Module 10: Analysis of Categorical Data Statistics (OA3102)

Transcription

1 Module 10: Analysis of Categorical Data Statistics (OA3102) Professor Ron Fricker Naval Postgraduate School Monterey, California Reading assignment: WM&S chapter Revision:

2 Goals for this Lecture Understand and be able to conduct tests for discrete contingency table data One-way chi-square goodness-of-fit tests Homogeneity Other distributions Two-way chi-square tests Independence Fixed row or column totals Revision:

3 Analyzing Categorical Data Many experiments result in, or data sets have, qualitative (often nominal) data Data can be summarized in terms of the number of observations that fall in each category Examples A survey asks sailors to select the top reason why they would leave the service INSURV rates ships and submarines as sat, deficient, or unsat In an operational test, equipment failure mode is recorded Revision:

4 Experiment Characteristics 1. There are n identical trials 2. The trials are independent 3. Trial outcome must be exactly one of k possible outcomes where for p i, the probability of outcome i, p p p k The data are the number of each outcome n n n n 1 2 k Revision:

5 One-Way Classification Each item classified into one (and only one) of k categories (cells) Denote counts as n 1, n 2,, n k with n 1 + n n k = n Population Random sample of size n Classify Category 1 Category 2 Cell frequency n 1 Cell frequency n 2 Category k Cell frequency n k Revision:

6 Testing Hypotheses So, we might have a hypothesis we want to test, such as H : p p, p p,..., pk pk 0 1 1,0 2 2,0,0 One approach: Look at how far off counts are from what is expected under the null, where E n How to decide what is far off? i np i Revision:

7 Chi-square Test Turns out that, assuming the null is true, X 2 k i1 k i1 (observed count - expected count) expected count ( n - np ) i np i i 2 ~ 2 2 Reject if statistic too large Assess too large using chi-squared distribution Revision:

8 A Couple of Notes To implement, first calculate X 2 statistic Then, after determining the appropriate degrees of freedom (), calculate either the rejection region or the p-value These problems will always be right tailed tests So, it s either RR X : X, 2 2 or p-value Pr( X ) For chi-squared approximation to hold, all expected counts need to be greater than 5 Revision:

9 Goodness-of-Fit Tests Have counts for k categories, n 1, n 2,, n k, with n 1 + n n k = n The hypotheses to be tested are H : p p, p p,..., p p 0 1 1,0 2 2,0 k k,0 H : at least one p p a i i,0 2 2 X ~ n 1 Under the null hypothesis, That is, the test statistic has a chi-squared distribution with n-1 degrees of freedom Revision:

10 Goodness-of-Fit Test for Homogeneity Null hypothesis is the probability of each category is equally likely: pi,0 1/ k, i 1,2,..., k I.e., the distribution of category characteristics is homogeneous in the population If the null is true, in each cell (in a perfect world) we would expect to observe npi,0 counts per cell n k Revision:

11 Example 14.1 A group of rats are tested for whether they have a preference for one of three doors Is there sufficient evidence to say that one they prefer one door over the others? Write out the hypotheses: Revision:

12 Example 14.1 (continued) Observed frequency Expected frequency Door 1 Door 2 Door Revision:

13 Example 14.1 (continued) Revision:

14 Doing the Test in R In R, use the chisq.test function Goodness-of-fit is the default if all you give it is a vector of counts For Example 14.1: Revision:

15 Example 14.2 The number of accidents Y per week at an intersection were recorded for n=50 weeks Assuming the weekly counts are independent, test the hypothesis that Y has a Poisson distribution (at =0.05) Write out the hypotheses: Revision:

16 Example 14.2 (continued) y n i p i,0 np i, or more 0 Revision:

17 Example 14.2 (continued) Revision:

18 Doing the Test in R As before, use the chisq.test function For goodness-of-fit tests other than homogeneity, must also give a vector of probabilities For Example 14.2: But here need to correct for degrees of freedom Revision:

19 Variable 2 Contingency Tables A contingency table gives counts of all pairwise combinations for two variables Variable 1 Category 1 Category 2 Category 3 Category 1 Category 2 Category 3 Revision:

