Chapter 13. Inferences for Tables

Transcription

1 Chapter 13 Inferences for Tables

2 Lesson 13-1 Test for Goodness of Fit

3 Test for Goodness of Fit To analyze categorical data, we construct two-way tables and examine the counts or percents of the explanatory and response variables

4 Two Way Table Proportion of M&M Colors 1.69 ounce bag of plain M&M Color Observed Count, O Expected Count, E O E 2 E Blue (57) = Brown (57) = Green (57) = Orange (57) = Red (57) = Yellow (57) = Total

5 Null and Alternative Hypothesis We want to compare the observed counts to the expected counts The null hypothesis is that there is no difference between the observed and expected counts. The alternative hypothesis is that there is a difference between the observed and expected counts.

6 Chi-Square Statistic The chi-square statistic measures how well the observed counts fit the expected counts, assuming that the null hypothesis is true. In order to determine whether the distribution has changes we need a way to measure the observed counts (O) from the expected counts (E) X 2 O E E 2

7 Chi-Square Distribution, 2 The distribution of the chisquare statistic is called the chi-square distribution, 2. This distribution is a density curve. The total area under the curve is 1. The curve begins at zero on the horizontal axis and is skewed right. As the degree of freedom increase, the sample of the curve becomes more symmetric.

8 Goodness of Fit Test Use the M&M chi-square statistic, find the probability of obtaining a X 2 value at least this extreme assuming that the null hypothesis is true. This is known as the Goodness of Fit Test Degree of freedom (n 1), where is the number of categories.

9 Goodness of Fit Test Using the M&M data chi-square statistic X 2 = Degree of freedom (n 1) = 6 1 = 5. If the M&Ms were distributed as claim by Mars Inc, an observed value of or higher would occur about 0.50% of the time.

10 Assumptions and Conditions Counted Data Condition Check to see if that the data are counts for categories of categorical variables. Randomization Condition Expected Cell Frequency Condition We should expect to see at least 5 individuals in each cell.

11 Example Chi-Square Test for Goodness of Fit Does your zodiac sign determine how successful you will be later in life? Fortune magazine collected the zodiac signs of 256 heads of the largest 400 companies. Here are the number of births for each sign: We can see some variation in the number of births per sign and there are more Pisces, but is it enough to claim that successful people are more likely to be under some signs than others? Births Signs 23 Aries 20 Taurus 18 Gemini 23 Cancer 20 Leo 19 Virgo 18 Libra 21 Scorpio 19 Sagittarius 22 Capricorn 24 Aquarius 29 Pisces

12 Example Chi-Square Test for Goodness of Fit If births were distributed uniformly across the year, we would expect about 1/12 of them to occur under each sign of the zodiac. That suggests 256/12, or about 21.3 births per sign. How closely do the observed numbers of births per sign fit this simple null model?

13 Example Chi-Square Test for Goodness of Fit Step 1 State appropriate null and alternative hypotheses for investigating the genetic model. H o : Births are uniformly distributed over zodiac signs. H a : Births are not uniformly distributed over zodiac signs Formal approach specifies the parameter values H o : P Aries = P Taurus =.=P Pisces H a : P Aries P Taurus. P Pisces

14 Example Chi-Square Test for Goodness of Fit Step 2 Verify conditions for performing inference in this setting. 1. Counted Data: we have counts of the number of executives in categories. 2. Randomization: we have convenience sample of of executives, but no reason to suspect bias 3. Expected cell frequency: The null hypothesis expects that 1/12 or the 256 births, or , should occur in each sign. These expected values are greater than 5

15 Example Chi-Square Test for Goodness of Fit Step 3 Calculate the test statistic and the P-value ( Obs Exp) ( ) ( ) X Exp p value

16 Example Chi-Square Test for Goodness of Fit Step 4 State the conclusion. There is sufficient evidence to fail to reject H o since, p- value = > = 0.05, and conclude that there is no evidence of non-uniform distribution of zodiac signs among executives.

17 Example Chi-Square Test for Goodness of Fit The p-value is the area in the upper tail of the 2 model for 12 1 = 11 degrees of freedom above the computed X 2 value. The p-value of says that if the zodiac signs of executives were fact distributed uniformly, an observed chi-square value of 5.09 or higher would occur about 93% of the time. P P X 2 2 ( χ ) P 2 ( χ 5.094) 0.926

18 Lesson 13-2, Part 1 Inference for Two-Way Tables

19 Two-Way Tables The first step in the overall test for comparing several proportions is to arrange the data in a two-way table. Example Many high school graduating classes determine the plans of the graduates. We might wonder whether the plans of students have stayed roughly the same over past decades or whether they have changed. Here is a summary table from one high school.

20 Two-Way Tables Each cell of the table shows how many students from a particular class (the column) made a certain choice (row). We might wonder about the changes in the choice of military service. The numbers for 1980 and 1990 look similar, until you notice the class sizes are quite different. Table Total College/Post-HS-Education Employment Military Travel Total

21 Two-Way Tables The 18 seniors in 1980 who chose military service were only about 18/453 = 3.97% of the graduating class. In 1990, the 19 seniors making the same choice represented 19/290 = 6.55% of the class. Because the class sizes change so much, we re better off examining the proportions for each class rather than counts Percentages of the class Total College/Post-HS-Education Employment Military Travel Total

22 Chi-Square Test of Homogeneity We already know how to test whether two proportions are the same. For example, we could consider whether the proportion of students choosing military service was the same between 1980 and 1990 with a two proportion z test. But know we have more than two groups. We want to test whether students choices are the same across all three graduating classes. The z-test for two proportions generalizes to a chisquare test of homogeneity.

