# MANOVA Basic Concepts

### Univariate case

One-way ANOVA investigates the effects of a categorical variable (the groups, i.e. independent variables) on a continuous outcome (the dependent variable). In one-way ANOVA, we have m random variables x1, …, xm (also called groups or treatments). For each group we have a sample, where we denote the jth group sample as {$x_{1j}$, …, $x_{n_jj}$}. Group j is said to have nj subjects in its sample. We also define $n = \sum_{j=1}^m n_j$.

Our objective is to test the null hypothesis H0: μ1 = μ2 = ⋯ = μm.

We use the following definitions for the total (T), between groups (B) and within groups (W) sum of squares (SS), degrees of freedom (df) and mean square (MS):

The test statistic F is defined as follows and has an F distribution with dfB, dfW degrees of freedom:

We reject the null hypothesis if F > Fcrit.

### Multivariate case

MANOVA also investigates the effects of a categorical variable (the groups, i.e. independent variables) on a continuous outcome, but in this case the outcome is represented by a vector of dependent variables.

We could simply perform multiple ANOVA’s, one for each dependent variable, but this would have two disadvantages: it would introduce additional experiment-wise error and it would not account for the correlations between the dependent variables. It is therefore possible that MANOVA shows a significant difference between the means while the individual ANOVA do not.

Also MANOVA can be used in place of ANOVA with repeated measures; in which case no sphericity assumption needs to be met when using MANOVA. In this case, you treat the repeated levels as dependent variables.

Definition 1: In One-way MANOVA, we have m random vectors X1, …, Xm (representing groups or treatments). Each Xj is a k × 1 column vector of form

where each xjp is a random variable.

For each random vector Xj we collect a sample {$X_{1j}$, …, $X_{n_jj}$} of size nj. We also define $n = \sum_{j=1}^m n_j$. Each sample Xij is a k × 1 vector of form

where each xijp is a data element (not a random variable), where index i refers to the subject in the experiment (1 ≤ i ≤ nj), index j refers to the group (1 ≤ j ≤ m) and index p refers to the position (i.e. dependent variable) within the random vector (1 ≤ p ≤ k).

Our objective is to test the null hypothesis H0: μ1 = μ2 = ⋯ = μwhere the μj are vectors

and so the null hypothesis is equivalent to H0μ1pμ2p = ⋯ = μmp for all p such that 1 ≤ p ≤ k. The alternative hypothesis is therefore H1: μr ≠ μj for some r, j such that 1 ≤ r, jm, or equivalently, μrp ≠ μjp for some r, j, p such that 1 ≤ r, j ≤ m and 1 ≤ p ≤ k.

Now we define the various means as in the univariate case, except that now these means become k × 1 vectors. The total (or grand) mean vector is the column vector

where

The sample group mean vector for group j is a column vector

where

Example 1: A new type of corn seed has been developed and a team of agronomists wanted to determine whether there was a significant difference between the types of soils that they are planted in (loam, sandy, salty, clay) based on the yield of the crop, amount of water required and amount of herbicide needed. Eight fields of each type were chosen for the analysis. Based on the data in Figure 1, determine whether there is a significant difference between the results for each type of soil condition.

Figure 1 – Data for Example 1 in standard form

We also calculate the total mean vector and group vectors (expressed as row vectors) in Figure 2:

Figure 2 – Total mean and group mean vectors

Here, for example, the total mean for yield (cell G10) is calculated by the formula =AVERAGE(B4:B35) . The other total mean values are calculated by highlighting range G10:I10 and pressing Ctrl-R.

