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## calculate total df in latin square design

I have a design where 4 blends of oil are being used on 4 cars and at 4 routes. 3 replications

so

cars = 3 df

route = 3 df

oil blends = 3df

replications = 2df

model df = 11

how i can caluctate total degree of freedom.

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## Re: calculate total df in latin square design

My apologies. I used a different design. Let's start over.

You have 3 factors, each at 4 levels. The Latin square design is a regular fractional factorial with 4^(3-1) = 16 runs. You replicated the  design twice for a total of 48 runs. There are 48 degrees of freedom in this data set.

So the model uses 1 degree of freedom as before to estimate the corrected total sum of squares. The estimates of the fixed effects use 3 df for oil blend, 3 df for car, and 3 for route or 9 df for the model. The df for the error sum of squares is then 48 - 1 - 9 = 38. Your book example also estimates the replicate as a fixed effect, presumably as blocking, so you would subtract another 2 df for 36. Are you using the replicates for blocking? If so, what is the nature of your blocks?

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3 REPLIES 3
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## Re: calculate total df in latin square design

You start with 4 blends x 4 cars x 4 routes x 3 replicates = 192 degrees of freedom assuming that the observations are independent. You lose 1 degree of freedom for the grand mean when calculating the corrected total sum of squares. Your message indicates that you will use a model with only 'main effects' (first order additive terms) for a total of 9 degrees of freedom. That leaves you with 182 degrees of freedom for the error sum of squares.

The Analysis of Variance report in Fit Least Squares confirms this accounting: Learn it once, use it forever!
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## Re: calculate total df in latin square design

Okay, then why book is showing something els e. Please see attached picture.

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## Re: calculate total df in latin square design

My apologies. I used a different design. Let's start over.

You have 3 factors, each at 4 levels. The Latin square design is a regular fractional factorial with 4^(3-1) = 16 runs. You replicated the  design twice for a total of 48 runs. There are 48 degrees of freedom in this data set.

So the model uses 1 degree of freedom as before to estimate the corrected total sum of squares. The estimates of the fixed effects use 3 df for oil blend, 3 df for car, and 3 for route or 9 df for the model. The df for the error sum of squares is then 48 - 1 - 9 = 38. Your book example also estimates the replicate as a fixed effect, presumably as blocking, so you would subtract another 2 df for 36. Are you using the replicates for blocking? If so, what is the nature of your blocks?

Learn it once, use it forever!
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