Question

1. You ask her what was the experimental unit and were the experimental units randomized. She is unfamiliar with these concepts. Please explain experimental unit and the need for randomization of the experimental units to the treatment combinations to her.

2. You decide to analyze the experiment as a 3 X 3 factorial arrangement using a completely randomized design and obtain the following ANOVA tables.

Analysis of Variance Table

Response: diastolic

          Df Sum Sq Mean Sq F value   Pr(>F)  

diet       2 322.83 161.42 4.2055 0.032819 *

drug       2 811.06 405.53 10.5655 0.001042 **

diet:drug 4 155.76   38.94 1.0145 0.427585  

Residuals 17 652.50   38.38                   

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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Anova Table (Type II tests)

Response: diastolic

          Sum Sq Df F value   Pr(>F)  

diet      379.50 2 4.9437 0.020307 *

drug      811.06 2 10.5655 0.001042 **

diet:drug 155.76 4 1.0145 0.427585  

Residuals 652.50 17                   

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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Anova Table (Type III tests)

Response: diastolic

            Sum Sq Df   F value    Pr(>F)   

(Intercept) 185050 1 4821.2144 < 2.2e-16 ***

diet           375 2    4.8838 0.021093 *

drug           796 2   10.3735 0.001136 **

diet:drug      156 4    1.0145 0.427585   

Residuals      652 17                       

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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

a. For this analysis, which of the SS methods would you choose? Explain why?

b. The lack of a sign diet X drug interaction indicates __________________?

Answer 1

1.a. Experimental unit :

First we need to define what is meant by an ‘experimental unit’. It may seem trivial, but when encountering an experimental situation this critical step is not always obvious. Consider a situation where someone wants to evaluate polluted stream water for its effect on fish lesions. They set up 2 aquarium, each with 50 fish. They randomly assign a water treatment (polluted vs. control) to each of the aquariua. After 30 days, they catch 10 fish from each aquarium and count the number of lesions. The treatment design is a single-factor design with 2 levels of water treatment, and a one-way ANOVA can be run on the data. But which one is the experimental unit?

Going back to our definition, the experimental unit is that which receives the treatment. In this case, we have applied a water treatment to each aquarium. The fish are not the experimental units. In order for individual fish to be experimental units, somehow the investigators would have to take one fish at a time and apply the treatment independently to each fish. This would be impractical from a logistics standpoint, and was not done. Instead, the water treatment levels were applied to entire aquarium, and so the experimental unit is an aquarium with 50 fish.

1.b. The principle of randomization involves the allocation of treatment to experimental units at random to avoid any bias in the experiment resulting from the influence of some extraneous unknown factor that may affect the experiment. In the development of analysis of variance, we assume that the errors are random and independent. In turn, the observations also become random.

The principle of randomization ensures this. The random assignment of experimental units to treatments results in the following outcomes. a) It eliminates the systematic bias. b) It is needed to obtain a representative sample from the population. c) It helps in distributing the unknown variation due to confounded variables throughout the experiment and breaks the confounding influence.

Randomization forms a basis of valid experiment but replication is also needed for the validity of the experiment. If the randomization process is such that every experimental unit has an equal chance of receiving each treatment, it is called a complete randomization.

2.a. Here in this analysis i would surely go for type-ii SS, because

i. Here the order do not matters between the drug and the diet, i.e, if we interchange drug:diet with diet:drug in the main model then the result would have been same, which is a major cons of type - i SS,

ii. Here the interaction effect of the drug as well as the diet is insignificant, concluding that there is no association between them, they are independent of each other. Which is a major pro for type - ii.

iii. It is a factorial design experiment so type ii is much more preferable than the other types,

iv.More over testing of any main effects in presence of interaction is a vague concept so type - iii is discarded.

2.b. The lack of sign of explaining the response by the interaction effects explains that the two main factor are independent to each other, and there is no entangled effect of diet and drug present over the diastole. We can now further carry out our analysis by observing the differential effects of the drugs and the diets on the diastole for each of the patients respectively.