The loam group mean for yield (cell G6) is calculated by the formula

=AVERAGEIF($A$4:$A$35,$F6,B$4:B$35) The other group mean values are calculated by highlighting range G6:I9 and pressing Ctrl-R and Ctrl-D. It can be useful to create a chart with the group means shown in Figure 2. To do this, highlight the range F5:I5 and then select Insert > Charts|Line and then Design > Data|Switch Row/Column. The result is shown on the left side of Figure 3. Figure 3 – Chart of group mean vectors The group mean vectors all look fairly similar (although as we will soon see there are significant differences). It seems that the loam and sandy mean group vectors are very similar and a bit different from the salty and clay group mean vectors which are also very similar. These distinctions are even more evident when we look at the group means minus the total mean (shown in Figure 4 below). We can create a chart of the group means minus the total mean by highlighting F14:I18 (from Figure 4) and selecting Insert > Charts|Line and then Design > Data|Switch Row/Column. The result is shown on the right side of Figure 3. Definition 2: Using the terminology from Definition 1, we define the following total cross products for p and q. When p = q, we have which is the total sum of squares (as in ANOVA) and measures the total variation in the pth dependent variable. When p ≠ q, we have the total cross-product terms, which measure the dependence between the pth and qth variables across all observations. The multivariate equivalent of the total sum of square is the total sum of squares and cross products, i.e. the SSCPT matrix, which is abbreviated a T, and is defined as Note that the diagonal terms are SS1, …, SSk. An alternative way of expressing T is as follows: If the sample data is expressed as a range R1 in the format shown in Figure 1 (what we will henceforth call the standard format) and R2 is the total mean row vector, then T can be calculated by =MMULT(TRANSPOSE(R1–R2),R1–R2) which for Example 1 is the array formula =MMULT(TRANSPOSE(B4:D35-G10:I10),B4:D35-G10:I10) The sample covariance matrix plays the role of MST since MST = SSCPT / dfT where the degrees of freedom is given by dfT = n – 1. In fact (using the supplemental array formula COV or COVP), we can calculate T as follows: T = COV(R1)*(n–1) = COVP(R1)*n Thus for Example 1, T can be calculated by the array formula =COVP(B4:D35)*COUNT(B4:B35) Definition 3: We define the hypothesis cross products for p and q as follows: We define the hypothesis sum of squares and cross products as the matrix H where Alternatively, H can be defined as To calculate H for Example 1, we first use the data in Figure 1 and 2 to derive the table in Figure 4, consisting of the differences between the group means and total means. Figure 4 – Group mean vectors minus total mean vector The value for loam × yield (cell G15) is calculated by the formula =G6-G$10. To obtain the other mean differences, highlight the range G15:I18 and press Ctrl-R and Ctrl-D. To obtain the count for loam enter the formula =COUNTIF($A$4:$A$35,F15) in cell J15. The counts for the other groups are obtained by highlighting the range J15:J18 and pressing Ctrl-D.

If the sample data is expressed as a range R1 in the standard format shown in Figure 1 and R2 is the column vector with the counts for each of the groups (as in range J15:J18 of Figure 4),  then H can be calculated by

=MMULT(TRANSPOSE(R1),R1*R2)

which for Example 1 is the array formula

=MMULT(TRANSPOSE(G15:I18),G15:I18*J15:J18)

Definition 4: We define the error (or residual) cross products for p and q as follows:

We define the error (or residual) sum of squares and cross products as the matrix E where

Alternatively E can be defined as

Property 1: T = H + E

Also for any p and q, CPT = CPH + CPE

Proof:

Observation: The three sum of squares and cross product (SSCP) terms play the role of the SS in ANOVA. The degrees of freedom terms are dfT = n – 1, dfH = m – 1, dfE = n – m. As usual dfT = dfH + dfE. How these terms are used to create the appropriate F test is more complicated than in ANOVA. We will look at this shortly.

Example 1 (continued): Calculate T, H and E.

Using the formulas described above, we can calculate T and H, as shown in Figure 5.  Using Property 1, we have that E = T – H = L4:N6–L9:N11.

Figure 5 – SSCP matrices for Example 1

Observation: We would now like to create an F-test by dividing H by E, as was done for ANOVA. The equivalent in the matrix world is to look at the matrix HE-1. We then reject the null hypothesis if H is “large” compared to E. For MANOVA we have the following four different measures which can used to determine whether H is large compared to E.

Definition 5:

Wilk’s Lambda:  Λ $\frac{|E|}{|H+E|}$

H is large compared to E when the numerator of the above is small compared to the denominator. Thus we reject the null hypothesis when Wilk’s Lambda is close to zero. By Property 1, .

Hotelling-Lawley Trace: $T_0^2$ = trace(HE-1)

H is large compared to E when Hotelling-Lawley Trace is large. In this case we reject the null hypothesis.

