2 way anova
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 This topic has 20 replies, 7 voices, and was last updated 17 years, 7 months ago by Slesh.

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January 21, 2005 at 5:13 pm #38153
I’m trying to run a 2 way anova for the first time
If I include in the analysis the interaction term, both the factors and the interaction are significant (p=0.000). If I do not include the interaction term, one of the 2 factors is not significant. I cant explain why. Is it maybe due to the fact that the interaction is significant and therefore both the factors are considered important for hierarchical reasons? Im not convinced of this explanation; can you give me a better one?
Thanks a lot
0January 21, 2005 at 5:34 pm #113836
K.SubbiahParticipant@K.Subbiah Include @K.Subbiah in your post and this person will
be notified via email.Hello There:
Please check the F values of the coefficients in the models. Also, check the MSE values in the models. My guess is that your experimental error (MSE) is aligned with the interaction term. That is why, in the first model, by including the interaction term, MSE would have lowered in value and therefore, made all the terms significant. This could be due to the lack of randomization in your experiment. Hope this helps. K.Subbiah0January 21, 2005 at 5:39 pm #113838Thanks K.Subbiah,
these are the F values and the MSE in case the interaction is included
Temp 3 4107.09 1369.03 471.06 0.000
Oven 3 142.59 47.53 16.35 0.000
Temp*Oven 9 601.53 66.84 23.00 0.000
Error 16 46.50 2.91
Total 31 4897.72
This is the result if interaction term is NOT included
Source DF SS MS F P
Temp 3 4107.09 1369.03 52.82 0.000
Oven 3 142.59 47.53 1.83 0.167
Error 25 648.03 25.92
Total 31 4897.720January 21, 2005 at 7:53 pm #113846
K.SubbiahParticipant@K.Subbiah Include @K.Subbiah in your post and this person will
be notified via email.Hello there:
Please check the residual distribution in the cases. If you have a distinct pattern, then randomization of the experimental runs will be the only solution. You have to run the experiment, in random order, and do the analysis again and see how the ANOVA results turn out. Why don’t you DoE experiment with 2×2 design. Please use multiple runs of the combinations with maximum possible randomization. Sometimes physical limitations will prevent the extent to which you can randomize. Therefore, balance things there. Good Luck and it will interesting to see the results and hope you can post it.0January 22, 2005 at 12:54 am #113861
BrookiepParticipant@Brookiep Include @Brookiep in your post and this person will
be notified via email.The answer here is that the interation Temp*Oven has a large practiacl contibution to the model so if you then try to remove the the interaction the varience must become error. This in turn will then affect the result as follows
MS is derived by: DF/SSFactor
F is derived by MSFactor/MSError
So you can see that if you increase the error term then the resultant F value is then smaler than F critical looked up in the F table. In this case a value of 2.32 therfore the factor of oven is no longer significant due to the large amount of error intoduced to the model.
The model looks OK as has been suggested just check for normality and ranomization of the residuals
0January 27, 2005 at 11:10 am #114082Doesn’t it indicate from the initial test that the interaction is very strong. Once you’ve removed the interaction from the model, that it stating that the temp is the only factor that plays a statistically in this ?
If the model is temperature sensitive, I assume that any other factors would also have a low pvalue or a high Fvalue. Is this correct ?
Thanks0January 27, 2005 at 8:07 pm #114106
BrookiepParticipant@Brookiep Include @Brookiep in your post and this person will
be notified via email.The contribution from each element is as follows
Temp = 84%
Oven = 3%
Temp*Oven = 12%
Error = 1%
So you can see that as you say that Temp is the main effect and that Oven has little effect. But the interation is important as it conributes 12% so if you just ignored Oven then the output would be unpredictable if no control of Oven was made.
As for other factors this model has very little error so it is likely that any other factors would have little effect as you rightly say.
Just a last thought are the factors of Oven and Temp in your experiment independant? If they are not that would explain the strong interation0January 28, 2005 at 1:48 am #114124Real stupid question,
How were you able to determine the percent contribution from each element?
Would greatly appreciate the explination0January 28, 2005 at 7:47 am #114132Thank you all for your replies,
considering that the Oven has a small effect compared to Temperature, I’m wondering if the factor Oven can ve considered just a blocking factor. In fact the different ovens simply heat up the components under analysis, and maybe there souldn’t be any reasonable difference among them.
