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.

**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

We use terms such as *x̄ _{i}* (or

*x̄*) as an abbreviation for the mean of {

_{i.}*x*}. Similarly, we use terms such as

_{ij}: 1 ≤ j ≤ c*x̄*(or

_{j}*x̄*) as an abbreviation for the mean of {

_{.j}*x*}.

_{ij}: 1 ≤ i ≤ rWe estimate the level means from the total mean for factor A by *μ _{i} = μ + α_{i}* where

*α*denotes the effect of the

_{i}*i*th level for factor A (i.e. the departure of the

*i*th level mean

*μ*for factor A from the total mean

_{i}*μ*). We have a similar estimate for the sample of

*x̄*=

_{i}*x̄*+

*a*.

_{i}Note that

Similarly we estimate the level means from the total mean for factor B by *μ _{j} = μ + β_{j} *where

*β*denotes the effect of the

_{j}*j*th level for factor B (i.e. the departure of the

*j*th level mean μ

_{j}for factor B from the total mean

*μ*). We have a similar estimate for the sample of

*x̄*=

_{j}*x̄*+

*b*.

_{j}As for factor A,

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

H_{0}: *μ _{1.} = μ_{2.} =⋯= μ_{r.} *(Factor A)

or

H_{0}: *μ. _{1} = μ._{2} =⋯= μ._{c} *(Factor B)

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

H_{0}: *α _{i} *= 0 for all

*i*

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

H_{0}: *β _{j}* = 0 for all

*j*

Finally, we can represent each element in the sample as *x _{ij} = μ + α_{i} + β_{j} + ε_{ij} *where

*ε*denotes the error (or unexplained amount). As before we have the sample version

_{ij}*x*=

_{ij}*x̄*+

*a*where

_{i}+ b_{j}+ e_{ij}*e*is the counterpart to

_{ij}*ε*in the sample.

_{ij}**Observation**: Since

**Definition 2**: Using the terminology of Definition 1, define

**Correction**: The term *x _{ij}* in the formula for

*SS*in the above table should not have a bar over it.

_{E}**Property 1**:

Proof: Clearly

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,

**Property 2**: If a sample is made as described in Definition 1, with the *x _{ij} *independently and normally distributed and with all (or ) equal, then

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* x _{ij} *independently and normally distributed.

If all *μ _{i} *are equal and all are equal then

If all *μ _{j}* are equal and all are equal then

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

**Property 3**:

**Observation**: We use the following 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.

**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:

H_{0}: 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

H_{0}: 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.

**Figure 3 – Key formulas for analysis from Figure 2**

The formulas for calculating *SS _{Row} *and

*SS*in Definition 2 involve taking squared deviations of the group means. E.g.

_{Col}*SS*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.

_{Row}**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.

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”.

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.

Could you please define two factor anova without replication.

This defined on the referenced webpage.

Charles

I want a correlation/comparison of yields(dependent variable) of brand of marketed tea with different temperature and power combination of ultrasonic machine as independent variable. How i can arranged may data for ANOVA. Kindly help me

Mohammed,

You haven’t provided enough information for me to answer your question.

Charles

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

Ed,

Because it is Anova w/o replication SSAB becomes the error term SSW (or SSE).

Charles

Sir

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

Sir

Sorry, you are right. The SSE formula is different with the textbook I read, but it is correct.

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

Should the Figure 2 labels read without replication?

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

The math here is not correct, possible typo on the greater than sign>

(or F = 2.63 > 4.76 = F-crit)

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

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.

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

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.

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!

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

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

Like any error value, the lower the value the better the fit of the model.

Charles

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

There are no interaction effects in two factor ANOVA without replication.

Charles

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

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

Charles

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

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

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

Hi Sean,

For ANOVA without replication choose the

Anova: two factorsoption. Then in the dialog box that appears insert 1 in theNumber of Rows per Samplefield.Charles

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.

Please explain better what you mean by “P and F value are not coming normal”.

Charles

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

Kelly,

You can use the usual post hoc tests for ANOVA with replication, except those that test the interaction (since you are assuming there is no interaction). Probably most useful are contrasts and Tukey’s HSD. An example of the Tukey’s HSD is given on the following webpage:

Question (3) of the Spring 2005 exam from http://www.math.utah.edu/~treiberg/M3081Finalsamplesoln.pdf

Charles

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?

