Two Factor ANOVA without Replication

To help introduce the basic concepts we start with the following example.

Example 1: A new fertilizer has been developed to increase the yield on crops, and the makers of the fertilizer want to better understand which of the three formulations (blends) of this fertilizer are most effective for wheat, corn, soy beans and rice (crops). They test each of the three blends on one sample of each of the four types of crops. The crop yields for the 12 combinations are as shown in Figure 1.

Data ANOVA without replication

Figure 1 – Data for Example 1

We interrupt the analysis of this example to give some background, after which we will resume the analysis.

Definition 1: We define the structural model as follows.

A factor is an independent variable. A k factor ANOVA addresses k factors.

A level is some aspect of a factor; these are what we called groups or treatments in the one factor analysis discussed in Basic Concepts for ANOVA.

In Example 1 there are two factors: blends and crops. The blend factor has 3 levels and the crop factor has 4 levels.

In general, suppose we have two factors A and B. Factor A has r levels and factor B has c levels. We organize the levels for factor A as rows and the levels for factor B as columns. We use the index i for the rows (i.e. factor A) and the index j for the columns (i.e. factor B). Thus we use an r × c table where the entries in the table are

image1297

We use terms such as i (or i.) as an abbreviation for the mean of {xij: 1 ≤ j ≤ c}. Similarly, we use terms such as j (or .j) as an abbreviation for the mean of {xij: 1 ≤ i ≤ r}.

We estimate the level means from the total mean for factor A by μi = μ + αi where αi denotes the effect of the ith level for factor A (i.e. the departure of the ith level mean μi for factor A from the total mean μ). We have a similar estimate for the sample of i = + ai.

Note that

image1308

Similarly we estimate the level means from the total mean for factor B by μj = μ + βj where βj denotes the effect of the jth level for factor B (i.e. the departure of the jth level mean μj for factor B from the total mean μ). We have a similar estimate for the sample of j = + bj.

As for factor A,

image1312

The two-way ANOVA will either test for the main effects of factor A or factor B, namely

H0: μ1. = μ2. =⋯= μr. (Factor A)

or

H0: μ.1 = μ.2 =⋯= μ.c (Factor B)

If testing for factor A, the null hypothesis is equivalent to

H0αi = 0 for all i

If testing for factor B, the null hypothesis is equivalent to

H0βj = 0 for all j

Finally, we can represent each element in the sample as xij = μ + αi + βj + εij where εij denotes the error (or unexplained amount). As before we have the sample version xij = + ai + bj + eij where eij is the counterpart to εij in the sample.

Observation: Since

image1319

It follows that
image1320

Similarly,
image1321

It is easy to show that
image1322

Definition 2: Using the terminology of Definition 1, define

ANOVA without replication MS

Correction: The term xij in the formula for SSE in the above table should not have a bar over it.

Property 1:

image1335 image1336

Proof: Clearly

image1337

If we square both sides of the equation, sum over i, j and then simplify (with various terms equal to zero as in the proof of Property 2 of Basic Concepts for ANOVA), we get the first result. For the second,

image1338

Property 2: If a sample is made as described in Definition 1, with the xij independently and normally distributed and with all \sigma^2_{i.} (or \sigma^2_{.j}) equal, then

image1341

Proof: The proof is similar to that of Property 1 of Basic Concepts for ANOVA.

Theorem 1: Suppose a sample is made as described in Definitions 1 and 2, with the xij independently and normally distributed.

If all μi are equal and all \sigma^2_{i} are equal then

image1343

If all μj are equal and all \sigma^2_{j} are equal then

image1344

Proof: The result follows from Property 2 and Theorem 1 of F Distribution.

Property 3:

image1345 image1346

Observation: We use the following tests:

ANOVA without replication tests

Recall that the assumptions for using these tests are:

  • All samples are drawn from normally distributed populations
  • All populations have a common variance
  • All samples were drawn independently from each other
  • Within each sample, the observations were sampled randomly and independently of each other

We now return to Example 1 and show how to conduct the required analysis using Excel’s Anova: Two-factor Without Replication data analysis tool.