Using a Balanced anova where the Oven is considered a random factor I get these results:
Analysis of Variance (Balanced Designs)
Factor Type Levels Values
Temper fixed 3 350 400 450
Oven typ random 4 A B C D
Analysis of Variance for Sigma Y
Source DF SS MS F P
Temper 2 1035.75 517.87 8.48 0.018
Oven typ 3 319.17 106.39 1.74 0.258
Temper*Oven typ 6 366.58 61.10 25.28 0.000
Error 12 29.00 2.42
Total 23 1750.50
In this case the Oven is not significant0January 28, 2005 at 7:50 am #114133Sorry for the bad format in the previous post
Analysis of Variance (Balanced Designs)
Factor Type Levels Values
Temper fixed 3 350 400 450
Oven typ random 4 A B C DAnalysis of Variance for Sigma Y
Source DF SS MS F P
Temper 2 1035.75 517.87 8.48 0.018
Oven typ 3 319.17 106.39 1.74 0.258
Temper*Oven typ 6 366.58 61.10 25.28 0.000
Error 12 29.00 2.42
Total 23 1750.500January 28, 2005 at 10:05 am #114139Ff,
You problem is timely given the recent discussion on the significance of main effects and their interactions.
Something to think about is that oven temperature is one thing, but the way components heat up and lose heat is quite another. I think you’ll find that the latter is related to the type of load and the performance of the oven.
Last but not least, you should be aware that equipment often fails, so that it is important to understand how to interpret data if and when it is not right. This is why previously I recommended drawing all the possible interactions for a two factor design and draw some distributions at each level.
If you assume the equipment is working properly and the data is homogeneous (same variance.) Then you will be able to understand why you can’t have an insignificant main effect and a significant interaction with that main effect. If you do then something is wrong!
Regards,
Andy0January 28, 2005 at 5:34 pm #114157
BrookiepParticipant@Brookiep Include @Brookiep in your post and this person will
be notified via email.The calculation as follows
SSfactor / SStotal
Substituting error for factor to understand the % error (unexplained variation) in the model
0January 28, 2005 at 6:05 pm #114159
BrookiepParticipant@Brookiep Include @Brookiep in your post and this person will
be notified via email.It looks like you have changed the model completely the overall degress of freedom has reduced from 31 to 23 so you must have removed some data from the model in which case the two tests are not comparable.
I think that in this model oven and temp are not truly independant factors my recomendation would be to run a one way ANOVA with oven as the factor and temp as the output that way you will be able find which oven has the most appropriate temp control and by looking at the TEV you will also see which oven has the least variation
Then you can either calibrate the ovens or apply an offset
0January 31, 2005 at 7:29 am #114206Brookiep,
you are right, I erroneously posted a wrong analysis where I had removed some data. This is the correct analysis with the Oven considered as a random factor:
Analysis of Variance (Balanced Designs)
Factor Type Levels Values
Temp fixed 4 1 2 3 4
Oven random 4 A B C D
Analysis of Variance for Sigma
Source DF SS MS F P
Temp 3 4107.09 1369.03 20.48 0.000
Oven 3 142.59 47.53 0.71 0.569
Temp*Oven 9 601.53 66.84 23.00 0.000
Error 16 46.50 2.91
Total 31 4897.720January 31, 2005 at 7:31 am #114207…… I must have some problems in pasting data from Minitab. I hope this time it works:
Analysis of Variance (Balanced Designs)
Factor Type Levels Values
Temp fixed 4 1 2 3 4
Oven random 4 A B C DAnalysis of Variance for Sigma
Source DF SS MS F P
Temp 3 4107.09 1369.03 20.48 0.000
Oven 3 142.59 47.53 0.71 0.569
Temp*Oven 9 601.53 66.84 23.00 0.000
Error 16 46.50 2.91
Total 31 4897.72
0January 31, 2005 at 3:27 pm #114234FF
Not withstanding any of the interesting post trying to solve your problem, I wanted to address your original question: “Why, when I remove the interactions from my model does the significance of my main effects change?” To answer the question, you need to go back to what an ANOVA is really telling you relative to the hypothesis you are making. If you start with a simple oneway ANOVA you can expand the concepts then to the twoway.
This was your original post:
I’m trying to run a 2 way anova for the first time
If I include in the analysis the interaction term, both the factors and the interaction are significant (p=0.000). If I do not include the interaction term, one of the 2 factors is not significant. I cant explain why. Is it maybe due to the fact that the interaction is significant and therefore both the factors are considered important for hierarchical reasons? Im not convinced of this explanation; can you give me a better one?