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

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.

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

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

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

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

Anila,

A p value of 2.76E-08 is a small value which is equal to .0000000276. This value is written in what is called scientific notation.

Charles

can u do anova for me….if yes , then reply to me at santoshpandey511@gmail.com

No. You will need to do this yourself. I am providing you the tools and explanations, but you need to do the work.

Charles

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

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

I’m using the Two Factor Anova data analysis tool and just noticed that the means for each of the categories in the two factors (the categorical “total” means as opposed to the means at each intersection of variables, if you will) are calculated by averaging the corresponding cells in the matrix of means, as opposed to either calculating them directly from the data or using a weighted average. This seems to be producing inaccurate results.

Madison,

The Two Factor Anova supports two input data formats: Excel format and standard (i.e. stacked) format. When Excel format is used, I believe that marginal means are based on the original data, but when standard format is used then the marginal means are the average of the group means, as you have observed.

For balanced models (when all interactions have the same number of elements), both approaches yield the same result. This is not the case for unbalanced models. For unbalanced models, you should choose the Regression option on the Two Factor Anova dialog box. This will use the standard format approach to calculating the marginal means. This is the preferred approach as described on the following webpage:

Unbalance Approach to Two Factor Anova

Charles

The distinction between balanced and unbalanced models was what I was missing, thank you! Your site is an excellent resource, and it is very much appreciated!

Hi Charles

I have done carried out a biofilm assay with two different nanoparticles with 6 different concentrations (0, 100, 200,300,400,500) without replication. Please let me know whether I can use two way ANOVA without replication for this dataset. I find significant difference betwee the two types of nanoparticles when I do the test. However, there is no significant differences between different concentrations. What I am confused is that the difference between columns (diff concentrations) is for both types of nanoparticles. However, when I replicate and do t-test assuming equal variance between control and different concentrations of nanoparticles, I find there is a significant difference. Please let me know which method and analysis is appropriate .

Niluka,

If I understand your scenario correctly, it seems that you can use Two Factor ANOVA without replication. The differences between the columns is for both types of nanoparticles (combined).

I don’t understand what you mean by “when I replicate and do t-test”. If you like, you can send me an Excel file with this data and analysis so that I can better understand what you are trying to do.

Charles

Hello,

Thank you for your great addin. I have been working with your Gage R&R feature. The way that I understand the number of categories is that it is the (stddev(part)/stddev(gage))*sqrt(2). Your Gage R&R report uses (stddev(part)/stddev(total))*sqrt(2). Am I misunderstanding a variable, or should the top formula be used?

Thanks

Neil,

I believe that I am using the formula with gage and not total variation.

Charles

Hi Charles, I want to know if which two ANOVA is appropriate to calculate the significant difference in the means of some data with two treatments across three different age groups

Ajibola,

You have 2 levels for the Treatment factor and 3 levels for the Age factor. Now the question for you is how many subjects do you have for each of the 6 combinations of Treatment x Age? If one then you need Two Factor ANOVA without Replication. If more than one then you need Two Factor ANOVA with Replication.

Charles

Could you tell me how i can determine my analyse method for my experiment, i have 9 chemicals materials every one was divided to three concentrations (50, 100 and 150%), if you can help me using SPSS or Excel

without replications number 10 without chemical materials (just one value)

Gh,

Sorry, but I don’t understand.

Charles

Gh,

What do you want to test?

Charles

i have 9 factors: every one has three levels without interactions among factors and the tenth factor without levels, what is the best method to analyse this experiment? thank you

Hi Charles

thanks for your amazing explanation.

I am doing my thesis and a bit confuse about analyse part.

Here is summary:

I have two different wheat varities(D and R ). each varitie went under three different level of fertilizer(1.2. 1.5 1.7) and then treatment R and D with the rate of 1.2 and 1.5 got three replicants and treatment R and D with the rate of 1.7 got four replicants.

here is my questation can I use Anova two way test with replicants or not???

