Homogeneity of Variances

Certain tests (e.g. ANOVA) require that the variances of different populations are equal. This can be determined by the following approaches:

  • Comparison of graphs (esp. box plots)
  • Comparison of variance, standard deviation and IQR statistics
  • Statistical tests

The F test presented in Two Sample Hypothesis Testing of Variances can be used to determine whether the variances of two populations are equal. For three or more variables the following statistical tests for homogeneity of variances are commonly used:

  • Levene’s test
  • Fligner Killeen test
  • Bartlett’s test

Using the terminology from Definition 1 of Basic Concepts for ANOVA, the following null and alternative hypotheses are used for all of these tests:

H0\sigma_1^2 = \sigma_2^2 = ⋯ = \sigma_k^2

H1: Not all variances are equal (i.e. \sigma_i^2 ≠ \sigma_j^2 for some i, j)

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46 Responses to Homogeneity of Variances

  1. Duygu says:

    Hi, I am comparing 7 groups defined under one independent variable. My problem is with the homogeneity of variance test as equal variance is one of the assumptions for the one way anova. Although my groups have equal sample size, they show unequal variance in the levene test (p-value:0.001). Can I still use Anova test or should I use Kruskall- Wallis H test. and the sample size is 20 for each group, there are 7 groups, the data is normally distributed but with unequal variance, please kindly guide me regarding choosing the correct anova and post hoc tests for this analysis. Many thanks.

    • Charles says:

      With unequal variances you should probably choose the Walsh test. A post-hoc test for unequal variances is Games-Howell.
      Charles

  2. Stan says:

    Good afternoon– I am trying to calculate the sample size required to test whether a sample variance is = 0.63 vs. H1: var < 0.63 with, say, 80% power, alpha=0.05. Could you give me some advice on how to calculate the sample size required for this test?
    Thanks! Stan

    • Charles says:

      This is a one sample variance test (using the chi-square distribution). You can find more information about this test at One Sample Variance Test.

      You can estimate the sample size by pressing Ctrl-m and choosing the Statistical Power and Sample Size data analysis tool. On the resulting dialog box choose the One sample variance and Sample size options.

      Charles

  3. Naza says:

    Dear Sir,
    I used Levene’s test to check homogeneity of variance for my study. The result shows that p-value of Levene’s test is 0.165, while p-value for F-test is 0.000. Can I conclude that the variances is equal?
    Thank you.

    • Charles says:

      A p-value of .165 indicates that there is no significant difference between the variances.
      I don-t know which F-test you are referring to.
      Charles

  4. ojirobe says:

    Good day sir! Pls. I am currently working on a multiple regression problem, but I don’t know how to test for the homogeneity of variances. I am using the spss package which has a built in levene’s test procedures. Is it expected of me that check for the equality of variances of the DV and all IV? When I tried comparing the means using oneway anova, I kept on getting response like: homogeneity of variances cannot be performed cos only one group has a computed variance and also because the sum of case weights is less than the number of groups. Any form of explanation or references given would be appreciated.
    Thank you in anticipation of a swift response

    • Charles says:

      Levene’s test is a good way to test for homogeneity of variances. If you are performing one-way ANOVA then you need to test the homogeneity of the k groups in the test. I don’t use SPSS and so I am not able to explain the error message you received.
      Charles

    • Kevin says:

      Hi there,

      I use SPSS, and have received an error that I believe is very similar to yours in wording. In my experience, it usually means that the two variables were entered incorrectly…the IV was labelled as the DV and vice versa. Try switching the group’s around and see if that doesn’t solve the problem.

      Kevin

  5. Kevin says:

    Charles,

    Great info here…could you possibly do a page on the Fligner Killeen test (nonparametric assessment of homogeneity of variance with >2 groups), using concrete numbers? Every “example” I can find either just gives the chi-square result without the computation details, or just uses the abstract symbols and summation notation to explain it…really confusing and infuriating! Why can’t they just use an example with raw numbers? Thanks again!

  6. brain matienga says:

    dear sir
    I am doing research on the effects of wet/dry and wet feeding troughs on feeding grower pigs . so i have 2 treatments wet feeding and dry feeding and 2 blocks that is males and female so i have 24 males and 24 females all pigs are of same age and genetic .Can you help me on the method of data analysis to use. I am measuring the feed in take daily and weighing pigs on entery and then once weekly for 8 weeks

    • Charles says:

      It all depends on what you are trying to test, but based on what you have described it sounds like a three factor Anova. One fixed factor with two levels for feeding type, one fixed factor with two levels for gender and one repeated factor with 9 levels for week. The dependent variable is weight.
      Charles

  7. niki says:

    helloo

    i conducted a two way anova and the levene’s test p value is 0.001, my study has a continuous dv (eating style) and two categorical iv’s
    Not sure what to do as ive used this analysis to show the effects of two other eating styles?

    • Charles says:

      Niki,
      Sorry, but I don’t understand your question. Please explain in a little more dtail, what you have done and what you want to accomplish.
      Charles

      • niki says:

        i have conducted a two way anova to see if there is a sig difference in the means of eating style scores when participants reported their stress levels (high vs low stress) and sleep quality (good vs poor). I have conducted this for both emotional and external eating style but when i conducted the analysis for restrained eating style the homogeneity assumption was violated what can i do?

        • Charles says:

          Niki,
          Look at the Dealing with non-heterogeneity of variances topic towards the end of the referenced webpage.
          Charles

  8. saif salim says:

    Dear sir please I need your help:
    1- 40 students are divided into two groups of 20 each [the control group (i.e CG) and the experimental group(i.e. EG)] they are considered equal in their level of English language study. The CG read version A of a text with certain rhetorical organization , on the other hand the EG read version B of the same text with different rhetorical organization . They are asked to read these two versions and recall information from them so the amount of information recalled and speed of reading spent are recorded . I want to test the following hypothesis ” To what extent will the change (in the rhetorical pattern) affect the ease of information recalled and speed of reading as well? How can I use ANOVA test please when the CG have two marks of amount of information recalled and marks of speed of reading and the EG have also two marks also

    • Charles says:

      It sounds like you have one independent variable Rhetorical organization (RO) and two dependent variables: Information recalled (IR) and Reading speed (RS). You can use MANOVA, or more simply Hotelling’s T-square as described on the webpage Hotelling’s T-square.
      Charles

      • saif salim says:

        Dear sir, I consider the students (i.e. CG and EG) as dependent variables since there is no difference between them and considered equal, the RO (i.e.the two versions of texts version A and B with different rhetorical organization) as independent variable since it is varied . However the IR and SR are the responses or marks gathered and not variables .Can I use ANOVA test to examine the responses i.e. the IR (amount of information recalled) then do the same procedures to examine the RS (speed of reading) with my thanks and regards sir.

        • Charles says:

          Saif,
          Sorry, but I still don’t understand the question. E.g. in the statement “I consider the students (i.e. CG and EG) as dependent variables since there is no difference between them and considered equal…”, I don’t understand why this statement would make CG and EG dependent variables. I really don’t understand the other statements either.
          Charles

  9. Sumit says:

    Can one perform t-test or ANOVA using CV if the variance between group/s is not similar? If yes, how does one do it? I am a statistics illiterate

    • Charles says:

      There is a version of the t-test which you can use when the variances are not similar. See the webpage two sample t-test with unequal variances.

      There are also substitute tests for ANOVA when the variances are not equal. See the Dealing with non-heterogeneity of variances topic on the referenced webpage.

      You use the abbreviation “CV”. What does this stand for?

      Charles

  10. Kennedy says:

    I love your website it is so useful in helping me solve statistical problems. I am a little confused about how to perform hypothesis testing when the observations are just given as one total without actually listing them separately – 125 observations (Southern States) and 132 observations (Northern States) with a sample mean of 87 and 88 respectively and a population variance of 7.0 and 6.2 respectively. Level of significance is .01 is there evidence that the workers in southern states are receiving less pay than workers in northern states?

    • Charles says:

      Since you have the population variances you can use a two sample test using the normal distribution, as described in Theorem 1 of Comparing Two Means.

