Then if the two classifications are independent of each other, a cell probability will equal the product of its respective row and column probabilities in accordance with the multiplicative law of probability.
Example of obtaining Column and Row Probabilities:
For example, the probability that a particular defect will occur in shift 1 and is of type A is (p1) (pA). While the numerical values of the cell probabilities are unspecified, the null hypothesis states that each cell probability will equal the product of its respective row and column probabilities.
This condition implies independence of the two classifications. The alternative hypothesis is that this equality does not hold for at least one cell.
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In other words, we state the null hypothesis as:
H0: the two classifications are independent, while the alternative hypothesis is;
Ha: the classifications are dependent.
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To obtain the observed column probability, divide the column total by the grand total, n denoting the total of column j as cj, we get.
Similarly, the row probabilities P1 P2, and P3 are estimated by dividing the row totals r1, r2, and
R3 by the grand total n, respectively.
The represent of the observed frequency of the cell in row i and column j of the contingency table by nij, we have;
Expected cell frequencies the represent of the observed frequency of the cell in row i and column j of the contingency table by nij, we have;
Estimated expected cell frequency when Ho is true. In other words, when the row and column classifications are independent, the estimated expected value of the observed cell frequency nij in an r x c contingency table is equal to its respective row and column totals divided by the total frequency.