Thanks a lot

Azin,

Yes, you can use two way ANOVA with replications. Since the number of replications differs, you need to use the regression version of the analysis. See

Unbalanced Anova

Charles

How do i interpret this :

ANOVA

Source of Variation SS df MS F P-value F crit

Rows 185.2422414 57 3.249863884 6.740541094 2.41869E-41 1.339297472

Columns 190.9956897 19 10.05240472 20.84968772 1.0899E-60 1.596133621

Error 522.1543103 1083 0.482136944

Total 898.3922414 1159

0.851643958

Since the p-values are almost zero, this means that there is a significant result for both the Rows and Columns factor.

Charles

Hi Charles,

I’m bit confusing, I’d appreciate your advice about this study.

I’m having a drug group and a placebo group, and I’m measuring lab parameters on two time points; at baseline and after two month. It’s an RCT.

Data are normally distributed, so I used an independent t-test to compare between groups.

I’ve been advised to perform a 2-way ANOVA ! after searching, it was revealed that such analysis needed a 2 categorical independent variables.

I’ve thought they maybe considered a one-factor repeated measure ANOVA?!

I’ve done it with time (2 level) and a Group as a fixed variable, and there was no sig. on between groups test, although the results from both ANCOVA (with pre-tests values as covariates) and independent t-test were significant between groups.

I’ve read that ANCOVA is considered better in pre-post treatment analysis especially in RCTs.

Can I even perform a 2-way ANOVA ? and if No, what Can I argue with? I’m sending it a reviewer who asked me to use a 2-way ANOVA.

Sorry for disturbing .

Mona,

If I have understood the situation correctly, you have two factors: a fixed factor Treatment (drug, placebo) and a repeated measures factor Time (baseline and 2 months later). I described this type of test on the following webpage:

http://www.real-statistics.com/anova-repeated-measures/one-between-subjects-factor-and-one-within-subjects-factor/

Charles

Hi again,

thank you very much for your reply. I do appreciate the help.

I’ve just tried to make it, it seems that ANOVA with replication needs same nember of samples, I have the treatment group 28 and the control group 29.

what should I do ?

thank you in advance,

Mona

See Unbalanced ANOVA

Charles

Hi there Charles! I’m presently having my thesis. I’m quite confused on what to do, tried to search over the internet then ended on your excellent site. Anyway, I’m doing a correlational study on self-concept, job satisfaction and perceived organizational support. I have two moderating variables, work tenure and length of professional experience. For the latter (length of professional experience), I have categorized it into 5 groups so I was advised to use ANOVA and the queries start their. Should I use the two factor ANOVA with replication or the one without? How should I do this? Thanks and more power!

This depends on whether or not you have data from multiple sources for each intersection cell.

More specifically, compare Figure 1 from the referenced webpage with Figure 1 from the following webpage

http://www.real-statistics.com/two-way-anova/two-factor-anova-with-replication/

Charles

Hi,

I have 2 IV (blue and red paper)and 2DV(anxiety and stress). The Dv’s are conducted pre and post the IV. I have been advised to conduct a 2×2 ANOVA . Would I start by adding the anxiety before and after looking at the coloured papers together or would I add the anxiety scores before looking at the paper and then add the scores for after and do this also for stress? And then how would I put this in SPSS because don’t i need to only add one DV for a ANOVA?.

Any help will be appreciated

Thanks

HH,

Sorry, but I don’t have a sufficiently clear idea of your situation to be able to give you an answer.

Charles

Hello sir,

I have 3 Inputs(independent variables) and 1 output(Dependent variables). Using ANOVA I have to determine 2 things here:

1) Which of the 3 inputs is dominantly affecting the output.

2) What in the uncertainty in the output value due to input values.

Can u please suggest me the procedure to do so.

Thanks.

It is difficult to answer for sure, but based on your description with three independent variables (factors), this could be a fit for 3-way ANOVA, although from your question perhaps you are looking for multiple regression.

Charles

Hi. I was looking for how to’s on 2 way anova and ended on your site. But I’m still having trouble with my data. I have 7 categories of problems on shipment and their frequency for 2015 and 2016. Will this suit my hypothesis that says there is no significant difference in the commpon problems of 2015 and 2016?