Pillai-Bartlett Trace: = trace(H(H+E)-1)

If H is large compared to E then this statistic will be large. Thus, we reject the null hypothesis when this value is large.

Roy’s Largest Root: Θ = largest eigenvalue of HE-1

Again we reject the null hypothesis if this statistic is large. The following alternative version of Roy’s Largest Root is also sometimes used:

$\frac{\lambda_p}{1+\lambda_p}$ where λp largest eigenvalue of HE-1

See Eigenvalues and Eigenvectors for the definition of eigenvalue.

Property 2: Λ = $\frac{1}{|I+HE^{-1}|}$

Proof: Note that E and H are both symmetric matrices, and so E-1 is also symmetric. This means that HE-1 = E-1H. But then E(I+HE-1) =  E(I + E-1H) = EI + EE-1H = E + H. By Property 2 of Determinants and Simultaneous Linear Equations, the determinant of a product of two square matrices is equal to the product of the determinants of each matrix, and so |E| ∙ |I + HE-1|=|E(I+HE-1)|= |E + H|. Thus

Example 1 (continued) test statistics: Calculate the above statistics for Example 1.

Figure 6 – Test statistics for Example 1

Referring also to Figure 5, HE-1 (range X4:Z6) is calculated by the array formula =MMULT(L9:N11,MINVERSE(L14:N16)) and HT-1 (range AB4:AD6) is calculated by the array formula =MMULT(L9:N11,MINVERSE(L4:N6)). Thus Wilk’s Lambda can be calculated by the formula =MDETERM(L14:N16)/MDETERM(L4:N6), Hotelling-Lawley Trace by =TRACE(X4:Z6) and Pillai-Bartlett Trace by =TRACE(AB4:AD6).

We will calculate Roy’s Largest Root in a moment.

Property 3: We can calculate the above metrics from the eigenvalues λ1, …, λk of HE-1 as follows:

Wilk’s LambdaΛ $\prod_{p=1}^k \frac{1}{1+\lambda_p}$

Hotelling-Lawley Trace$T_0^2 = \sum_{p=1}^k \lambda_p$

Pillai-Bartlett Trace: $\sum_{p=1}^k \frac{\lambda_p}{1+\lambda_p}$

Proof: By Property 5 of Eigenvalues and Eigenvectors, the eigenvalues of IHE-1 are 1+λ1, …, 1+λk, and so by Property 1a of Eigenvalues and Eigenvectors, |I + HE-1| = $\prod_{p=1}^k (1+\lambda_p)$. It now follows by Property 2 that

By Property 1b of Eigenvalues and Eigenvectors, for any matrix A, trace A = sum of its eigenvalues, and so $T_0^2$ = trace(HE-1) = $\sum_{p=1}^k \lambda_p$

Finally, H + E = E + H = IE + HI = IE + H(E-1E) = IE + (HE-1)E = (I + HE-1)E.

Thus, H(H + E)-1 = H((I + HE-1)E)-1 = HE-1(I + HE-1)-1. Now by Property 5 of Eigenvalues and Eigenvectors, if λ is an eigenvalue of HE-1 then 1 + λ is an eigenvalue of I + HE-1 and by Property 4 of Eigenvalues and Eigenvectors, $\frac{1}{1+\lambda}$ is an eigenvalue of (I + HE-1)-1. Thus

And so we see that $\frac{1}{1+\lambda}$ is an eigenvalue of (HE-1)(I + HE-1))-1 = H(H + E)-1. Thus by Property 1b of Eigenvalues and Eigenvectors= trace(H(H+E)-1) = $\sum_{p=1}^k \frac{\lambda_p}{1+\lambda_p}$

Observation: The Pillai-Barlett Trace is similar to SSB/SST, which is the percentage of the variance explained by the model, which is similar to R2. It is the most conservative of the tests but is most robust in cases of violation of the assumptions at least for balanced models.

Wilk’s Lambda is similar to SSE /SST = 1 – R2. It is the most commonly used of the tests.

The Hotelling-Lawley Trace is similar to SSB/SSE which is the F-test used in univariate ANOVA. This is the least conservative of the three tests.

Example 1 (continued): Calculate the eigenvalues of HE-1 and the values of the test statistics based on Property 3.

Figure 7 – Test statistics using eigenvalues for Example 1

Here the eigenvalues of HE-1 (range X15:Z15) are calculated by the Real Statistics array formula eVALUES(X11:Z13). The Wilks test statistic is then calculated by the array formula =1/PRODUCT(1+X15:Z15), Hotelling’s Trace by the ordinary formula =SUM(X15:Z15), Pillai’s Trace by the array formula =SUM(X15:Z15/(1+X15:Z15)) and Roy by =X15. Note that the results obtained in Figure 6 are the same as those in Figure 7.

Property 4 (Wilk’s Lambda Test)

Let

If the null hypothesis is true, then

Property 5 (Hotelling-Lawley Trace Test)

Let

s = min(k, m -1) = # of non-zero eigenvalues in HE-1

If the null hypothesis is true, then

Note that df1s ∙ max(k, m – 1)

Property 6 (Pillai-Barlett Trace Test)

Under the same assumption as Property 5

Observation: For large samples the above three measures are quite similar. For small sample, Pillai-Barlett Trace Test is the preferred method since it is less vulnerable to violations of the assumptions.

Example 1 (continued): Calculate the various test statistics and determine whether to reject the null hypothesis.

Using the values calculated previously for each test statistic and applying Properties 4, 5 and 6, we obtain the results shown in Figure 8.

Figure 8 – Test results for Example 1

Since p-value < α = .05 for all the tests, we conclude that there is a significant difference between the four groups.

### 58 Responses to MANOVA Basic Concepts

1. Eqbal mahmood says:

Hell thanks for your helpI have data with male and female of animals
With 4 independent variable can I use MANOVA CRD
The significant between the sex

• Charles says:

Mahmood,
If you have 4 dependent variables, then you could use MANOVA. In fact with only two independent variables (male and female), you can use Hotelling’s T-square Test.
Charles

2. eqbal mahmood says:

hello thank you for your help
please do you have new MANOVA ONE WAY data I NEED IT PLEASE SEND TO ME

• Charles says:

There is no new MANOVA one-way data.
Charles

3. eqbal mahmood says:

hello
is there EXAMPLE unbalaced manova CRD

4. kyriacos ktenas says:

hi
i am looking into investigating causalities in rail project costs. costs would be disagreggated into subcategories reflecting different subworks. would manova be appropriate?

• Charles says:

Sorry, but you haven’t provided enough information for me to respond.
Charles

5. Myke John says:

Thanks prof for this useful insight. Please i’m researching on effects of cement stabilization on geotechnical properties of expansive soils with the % of cement added which is an interval continuous variable (IV) , and the various properties ( liquid limit, plastic limit, plastic index, linear shrinkage, max dry density, optimum moisture content and california bearing ration, CBR) as the response variables. Would MANOVA be appropriate?, and if yes..would it be a repeated measures or otherwise? Pls put me through on this.
Thank you.

• Charles says:

John,
Based on the limited information that you have provided it seems like MANOVA could be a good choice. I don’t see any repeated measure factor in your list, but perhaps there is one.
Charles

6. Jack Knight-Scott says:

Do you have instructions for expanding the MANOVA to a two-way or three-way?

• Charles says:

Jack,
Sorry, but not yet. I expect to do this later this year.
Charles

7. Jack Knight-Scott says:

In the example, you conclude there is a significant difference between the four groups. The next step would be to determinewhich groups. How does one do ad-hoc comparisons under manova?

• Charles says:

Jack,
Charles

8. K.Stephen Choi says:

Hello Charles,

I have installed the data analysis add-in module that you have provided on this website.

On Manova function,
I keep getting an error on the Input range field.
Whatever data range I put in the range becomes absolute range with that dollar sign. Thereby the result is also errored. I tried manually correct it by deleting the \$ signs, but it then doesn’t work. I have Win 10 and MS Office 365.

Is this with my computer? or something else?

Any help is greatly appreciated.

• Charles says:

Since this is the input range, I can-t see why there is a problem using absolute addressing. What is the problem_
Charles

9. L.A. says:

Good day sir!

Is MANOVA a suitable test for RCBD experiments? I was planning on testing some 14 factors, all of which are agronomic data (e.g. plant height, biomass, grain yield) and I’m trying to figure out on how to add the blocks as an additional factor. Thanks in advance!

10. Chris says:

Hi Sir,

First of all, it’s really amazing that you’ve put all this up for free and I really appreciate it. I noticed that you mention a “COV” and “COVP” function in this page and the workbook containing this example seems to reference it, but there isn’t actually a COV or COVP function in excel or, it seems, previous versions. Could I clarify if these are related to the COVARIANCE.P function in excel or if they are custom functions?

Many thanks,
Chris

• Charles says:

Chris,
Charles

11. Anis says:

Hello Sir, we need your help. Could you please show us how the steps to calculate E ? Actually we already found H. But then stuck to get E. Hope get respond from you Sir. Thanks.

• Charles says:

Hello Anis,
To calculate E you need to use Definition 4 on the referenced webpage. It is not so different from the calculation for H and T. If you are having problems, you can download the Multivariate Examples file which contains all the multivariate calculations shown on the website. This is available for free download at Examples Workbooks.
Charles

12. Jose says:

How do you determine the degrees of freedom when reporting the results of an MANOVA?

I have a data set with 2 independent variables and 2 dependent variables. One of the independent variables has 2 levels and the other one 3 levels). In total, I have 62 subjects.

• Charles says:

Jose,
This is esplained on the MANOVA webpages.
Charles

13. Livison says:

Halo Prof.. Please help me here i want to analyse the effects of sustainable building materials on housing construction. My dependant variable are cost of construction, Running costs and Waste reduction. How can i go about it?

• Charles says:

Hello Livison,
It sounds like this could be done via MANOVA or regression, but you haven’t provided enough information for me to give a definitive answer.
Charles

14. marydina balabis says:

Mr. Charles,

Good Day sir!

May I ask your help regarding our research. If MANOVA is applicable to our study. Our IV is Smoking cigar and two DV are Labor productivity of workers and productivity of the firm.

Thank you sir.

• Charles says:

This has the form of MANOVA. Are the assumptions met?
Charles

15. CWVallsr says:

How come in figure 2 your mean for G6 (loam, yield) is 69.7125 when it’s supposed to be 69.7088? Yeah, I copied your data and that’s the mean I got. Am I missing something?

• Charles says:

I just recalculated the mean and still came up with the value of 69.7125. If you send me an Excel file with your calculation, I will check it out.
Charles

16. VALENTINO says:

eeehhhh gilllll mas entendible tenias que hacerlo ooohhhhh

17. juan says:

Hello Charles,
Could you help me please to choose the correct method to analyze the following data:
I carried out an experiment in which 15 subjects had to assess 3 audio reproduction systems evaluating 7 characteristics (variables). So they did a comparative listening task of the 3 systems evaluating the 7 characteristics with continuous scales. One of the variables (the variable seven) is supposed to depend on the other six. I want to know the influence of the audio system over the variables and the relationship of the variable seven to the other six in regards to the three systems.
Thanks,
Juan

• Charles says:

Juan,

If by “variable seven is supposed to depend on the other six [variables]” you mean that there is a correlation between variable 7 and the others, then MANOVA seems to be the way in order to go to determine whether the 3 audio reproduction systems have the same characteristics.

Can you please describe in more detail what you mean by “I want to know … the relationship of the variable seven to the other six in regards to the three systems”?

Charles

• juan says:

The variable seven is related to the preference of the subject in relation to the type of system. So, I want to know how this preference is associated with the other six variables.

When you say that MANOVA seems to be the way… Do I use MANOVA with repeated measures ? Can I do that with real statistic tool ? I saw that there is an option of repeated measures in MANOVA but I’m not sure if it applies for my case.

Juan

• Charles says:

Juan,

Sorry, but I don’t completely understand the problem. When you speak about “the relationship of the variable seven to the other six variables” it sounds like a regression problem.

Part of what you described sounds like a repeated measures MANOVA, which is supported in the Real Statistics Resource Pack.

Charles

• juan says:

Thanks Charles,

Actually I want to know firstly if there is a difference between systems concerning the variable 7 (preference) and then I want to know how the other 6 variables are related to this “preference”.

Thanks again,
Juan Pablo

• Charles says:

Juan,
To determine whether variable 7 is different from each of the other 6 variables can be done via MANOVA or 6 separate Hotelling’s T-square tests.
To determine how the 6 other variables are related to the 7th can be done via regression.
Charles

18. Mary says:

Hello Sir,
I study Corrosion Science, and as a field of interest, Microbial Corrosion. In my study, I focused on three factors affecting corrosion of metals: Salinity,Velocity and presence of specific micro-organism in the medium. As it seems, growth of micro-organism is also dependent of salinity and velocity, so in order to calculate the synergistic effect of these three factors on corrosion, i need to design a proper form of MANOVA matrix, which includes the effect of two independent variables “salinity and velocity”on the dependent variable”growth of bacteria”, and at the same time, calculate the effect of these three variables on corrosion rate.
In the study I used two levels for each variable : low level and high level…. so I assume, the size of samples would be n=2 ?

• Charles says:

Mary,
Clearly your sample size should be a lot more than 2. Perhaps you mean k = 2.
Charles

• Mary says:

yes k=2, but the first problem stands…

• Charles says:

Mary,
I didn’t see a statement of another problem. Based on the info that you provided, it looks like you can use MANOVA as described in Example 1 of the referenced webpage. The data will have two columns since you have only two dependent variables. Since you only have two dependent variables, you can also use Hotelling’s T-square Test. There are Real Statistics data analysis tools for both tests.
Charles

Hi. I would like to ask a question.

What is the equation/formula that I can use under MANOVA to find differences between dependent variables and three (3) independent variables?

I cant seem to find the answer, yet.

• Charles says:

There are multiple ways for me to interpret your question. Can you give me a concrete example of what you are looking for based on Example 1 of the referenced webpage?
Charles

20. Abubakar says:

If the MANOVA analysis shows that the null hypothesis can be rejected, how do we know exactly WHICH groups are different (For example, maybe clay, loam, and sandy are similar, but salty is very different from the others)? Would we perform pairwise Hotelling’s on each of our groups?

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22. mohsen says:

Thanks a lot sir, it was so helpful 🙂

23. Omar says:

hi. I just wanna ask if i can use MANOVA if i have 5 experimantal groups and is observed in 5 different hours.

24. waleed says:

ohhhhhhhhh

thanks thanks
i need answer for this question: in MANOVA analysis if hotelling test and Wilk’s Lambda test and Pillai-Bartlett Trace are not significant and there is one significant for any dependent variable what i can do??????

• Charles says:

Waleed,
It sounds like the MANOVA test supports the null hypothesis. No problem with that; you just need to report the result. I don’t completely understand what you mean by “there is one significant for any dependent variable”.
Charles

25. RGC says:

Hi, Thanks for making this available. Do you have a repeated measures MANOVA? ~R

• Charles says:

Hi RGC,
Repeated measures MANOVA is not yet supported, but I expect to add it later this year.
Charles

26. Colin says:

Sir

I just want to make sure in the example “loam, sandy, salty, clay” are categories and “yield of the crop, amount of water required, amount of herbicide ” are dependent variables, is that right?

Colin

• Charles says:

Colin,
“loam, sandy, salty, clay” are the categories of the dependent variable and “yield of the crop, amount of water required, amount of herbicide ” are independent variables.
Charles

• Roman says:

Sir,
I suspect my question is a bit tiresome, but I honestly tried to understand why “loam, salty, sandy and clay” should be dependent. By definition, “MANOVA … investigates the effects of a categorical variable (i.e. clay, loam etc. in our case) on a continuous outcome … represented by a vector of dependent variables (i.e. water, herbicide, yield in our case)”. My logic is that the type of soil, being INdependent, determinates the yield, the amound of water and herbicide needed, and not the other way round. Please help me get this right. Thank you.

• Charles says:

Roman,

You are correct: “loam, salty, sandy and clay” are the values of the independent variable “soil type”. The dependent vector consists of “water, herbicide, yield”.

Please let me know if you saw the opposite on the referenced webpage.

Charles

• Rachel Beckett says:

You say it yourself:
Colin,
“loam, sandy, salty, clay” are the categories of the dependent variable and “yield of the crop, amount of water required, amount of herbicide ” are independent variables.
Charles”