      The null hypothesis is mean1 >= mean2 (these are population means). The test statistic is z = (m1-m2)/stdev, where m1-m2 = 88-87 = 1 (sample means) and stdev = sqrt(var) where var = v1^2/n1 + v2^2/n2 = 6.2^2/132 + 7.0^2/125. If NORMSDIST(z) > .99 then you reject the hypothesis that the the workers receive the same pay. This is a one tailed test. If you want a two-tailed test you need to replace .99 by .995.

      If instead of the population variance you had the sample variances you would use Theorem 1 of Two Sample t Test instead.

      Charles

  11. Alfiya says:

    Dear Sir,
    I am doing the One-Way ANOVA analysis. My p-value =0,233 for Levene’s test. Since my data was not normally distributed I transformed it. Do I need to perform Levene’s test again for transformed data? (I have tried and p-value is less then 0.05)
    Thank you
    Alfiya.

    • Charles says:

      Alfiya,
      Since you will be testing the transformed data you need to make sure that the assumptions are satisfied on the transformed data. Since homogeneity of variance is an assumption for One-way ANOVA this assumption needs to be verified for the transformed data. Levene’s test is a way of checking this.
      Charles

  12. Lucy says:

    Dear Sir,
    I am analyzing a field experiment on 4 maize varieties. The varieties were replicated three times in one location. Should I examine the homogeneity and normality tests?

    Lucy

    • Charles says:

      Lucy,
      It really depends on what you are trying to test. The ANOVA tests will require homogeneity of variance and normality, but they can be quite forgiving even if these assumptions aren’t completely satisfied.
      Charles

  13. umar iqbal says:

    sir i want to know how do i find the relationship between export growth and variation between real and nominal exchange rate based on the measure of 3 months and 6 months i have collected data but now i m confused how do i apply non-parametric test on it and which test…

  14. praveen kumar says:

    I want do homogeneity test for two variances please tell me to do the test

    • Charles says:

      Levene’s test can be used for two variance. You can use the LEVENE function as described on the referenced webpage.
      Charles

  15. Valerian says:

    hi! i want to ask on the interpretation of Bartlet’s test on the Gen stat discover program for ANOVA, i do fail to interpret it

  16. sonia says:

    sir i wanted to ask why homogeneity of variance is so important?please tell me in some points…

    • Charles says:

      Sonia,
      Homegeneity of variances is a requirement for many of the most used statistical tests, including ANOVA. Fortunately most such tests are pretty foriving and as long as the variances are not too unequal the tests give pretty accurate results, but when the requirement is sufficiently violated then the results of these tests can be quite unreliable.
      Charles

  17. Deborah says:

    Hi, I just wanted to ask, what happens if your levene’s test is positive so homogeneity of variance cannot be assumed in a factorial independent measures ANOVA. I know that you have to change the significance to p=0.01 instead of 0.05 (or something along those lines) but what do I do in terms of SPSS? I have run the test as normal but I don’t know how I am supposed to interpret my results considering levene’s positivity.

    Please help!!! Thank you

  18. Colin says:

    Sir

    Will you add a real statistics function for “Bartlett’s Test” ?

    Colin

    • Charles says:

      Colin,
      Bartlett’s Test is also called Box’s Test. This is already included in the Real Statistics Resource Pack (see multivariate statistics portion of the website).
      Charles

  19. Sriya says:

    Hi,

    Can you please let me know what transformation method I should be using if both standard deviation to means and means to variances are not proportional? There is no strong correlation for both?

    Thanks.

    • Charles says:

      Sriya,
      There is no easy answer to your question. It all depends on your data. There are an unlimited number of transformations as well (1/x, x^2, etc.). It also may turn out that a particular transformation creates more problems than it solves.
      Charles

  20. Ned from Norn Iron says:

    Many thanks for this… The easy to follow guide to Levene’s and Bartlett’s included in your download is just what I needed to sort out a tricky analytical problem…

    • admin says:

      Ned,
      I am very pleased that the site has been useful for you. I hope that you will use it again in the future.
      Charles

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