Hotelling’s T-square Test with Unequal Covariance Matrices

Univariate case

When the variances of the two populations are unequal (as indicated by notably unequal sample variances), we use a modified version of the t-test. In particular we use the following t-statistic


We now test the null hypothesis H0: μx = μy using the fact that tT(m) where m is defined as


(see Two Sample t Test with Unequal Variances). As we have seen several times now, this is equivalent to


where t2 can be expressed as:


where =  – ȳ and µz = µx – µy.

Multivariate case

We now look at a multivariate version of the problem, namely to test whether the population means of the k × 1 random vectors X and Y are equal, i.e. the null hypothesis H0: μX = μY, under the assumption that the covariance matrices are not necessarily equal.

Definition 1: The modified two sample Hotelling’s T-square test statistic is


Observation: Note the similarity between the expression for T2 and the expression for t2 given above. Also note that if  nX = nY, then this definition of T2 is equivalent to that in Definition 1 of Hotelling’s T-square for Independent Samples.

Theorem 1: For nX and nY sufficiently large, T2 ~ χ2(k).

Observation: For small nX and nY, T2 is not sufficiently accurate and a better estimate is achieved using the following theorem

Theorem 2: Under the null hypothesis,


where n = nX + nY   1 and m is defined as follows:



If F > Fcrit then we reject the null hypothesis.

Example 1:  Repeat Example 1 of Hotelling’s T-square for Independent Samples using the data in Figure 1.

Hotellings T2 data 3

Figure 1 – Data for Example 1

Once again, we employ Box’s test, obtaining the results shown in Figure 2.

Box's test Hotelling Excel

Figure 2 – Box’s Test for Example 1

This time we see that p-value < α = .05, and so we conclude that there is evidence that the covariance matrices are unequal (or that the data is not multivariate normally distributed), although we have somewhat forced the issue since usually a significance level of α = .001 instead of α = .05 is used for Box’s Test.

As a result we will use the T2 test with unequal covariance matrices. This analysis is shown in Figure 3.

Hotellings T2 unequal covariances

Figure 3 – Analysis for Example 1

We conclude there is a significant difference between the drug and the placebo in treating the symptoms.

Confidence intervals

The simultaneous 1 – α confidence interval for μi is given by the expression


For n sufficiently large, we could use the following expression instead


Once again we can use Bonferroni confidence intervals instead.


13 Responses to Hotelling’s T-square Test with Unequal Covariance Matrices

  1. Kyunghoon Lee says:

    Dear Prof. Charles Zaiontz,

    I found your website as I searched for multivariate versions of t-test for unequal covariance. Your description really helps me, but I’d like to see reference papers about formulations on the multivariate case of Hotelling’s T-square test with unequal covariance matrices. I’d appreciate it if you’d help me with it.

    Best regards,
    K. Lee.

  2. Kay says:

    How is the calculation for p-value made in this case, if n_x and x_y are different?

    • Charles says:

      The calculation for the p-value shown on this webpage does not require that n_X = n_Y. In fact in the Example given n_X is not equal to n_Y. The calculation is shown in Figure 3 (using Theorem 2).

  3. Garrett says:


    I love this website. Thank you.

    I’m currently dealing with a zero inflated dataset. I’m applying the Hotelling’s T2 test to these data and was wondering if there are issues with having a lot of zeros in the data.

    I’m comparing two sites (control and experimental) using fish densities at 1 m depth increments (total of 15). The sample size for each 1 m depth increment is 50. So a matrix that is [50,15].

    • Charles says:

      Thanks for letting me know that you love the website.
      Regarding your question, unfortunately I don’t have any experience with Hotelling’s T2 test with zero-inflated data, and so I am not able to answer your question.

  4. Mladen says:

    This is what I was looking for. Thank You for Your work.

  5. Lakyn says:

    Dear Charles,
    Thank you very much for this website, it helps a ton in helping me understand Hotelling’s test.

    I was wondering whether you can explain to me why Hotelling’s test uses the F distribution? I cannot seem to connect the test with the F distribution.

    Thank you a lot for your help,

    • Charles says:


      To give a precise answer to your question would require the proof of Theorem 2 on the referenced webpage, which is too technical for our purposes. One way to motivate why the F distribution might be involved:

      – the univariate version of the Hotelling’s test is the t-test, which uses the t distribution
      – the t distibution can be expressed via the F distribution since if t has distribution T(df), then t^2 has distribution F(1,df)


  6. Ensia says:

    Dear Prof. Charles Zaiontz,
    I studied the Hotelling test in order to evaluate if a unique sample (I have just one value for each variable, so n_x=1) could belong to a population, whose I have n_y values of the different variables. In your opinion, is the Hotelling procedure with unequal covariance matrix adapt? How can I overcome to the problem of obtaining zero to the denominator (probably saying that I have 2 equal observations of the p variables)? If in your opinion this is not the right procedure, can you suggest me a more adapt one, please?

    Thanks for your attention.

    Best regards

    • Charles says:

      Are you saying that your sample consists of just one element in each group? In this case, you should expect much no matter what statistical test you use.

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