Example 1 (continued): The output from the data analysis tool is shown in Figure 2.

ANOVA without replication Excel

Figure 2 – Two factor ANOVA without replication data analysis tool

There are two null hypotheses: one for the rows and the other for the columns. Let’s look first at the rows:

H0: there is no significant difference in yield between the (population) means of the blends

Since the p-value for the rows = .0068 < .05 = α (or F = 12.83 > 5.14 = F-crit) we reject the null hypothesis, and so at the 95% level of confidence we conclude there is significant difference in the yields produced by the three blends.

The null hypothesis for the columns is

H0: there is no significant difference in yield between the (population) means for the crop types

Since the p-value for the columns = .1446 > .05 = α (or F = 2.63 < 4.76 = F-crit) we can’t reject the null hypothesis, and so at 95% level of confidence we conclude there is no significant difference in the yields for the four crops studied.

Observation: Although the analysis in Figure 2 was produced automatically by Excel’s data analysis tool, the same result can be produced using Excel formulas, just as we were able to do in Basic Concepts of ANOVA for one-way ANOVA. The most interesting cells are the ones corresponding to the four sum squares. We show how to calculate the values for each of those cells in Figure 3.

Key formulas ANOVA

Figure 3 – Key formulas for analysis from Figure 2

The formulas for calculating SSRow and SSCol in Definition 2 involve taking squared deviations of the group means. E.g. SSRow can be calculated via the formula =DEVSQ(I6:I8)/H6. Alternatively we can take squared deviations from the sums of each group, as is done in Figure 3.

Real Statistics Excel Capabilities: The Real Statistics Resource Pack contains a number of supplemental functions and the Two Factor ANOVA data analysis tool which support Two Factor ANOVA without Replication. You can get more information about these in Two Factor ANOVA with Replication.

44 Responses to Two Factor ANOVA without Replication

  1. Erik van Rensbergen / Flanders / Belgium says:

    Charles,
    Which test can I use when the assumptions for using a Two Factor Anova, especially the one of a common variance, are not met? I understand that the Kruskal Willis test can only be used in place of a One Way Anova.
    Thank you again,
    Erik

    • Charles says:

      Erik,
      You can use Scheirer-Ray-Hare as a substitute for Two Factor ANOVA with Replication. This test has limited power, but it is a possible approach.
      Charles

  2. ANILA AJAYAN says:

    Hello Mr Charles.
    Can you please tell me why I get a p value 2.76E-08 while performing Two way ANOVA without replication between Season and Species Density?
    Thank you

    Have A Nice Day
    Anila Ajayan

  3. Hilary says:

    I am attempting to perform this ANOVA, but my variance results column has #DIV/0! for all values, leading to #NUM in my P-value and F-crit value boxes. What is the problem with my data?

    • Charles says:

      Hilary,
      If you send me an Excel file with your data and calculations and I will try to figure out what is going on. You can get my email address on the webpage
      Contact Us
      Charles

      • L.A. says:

        I’m experiencing a similar problem. This time, I can’t do the Tukey HSD test follow-up for two factor anova because the variances has #DIV/0! for all values. Hope you can assist as well.

        • Charles says:

          You need to fill in the contrast column (labeled c) in the output with 1 for one group and -1 for another group. In this way you compare two groups.
          Charles

      • Zahra says:

        Hi Charles,

        I have the same problem. Since my data doesn’t have replication, the variance of response in every interaction of the factors cannot be calculated.. so it resulted #DIV/0!. Could you please give me any suggestion?

        Thank you

        Regards,
        Zahra

        • Charles says:

          Hi Zahra,
          Sorry, but when you say that you “have the same problem”, whom are you referring to?
          In the case where there is no replication, there is no interaction factor, and so you cannot analyze it. You can, however, analyze the two main factors, as described on the referenced webpage.
          Charles

  4. Kelly says:

    Which post hoc test should I use in the excel toolpack when I find significance in the results of the two factor anova without replication?

    Thanks

  5. gupreet says:

    hi I m doing two way anova with replication but results for P and F value are not coming normal. My data is having 6 columns and 24 rows for each column. I am confuse what is happening.

  6. Sean says:

    HI Charles

    Many thanks for your excellent website.

    I am having trouble finding the ‘ANOVA: two factor without replication’ tool in the data analysis toolkit. I go to the ‘analysis of variance section’ and check the the ‘Anova: two factors’ box but there is then no option for ‘without replication’. Hence I cannot understand how to arrive at the output for this example. (similarly with the example for Anova: 2 factor with replication).

    I hope you can help.

    Regards

    • Charles says:

      Hi Sean,
      For ANOVA without replication choose the Anova: two factors option. Then in the dialog box that appears insert 1 in the Number of Rows per Sample field.
      Charles

  7. Md.Abdul Kader says:

    How can I do multiple comparison like LSD, DMRT between treatment by this two way annova without replecation.

    • Charles says:

      I don’t support these follow up tests at present. I don’t support LSD because I find other tests are better.
      Charles

      • Md.Abdul Kader says:

        Can you tell me, name of some others test? Which one is better to draw conclusions.Thank You.

        • Charles says:

          Which follow up test is best depends on a number of things (equal sampler size or not, homogeneity of variances or not, etc.). Generally I use Tukey’s HSD post-hoc test for ANOVA with replication. See the following webpage:
          Unplanned Comparisons

          I have not thought about what sort of post-hoc tests are appropriate for ANOVA without replication.

          Charles

  8. pt says:

    Because the degree freedom of error is equal to zero when trying to calculate the interaction effect,so we can conclude that their are no interaction effects in this case? Is that right?

    Thank you

  9. rolly says:

    what does it mean by error values that come out in ANOVA table (2 way without replication)…how to interpret it?

  10. Dave Leet says:

    Nice work and thanks.
    You have written:
    rows = .0068 < 05 = α
    It ought to be:
    rows = .0068 05 = α
    should be:
    columns = .1446 > 0.05 = α

    (Or you could write 0.05 as .05, same thing)
    Thanks again.

    • Dave Leet says:

      Posting the above comment also dropped the decimal on my first example for the correction. Must be something in the way the HTML is conveyed.

      Test:
      05 needs to be written either 0.05, or .05.

      Strange!

    • Charles says:

      Dave,
      Thanks for catching this typo and for helping improve the accuracy of the website. I have now revised the webpage to include the decimal point.
      Charles

  11. boyu says:

    I use Analysis of variance-two factors Appear number of rows per sample must be a positive integer. what is this please teach me thanks.

    • Charles says:

      In a Two factor ANOVA there are two factors, which I will call Row and Column. Suppose the Row factor has 3 levels and the Column factor has 4 levels. If say there are 240 elements in the sample, with 20 elements in each combination of Row and Column levels (3 x 4 x 20 = 240). The value for number of rows per sample = 20.
      Charles

  12. Chris says:

    The math here is not correct, possible typo on the greater than sign>
    (or F = 2.63 > 4.76 = F-crit)

    • Charles says:

      Chris,
      Thanks for catching this typo. It should indeed state (or F = 2.63 < 4.76 = F-crit). I have now corrected this mistake on the referenced webpage. Thanks again for bringing this to my attention. Charles

  13. Jeff says:

    Should the Figure 2 labels read without replication?

    • Charles says:

      Jeff,
      Yes, you are correct. The caption for Fiure 2 should read “without replication”. Thanks for catching this typing mistake. I have now corrected the caption on the webpage.
      Charles

  14. zp says:

    very nice website! It is good to learn Stats with easy-to-use samples.

  15. Colin says:

    Sir
    I think there is a mistake about SSE in the table of definition 2. It may be a typo.

  16. Ed says:

    What happens to SSAB in the two factor without replication? Why is it not shown.

  17. J says:

    Either figure 1 is incorrect or the opening paragraph, from figure 1 there are 4 crops and 3 blends, where as your opening paragraph states “four blends on one sample of each of the three types of crops”.

    • Charles says:

      Thanks J for finding the typo. The figure is correct but the opening paragraph is not. It should state “three blends on one sample of each of the four types of crops”. The website has now been corrected. Thanks again for catching the error. Charles.

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