The following data were collected on protein levels in minnow larvae which had b
ID: 3200971 • Letter: T
Question
The following data were collected on protein levels in minnow larvae which had been exposed to particular levels of zinc and copper, which are toxic pollutants (Data from D.A.J. Ryan, J.J Hubert, and J. Parrott, Biometrics 48: 155.162, 1992) Suppose that these have been stored in a data file "minnows dat copper Zinc Protein 201 375 186 750 173 1125 110 1500 115 37.5 0 202 37.5 375 161 37.5 750 172 37.5 1125 138 37.5 1500 133 75 75 375 165 75 750 148 75 1125 143 75 1500 123 112.5 188 112.5 375 172 112.5 750 157 112.5 1125 115 112.5 1500 108 150 133 375 125 150 150 750 184 150 1125 135 150 1500 114 Suppose that the following analysis was run: minnows read. table( ''minnows ,dat header T results2 lm( Protein Copper Zinc, data minnowsExplanation / Answer
Here the two factors are Copper ad Zinc with each being assigned 5 levels which can be shown as below
The underlined part is the level for each factor .The total levels for each factor is 5.Thus the degree of freedom for factor Copper is (5-1=4) and degree of freedom for factor Zinc(5-1=4) .The error degree of freedom is (5-1)*(5-1)=16.Thus the total degree of freedom is 16+4+4=24 which is the same as n-1=25-1=24.
Considering the two way ANOVA with interaction where the interaction degree of freedom will be (5-1)*(5-1) =16 with the error degree of freedom being 0 which is not possible.It means we have no replication in each factor which is two way ANOVA without interaction.Thus,when the experiment is carried out more time under each cross tab cells then the interaction terms will play a role.Here,Two way ANOVA without interaction is possible.
Output for reference
Zn 0 375 750 1125 1500 Cu 0 201 186 173 110 115 37.5 202 161 12 138 133 75 204 165 148 143 123 112.5 188 172 157 115 108 150 133 125 184 135 114Related Questions
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