Meliza,

I need more information to answer your question. It might be that you need to use Hotelling’s T-square test since you have multiple categories.

Charles

Is it possible to perform single factor ANOVA without replication?

Manvik,

That would mean that you have only one sample element for each group. No you can’t use ANOVA in this case.

Charles

I have two variable independent and one dependent variable. should i use Anova: Two-factor Without Replication to find the significant of the variable?

Linda,

Whether you use ANOVA or some other test depends on what you are trying to test. I am not sure what you mean by “find the significan[ce] of the variable”.

Whether you use the Two Factor ANOVA with Replications or without Replications depends on whether you have replications.

Charles

Hi Charles,

Thank you so much for this clear and concise explanation.

I have a set of biodiversity data with two independent variables – distance (continuous) and location (categorical). For example, at 10 metres, the mean biodiversity is 0.2 at Location A and 0.5 at Location B. I would like to find out if the there is a significant difference between the two locations, and if the distance is a factor. Would the Anova Two Factor Analysis be suitable?

Anna,

I would need more information to say for sure, but it does seem like two factor Anova is suitable.

Charles

Hi Charles,

The table in property 3 in the row “Effect of factor A” suggests statistical test “MS(A)/MS(W) ~ F(df(A),df(E))”. What is MS(W)? Shouldn’t it be MS(E) instead? Same applies for the “Effect of factor B”…

Thanks,

Vitali

Vitali,

Yes, you are correct. I have now changed the table on the webpage. Thanks for your help in improving the Real Statistics website.

Charles

Hello Charles,

I am currently writing a Biology essay on comparing the antimicrobial effect of different types of chili (habanero, bell pepper and cayenne) on the growth of E.coli (measured by change in optical density). A part of my research question is to determine the chili with the most significant effect on reducing E.coli growth. I have already used the ANOVA on my data, which showed that there was a significant difference in the change in OD between the chilis. Do you think there is a statistical test that can rank the chili with the most significant effect for me?

I would appreciate it if you could help me solve this problem.

Best regards,

Jamie

(P.S. Thank you so much for this helpful website!)

Jamie,

If the one with the highest effect is significantly different from the others (e.g. by using the Tukey HSD post hoc test), then that is the one with the largest effect.

Charles

I have done experiment studying microclimate variables as influenced by planting density ( 4 levels) and pruning ( 2 levels) along with a control ( open)- forming a (4 x 2)+1 experiment. There are no replications. But I have data collected for 52 weeks. What would be the ideal method of analysis here

Santhosh,

What hypotheses are you trying to test?

Charles

can we made this data or any 2 way anova without replication its self in SPSS?

Sorry, but I don’t understand your question.

Charles

Hi

I really appreciate if you can help me resolve the following issue. In my wood preservation study, I have two independent variables which are the duration of treatment (3, 7 and 21 days) and the acetic acid concentration (0.5, 1, 1.5 and 2.0 Molar). I also had a control (untreated) in which I did not expose to acid. I have tested these two factors and recorded the mass loss of wood after fungal decay. How do I perform two-way ANOVA to describe if there is any significant difference among the two factors in relation to control?

Syazwan Azmi,

If I understand correctly, you have one fixed factor Concentration (with levels 0, .5, 1.0, 1.5, 2.0) and one repeated measures factor Duration (3, 7, 21). If so, you can use the mixed repeated measures version of ANOVA, as described on the following webpage:

http://www.real-statistics.com/anova-repeated-measures/one-between-subjects-factor-and-one-within-subjects-factor/

Charles

Ho Charles

If I have one factor with 4 levels and without replication. How can I do my statistical analysis ? And also I need to compare the mean of 4 levels of this factor between them. Thank you for your collaboration.

Mohamed,

If you don’t have any replications, then you only have 4 data elements, one for each level. The mean of each level is therefore the same as the single data element for that level. You can’t do much with this, and certainly not ANOVA.

Charles

Thank you very much for your reply

I need to clarify something. I have 8 observations for each level.

Mohamed,

You can use one-way ANOVA. See

http://www.real-statistics.com/one-way-analysis-of-variance-anova/

Charles

Hello again Charles,

Very good refresher! It is always surprising to see how we can make inferences with very small samples…

Why aren’t we looking at interactions in the case of ANOVA without replication?

