Topic 21 Goodness of Fit

Transcription

1 Topic 21 Goodness of Fit Contingency Tables 1 / 11

2 Introduction Two-way Table Smoking Habits The Hypothesis The Test Statistic Degrees of Freedom Outline 2 / 11

3 Introduction Contingency tables, also known as two-way tables or cross tabulations are a convenient way to display the frequency distribution from the observations of two categorical variables. For an r c contingency table, we consider two factors A and B for an experiment. This gives r categories A 1,... A r for factor A and c categories for factor B B 1,... B c 3 / 11

4 Two-way Table Here, we write O ij to denote the number of occurrences for which an individual falls into both category A i and category B j. The results is then organized into a two-way table. B 1 B 2 B c total A 1 O 11 O 12 O 1c O 1 A 2 O 21 O 22 O 2c O A r O r1 O r2 O rc O r total O 1 O 2 O c n where O i, i = 1,..., r are the row marginals, O j, i = j,..., c are the column marginals, and n is the number of observations. 4 / 11

5 Smoking Habits Returning to the study of the smoking habits of 5375 high school children in Tucson in 1967, here is a two-way table summarizing some of the results. student student smokes does not smoke total 2 parents smoke parent smokes parents smoke total / 11

6 The Hypothesis For a contingency table, the null hypothesis we shall consider is that the factors A and B are independent. To set the parameters for this model, we define p ij = P{an individual is simultaneously a member of category A i and category B j }. Then, we have the parameter space Θ = {p = (p ij, 1 i r, 1 j c); p ij 0 for all i, j = 1, r i=1 j=1 Write the marginal distribution c p i = p ij = P{an individual is a member of category A i } and p j = j=1 r p ij = P{an individual is a member of category B j }. i=1 c p ij = 1}. 6 / 11

7 The Test Statistic The null hypothesis of independence of the categories A and B can be written H 0 : p ij = p i p j, for all i, j versus H 1 : p ij p i p j, for some i, j. The null hypothesis p ij = p i p j can be written in terms of observed and expected observations as E ij n = O i O j or E ij = O i O j. n n n As before, the appropriate G 2 statistic follows from the likelihood ratio test criterion. The χ 2 statistic is a second order Taylor series approximation to G 2. G 2 = 2 r c i=1 j=1 O ij ln E ij O ij r c i=1 j=1 (O ij E ij ) 2 E ij = χ 2. 7 / 11

8 Smoking Habits For the data set on smoking habits in Tucson, we find that the expected table is student student smokes does not smoke total 2 parents smoke parent smokes parents smoke total For example, E 11 = O 1 O 1 n = = / 11

9 Degrees of Freedom To determine the degrees of freedom, start with a contingency table with no entries but with the prescribed marginal values. B 1 B 2 B c total A 1 O 1 A 2 O 2... A r O r total O 1 O 2 O c n The degrees of freedom is the number of values that we can place on the table before all the remaining values are determined. Note that we can fill c 1 values in each of the r 1 rows before the remaining values are determined. Thus, the degrees of freedom is (r 1) (c 1). Exercise. Determine the number of degrees of freedom and compute the χ 2 statistic for the example on smoking habits. 9 / 11

10 To perform the χ 2 test in R, Performing the Test > smoking<-matrix(c(400,416,188,1380,1823,1168),nrow=3) > smoking [,1] [,2] [1,] [2,] [3,] > chisq.test(smoking) Pearson s Chi-squared test data: smoking X-squared = , df = 2, p-value = 6.959e / 11

11 We can look at the residuals for the entries in the χ 2 test as follows. > smokingtest<-chisq.test(smoking) > residuals(smokingtest) [,1] [,2] [1,] [2,] [3,] Introduction O ij E ij Eij Exercise. Make three horizontally placed chigrams that summarize the residuals for this χ 2 test in the example above. Use this to explain the sources of the major contribution to the χ 2 statistic. 11 / 11