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L10 Analysis of Variance

# L10 Analysis of Variance - Multivariate Analysis Dealing...

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Multivariate Analysis. Dealing with more than one variable. In all we have done so far we have only studied the statistical analysis of one variable. The last part of the course is going to deal with the case where we have to consider more than one variable. Sometimes the additional variables will be random, sometimes they will not, but in all cases we will be concerned with the nature of the relationships between the variables. Mostly we shall concentrate on just two variables and generalize to the case of many. First we shall consider how to detect whether or not two variables are independent. A Goodness of Fit Test of Independence. Imagine two characteristics that an individual may have, say hair colour and eye colour, we may think of them as two variables and it is of interest to establish whether or not the variables are distributed independently across individuals. Suppose the entire range of hair colour is divided into c mutually exclusive and exhaustive categories numbered j = 1,..,c and similarly the entire range of eye colour is divided into r mutually exclusive and exhaustive categories numbered i = 1,..,r. The true probabilities that a randomly selected person has a particular hair colour / eye colour combination can be arranged on an r x c grid whose rows correspond to the eye colour categories and whose columns correspond to the hair colour categories as in Table 1 below. The p ij ’s are joint probabilities of having a particular hair / eye colour combination and taken together they correspond to the joint probability distribution of the eye colour-hair colour combinations. The row sums of probabilities (p i. ’s) represent the marginal probabilities of having the i’th eye colouring regardless of which hair colour an individual has and taken together they constitute the marginal probability distribution of eye colouring. Similarly the column sums (p .j ’s) are marginal probabilities of having the j’th hair colouring regardless of which eye colour an individual has and taken together they constitute the marginal probability distribution of hair colouring. Hence all of the p ij ’s sum to one as does the sum of the p i. ’s and the sum of the p .j ’s.

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Table 1. General probability structure. hair colour j=1,..,c Row sums p 11 p 12 p 1c p 1. p 21 p 22 p 2c p 2. p ij p i. p r1 p r2 p rc p r. eye colour i=1,..,r Col sums p .1 p .2 p .j p .c 1 Table 2. Independent Probability Structure hair colour j=1,..,c Row sums p 1. p .1 p 1. p .2 p 1. p .c p 1. p 2. p .1 p 2. p .2 p 2. p .c p 2. p i. p .j p i. p r. p .1 p r. p .2 p r. p .c p r. eye colour i=1,..,r Col sums p .1 p .2 p .j p .c 1
The Theoretical Implication of Independence. Back in chapter 2 we observed that, if two events A and B were independent, their joint probability was equal to the product of their marginal probabilities so that P(A B) = P(A)P(B) under independence. If this is true for hair colour and eye colour then P(i’th eye colour and j’th hair colour) = p ij = P(i’th eye colour regardless of hair colour) x P( j’th hair colour regardless of eye colour) = p i.

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