Thanks!

Fred

Fred,

In ANOVA without replications there are no interactions to analyze.

Charles

OK, I need to get my head around as to why interactions are not considered in this case, will do that later 🙂

Going back to sample sizes, can the ANOVA assumptions of normality and equal population variances be verified when there is only 3 or 4 values per level?

Thanks again Charles,

Fred

Fred,

1. In the case of ANOVA without replication, there is only one value for each interaction. You can’t even calculate a variance in this case.

2. Yes, you can verify the ANOVA assumptions, but with such small samples I would be cautious about any conclusions reached. In any case, the statistical power of such tests will be very low.

Charles

But where variance is used to estimate the interaction effect?

There is no interaction effect in the case of two factor ANOVA without replication.

Charles

I am very curious to know, how to read the final result table for Two Way ANOVA with / without replication, in Excel.

For One Way ANOVA, as it is very clear if all the factors have same population mean or not, what is the simplest interpretation for Two Way ANOVA (With / Without replication).

Please guide.

Vivek,

The interpretation is similar to that for one-way ANOVA except that now you can test the significance of factor A and factor B. There is no interaction factor.

Charles

Nice post! I have a small question. Suppose I want to perform a factorial analysis with three independent factors (Time: 24 and 48 hours; Temperature: 27, 35 and 43 degrees; and Concentration: 0.6, 0.8, 1.0, 1.2 and 1.4). This is a model without replication, so I get 30 observations. I ran an ANOVA specifying the model and I got some order two interactions significant. How is the homogeneity of variance assumption tested in this car? As you say, I have only one observation per cell so Levene’s test for example won’t say anything. Can I still trust the outputs from the ANOVA tests?

Thank you.

Alvaro,

With no replications, there is only one observation for the three factor interaction, but there are more than one observations for the pairwise interactions, and so you can use Levene’s test (or other approaches to determine homogeneity of variances). You can see this better by looking at Figure 2 of the following webpage

http://www.real-statistics.com/two-way-anova/three-factor-anova-without-replication/

Charles

Charles,

Thank you for the quick answer. But Levene’s test in my case doesn’t return any p-value. I have only, as I explained, three factors and two interactions between 2 factors. When running the homogeneity test (in SPSS), the output says that the degrees of freedom of the Levene’s test statistic are 29 and 0 and hence not returning any value. How is that possible? Should then I run the test several times, for two factors and their interactions?

Alvaro

Alvaro,

If you don’t have any interaction terms (i.e. one cell per interaction), then you can’t analyze interactions using ANOVA. With two interactions, you can calculate variances and so can see whether homogeneity of variances holds. You won’t be able to use Levene’s test though. You can see whether the largest variance is fairly similar to the smallest variance. Usually a ratio of 3 or 4 to 1 is acceptable, but I am not sure whether this is so with such a small sample.

Charles

If i want to detect the presence of a single contaminant in biscuits taken from nine towns of a city. from each town, 6 samples of biscuits would be collected. in this case whether a one – way or a two-way anova should be applied?

Sorry, but I don’t understand your scenario. It doesn’t appear that you are testing a hypothesis, which is what ANOVA is used for.

Charles

Hi, Sir Charles!

I just need some help with my data. I have done two-way ANOVA with replication on my data wherein, briefly, I am trying to compare the length of shoots of corn and anuang after application of mimosine extract. Based on the anova, the sample, the column and the interactions all yielded Pvalue<0.05. I am quite confused what follow up test should i use to determine where exactly the difference is. What should I do?

Ecarg,

There are many possible tests, but the most commonly used are contrasts and Tukey’s HSD. These are both supported in the Real Statistics website and software. See the following webpages:

http://www.real-statistics.com/two-way-anova/follow-up-analyses-for-two-factor-anova/tukey-hsd-after-two-factor-anova/

http://www.real-statistics.com/two-way-anova/follow-up-analyses-for-two-factor-anova/contrasts-two-factor-anova/

Since you have a significant result for the interaction, you should probably start by using Tukey’s HSD on the interaction.

Charles