You followed with this data analysis on Friday the 21st:
These are the F values and the MSE in case the interaction is included
Temp 3 4107.09 1369.03 471.06 0.000
Oven 3 142.59 47.53 16.35 0.000
Temp*Oven 9 601.53 66.84 23.00 0.000
Error 16 46.50 2.91
Total 31 4897.72
This is the result if interaction term is NOT included
Temp 3 4107.09 1369.03 52.82 0.000
Oven 3 142.59 47.53 1.83 0.167
Error 25 648.03 25.92
Total 31 4897.72
In a oneway ANOVA you hypothesize (null) that the level of your treatments do not have a significant effect on your mean output. The alternative is therefore. that at least one of the treatment levels is different (ie gives a different mean output). The technique in the ANOVA is to compare the variation of the treatment means around the grand mean of your data, to the variation in the individual elements in a treatment to the mean of that treatment. The ratio of these variances follows an F distribution and the probabilities of attaining any ratio can be calculated relative to the degrees of freedom in the data. So what does that mean? Basically its saying that if you look at the total variation in data, separate it between what is associated with the treatments and what is just random or error, then if the variation from treatments is approximately equal to that of the error, the treatments do not effect the output.
Now moving to the twoway scenario, if you can separate the variation caused by each set of treatments (the two variables and the interactions) then you can determine the probabilities for each. When you drop the interaction term from your model, you put the variation and degrees of freedom associated with interaction back into the error term and inflate its value. By increasing the denominator in the ratio of variances you lose sensitivity to see significance of the treatment effect you are testing and therefore alter your decision based against the null hypothesis. That is why your oven drops out when you dont account for the variation associated with the interaction. When faced with things like this, go back to what you do know (how the ratio of variance works in a oneway works) and expand the concept.0April 14, 2005 at 7:28 pm #117765hi
i need to doa a 2 way anova test on concentrations of NO3 in soil. there are four categories desert, argricultural, desert converted to residential and agricultural converted to residential.i have a stat software “minitab” any suggestions?
thank youslesh0April 14, 2005 at 9:06 pm #117771Are you sure you want 2way ANOVA? Are you trying to look at the NO3 levels in each of the 4 soils, or are soils and amount of NO3 producing another variable?
My suggestion is to use MINITAB help for both oneway and twoway ANOVA. Just open them under Stat>ANOVA>One Way… , then click on the “Help” tab in the lower left corner. This has a description of the method, an example you can run for yourself from a file (click the “example” link at the top of the page), and how to interpret the results.0April 15, 2005 at 2:12 pm #117829hi ftsbb,
maybe you are right, let me give you the details.
as stated earlier there are four categories:
1. agrarian (unchanged)
2. desert (unchanged)
3. agrarian ( converted to residential)
4. desert ( converted to residential)
there is a obvious difference in NO3 content between agrarian and the desert categories. in the order of high to low :
agrarian ( converted to residential) –> highest
agrarian (unchanged)
desert (unchanged)
desert ( converted to residential)–> lowest highest
agrarian (unchanged)
desert (unchanged)
desert ( converted to residential)–> lowest highest
agrarian (unchanged)
desert (unchanged)
desert ( converted to residential)–> lowest lowest lowesti would like to see if the past usage of land has any effect in the present category. do you think one way anova is sufficient. to see the legacy effect do i multiply present usage and past usage to make a new column? and see its effect?
thank you
slesh0April 15, 2005 at 2:40 pm #117837Did you look at the help options in MINITAB? Did this not help? I’m assuming you didn’t even look at them…
I’m still not sure I understand your actual data set, but I’ll give it a stab. Let’s try another approach: forget about statistics & ANOVA for a minute. Have you looked at a box plot of the data? For example, use the data from the following groups:
Agri unchanged
Agri converted
Desert unchanged
Desert converted
If one of the box plots is wildly different than the others, it should give you a hint about differences caused by previous soil use. Again, MINITAB has all kinds of help on box plots. I’m always a fan of using commonsense tools before diving into statistics.0April 15, 2005 at 7:32 pm #117873Hi FTSBB,you were right it’s more appropriate to go with simple appraches like box plot than to jump into hardcore statistics. this is my first attempt using minitab, or any statistical software for that matter so it is quite confusing… and to make things worse i do not have an indepth statistical background either.. but i need to this this analysis anyhow..
thank you for your time and help
i’ll be posting more messages if i run into more problems.
thanks
slesh0 
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