20 Some Notation for Contingency Tables Table has r rows and c columns Observed cell counts are n ij, with Denote row sums: Denote column sums: r c i1 j1 c r n, i 1,..., r i n ij n j1 r c n, j 1,..., c j ij i1 ij Revision:

21 Chi-square Test for Independence Independence means the probability of being in any cell is the product of the marginal row and column probabilities Variable 2 Row 1 Variable 1 Column 1 Column 2 Probability an obs is both in Row 1 and Column 2 p i=1 x p j=1 p i=1 x p j=2 Pr(Row 1) = p i=1 Row 2 p i=2 x p j=1 p i=2 x p j=2 Pr(Row 2) = p i=2 Revision: 3-12 Pr(Col 1) = p j=1 Pr(Col 2) = p j=2 Probability that a random obs is in Row 2 21

22 The Hypotheses The hypotheses to be tested are: H H 0 a Mathematically: H : p p p, i 1,2,..., r; j 1,2,..., c 0 : The variables are independent : The variables are not independent ij i j H : p p p, for some i and j a ij i j However, the probabilities are not observed, so what to do? Revision:

23 Estimating the Probabilities Since the probabilities are not observed, they must be estimated from the data (under the assumption of independence) Variable 1 Column 1 Column 2 Variable 2 Row 1 Row 2 pˆ pˆ pˆ pˆ pˆ pˆ 11 i1 j1 12 i1 j2 pˆ pˆ pˆ ˆ ˆ ˆ 21 i2 j1 p22 pi2 p j2 pˆ r / n i1 1 pˆ r / n i2 2 pˆ c / n j1 1 pˆ c / n j2 2 23

24 Chi-square Test Statistic Test statistic: X r c n ij - E n ij r c n ij - rc i j n E n rc i1 j1 i1 j1 i j ij since, under the null, the expected count is calculated as ˆ ˆ ˆ r c i j i j E nij npij npi p j n n n n n rc Revision:

25 Conducting the Test Proceed as with the goodness-of-fit test Except degrees of freedom are ( r1)( c1) Large values of the chi-square statistic are evidence that the null is false As before, these problems are right tailed tests So, it s either or p 2 2 -value Pr( X ) RR X X r 1c :, r 1 c 1 For chi-squared approximation to hold, the expected counts in all the cells must be > 5 Revision:

26 Example 14.3 Data collected on 1,000 people Whether or not they got the flu Whether they got none, one, or two flu shots We want to know whether the data indicate a dependence between getting the vaccine and not getting the flu Write out the hypotheses: Revision:

27 Example 14.3 (continued) Status No Vaccine One Shot Two Shots Total Flu No flu Total ,000 Revision:

28 Example 14.3 (continued) Revision:

29 Doing the Test in R Again, use the chisq.test function Chi-squared test of independence is the default if you give the function a matrix For Example 14.3: Revision:

30 Contingency Tables with Fixed Row or Column Totals Sometimes, desirable to pre-specify a fixed quantity for either row or column categories Idea: Each row (or column) is a population and proportion that falls in each category is the same Essentially, asking whether the distribution with some set of categories for two or more populations is the same or not Sometimes called chi-squared test of homogeneity Good news: Calculation is exactly the same as test for independence! Revision:

31 Example voters were polled in each of 4 wards to determine fraction favoring candidate A Do the data present sufficient evidence to indicate that the fraction of voters favoring candidate A differ among the wards? Write out the hypotheses: Revision:

32 Example 14.4 (continued) Opinion Ward 1 Ward 2 Ward 3 Ward 4 Total Favor Not favor Total Revision:

33 Example 14.4 (continued) Revision:

34 Doing the Test in R Again, use the chisq.test function Since the tests are equivalent, just use the chi-squared test of independence For Example 14.4: Revision:

35 What We Have Just Learned Discussed tests for contingency table data One-way chi-square goodness-of-fit tests Homogeneity Other distributions Two-way chi-square tests Independence Fixed row or column totals Revision:

36 Homework WM&S chapter 14 Optional practice problems: 4, 12, and 13 Extra credit: None Hints Exercise 12: Use dpois in R to calculate the necessary Poisson probabilities Exercise 13: Use the chisq.test function in R for part (c) Revision: