this person will
be notified via email.It means the spread of one of your data sets (as you indicated) is much larger than the others, thus negating the assumption of equal variances. You may want to try transforming your data (all sets) to achieve equal variances, and then repeat ANOVA. Another analysis worth considering is Analysis of Means (ANOM). ANOM assumes data is collected rationaly – thus std. dev calculation is different, not randomnly which ANOVA does. Try ANOM on your untransformed data, and ANOVA on your transformed data and see what you get.
0June 3, 2008 at 9:18 pm #172534Transform the data? Are you nuts?
0June 3, 2008 at 9:18 pm #172535Thanks.
What if I did a test of 2 variances using the smallest and largest of the 3 variance values and get a p value greater than .05? Would this approach be legitimate?0June 3, 2008 at 9:34 pm #1725362 variances using the smallest and largest of the 3 will destroy your credibility !
0June 3, 2008 at 9:58 pm #172538
TaylorParticipant@ChadVader Include @ChadVader in your post and this person will
be notified via email.From my experience your operator to part interaction has variance, This could be due to carry over effects of measuring the parts repeatably. As a suggestion you may want to try a KruskalWallis test which does not assume normality.
0June 4, 2008 at 4:28 am #172542Hi Dan,
Why you want to conduct ANOVA?
To use one way ANOVA, the following conditions MUST be present:Normally distributed
The samples must be independent of each other
Each population must have the same variance
The conclusion based on ptest value will be misleading if the assumptions are violated. Try to use non parametric test (Kruskall wallis, Mods median, etc)
Keep learning!0June 4, 2008 at 10:18 am #172545Dan, Why don’t you give us the ANOVA table? It would help to understand. When you say the one variance is 3 times the others, you do not indicate any statistical test of homogeneity. Was one done.The assumption of equal variance is because we are using that assumption to conduct analysis of the sums of squares between and within groups. With that assumption, any difference in variance can only be attributed to a difference in means. In your case, no difference in means was found, So even though one group may have larger variance, your null hypothesis still holds. The danger of violating the variance assumption lies in false differences being detected. This does not appear toi have happened. The underlying data do NOT have to be normally distributed. The residuals need to be. Have you conducted any residual analysis?
0June 4, 2008 at 5:03 pm #172550Dan,
I assume that you are trying to determine if the means of the 3 groups are “equal” or if “at least 1 is not equal to the others”. This is usually the case with one way anova with 3 or more treatments. (or should I say..always the case)
Your observation about the 3X variance within one of the groups is a significant finding, and you should be delving into the root cause of this finding! The statement about the pvalue seems inconsistent. Was this related to the test for equal variances? (if so, you would fail to reject the null…hence the variances would not be deemed to be different)
You might want to explain the basis for your derived pvalue. Did you ignore the unqual variances and proceed with the Ftest for comparison of means? If so, the inflated variance within the one group is probably the cause for the high pvalue. (SS within is very large ….F ratio is low….p value is high)
It would be helpful if you explained how the 3 groups were formed. Is this a shift to shift study (first, second, third) or something along these lines?
Bottom line..forget about making the statistical test work and find out the cause of the variation.
HACL’s 2 cent for the day (That’s all I can afford after going to the gas pump)0June 4, 2008 at 5:39 pm #172551Since there are only three sample sets that are normal, you might also do a paired comparison ttest with unequal variances to determine if any of the paired means are different. The degrees of freedom calculation by hand is a bear but most software packages handle that for you. Hope this helps.
Bob0June 6, 2008 at 5:48 pm #172630
BulusuParticipant@ramachandrudu Include @ramachandrudu in your post and this person will
be notified via email.In Anovaone way, we seek the equality of the 3 means under the assumption that their variances are same. When the variance of one sample is significantly high, that itself is information.
Omit that sample, and go ahead with testing for equality of means for the remaining 2 samples.0 
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