23 Chi-Square Test of Homogeneity Why chi-square? It turns out that the statistic is identical to the chisquare statistic for goodness-of-fit that we just saw. The goodness-of-fit test compared counts with a model. But here we re asking whether choices have changes, so we have no model. We find the expected counts for each category directly from the data.

24 Chi-Square Test of Homogeneity As a result, we count the degree of freedom slightly different as well. df = (R 1)(C 1) where R is the number of rows and C is the number of columns. The term homogeneity means that things are the same. Here, we ask whether the post-high school choices made by students are the same for these graduating classes.

25 Null and Alternative Hypothesis The null hypothesis is that there is no difference between the row variable and column variable. The alternative hypothesis is that there is some difference between the row variable and column variable.

26 Assumptions and Conditions Counted Data Condition The data must be counts. You can t do a test of homogeneity on proportions, so we have to work with counts of graduates given in the first table. Also, you can t do a chi-square test on measurements. Example If we recorded GPAs for these same groups over the time span, we wouldn t be able to test whether the mean GPAs had changes using this test.

27 Assumptions and Conditions Randomization Condition Often when we test for homogeneity; we aren t interested in some large population, so we don t really need a random sample. We would need a random sample if we wanted to draw more general conclusions. Example Graduating choices made by all high school students in these graduation years.

28 Assumptions and Conditions The null hypothesis can be thought of as a model in which the counts in the table are distributed as if each student chose a plan randomly according to the overall proportions of choices, regardless of the student s class. As long as we don t want to generalize, we don t have to check the randomization condition.

29 Assumptions and Conditions Expected Cell Frequency Condition We must be sure we have enough data for this method to work. The expected count in each cell must be at least 5.

30 Expected Count The expected count in any cell of two-way table when H o is true is Expected count = Row total x Column total Table total To calculate the expected counts, multiply the row total by the column total, and divide by table total.

31 Chi-Square Test Statistic The chi-square statistic is the sum over all r x c cell in the table. X O E 2 2 Remember that the chi-square statistic measures of how far the observed counts in a two-way table are from the expected counts. The degrees of freedom is (R 1)(C 1). The p-value is the area to the right of the X 2 statistic under the chi-square density curve. E

32 Example Chi-Square Test of Homogeneity We have counts of 1058 students observed in three different years a decade apart and categorized according to their post-graduation activities. Table Total College/Post-HS-Education Employment Military Travel Total Is the post-high school choices made by students the same for these graduating classes?

33 Example Chi-Square Test of Homogeneity Step 1 Identify population of interest. State the hypothesis in words and symbols. H o : The post-high school choices made by classes of 1980, 1990, and 2000 have the same distribution H a : The post-high school choices made by classes of 1980, 1990, and 2000 do not have the same distribution

34 Example Chi-Square Test of Homogeneity Step 2 Choose the appropriate inference procedure and verify conditions for it use. Conditions: 1. Counted Data We have counts of the number of students in categories. 2. Randomization Condition We don t want to draw inferences to other high schools or other classes, so there is no need to check for a random sample.

35 Example Chi-Square Test of Homogeneity 3. Expected cell frequency condition The expected value are (shown below) are all at least 5. Table Total College/Post-HS-Education 320/ / / Employment 98/ / / Military 18/ / / Travel 17/ /6.58 5/ Total (853) (24)

36 Example Chi-Square Test of Homogeneity Step 3 Carry out he procedure. 2 ( Obs Exp) x Exp nd Matrix Observed 2 nd Matrix p value Expected

37 Example Chi-Square Test of Homogeneity Step 4 Interpret your results in context of the problem. There is sufficient evidence to reject the H o since p- value = < = 0.05, and conclude that the choices made by high school graduates have indeed changed over the two decades examined.

38 Example Chi-Square Test of Homogeneity The shape of a 2 model depends on the degrees of freedom. A 2 model with 6 df is skewed right to the high end. The p-value is very small, indicating that the pattern we would see would be very unlikely to occur by chance were post high school choices homogeneous. 2 P P( χ 72.77)

39 Lesson 13-2, Part 2 Inference for Two-Way Tables

40 Independence A study from the University of Texas Southwestern Medical Center examined whether the risk of hepatitis C was related to whether people had tattoos and to where they got their tattoos. Hepatitis C causes about 10,000 deaths each year in the United States, but often lies undetected for many years after infection. Hepatitis C No Hepatitis C Total Tattoo, parlor Tattoo, elsewhere None Total

41 Contingency Tables These data differ from the kinds of data we ve considered before in this chapter because they categorize subjects from a single group on two categorical variables rather than only one. The categorical variables are hepatitis C status ( Hepatitis C or No Hepatitis C ) and tattoo status ( Parlor, Elsewhere, or None ). Contingency Tables Are tables that display categorize counts on two (or more variables so that we can see whether the distribution of counts on one variable is contingent on the other.

42 Chi-Square Test for Independence The natural question to ask of these data is whether the chance of having hepatitis C is independent of tattoo status. Here, this means the probability that randomly selected patient has hepatitis C should change conditional on learning the patient s tattoo status. The test to use here is the chi-square test for independence.

43 Chi-Square Test for Independence If hepatitis is independent of tattoos, we d expect the proportion of people testing positive for hepatitis to be the same for the three levels of tattoo status. This sounds a lot like the test for homogeneity. In fact the calculations are the same. The difference is that we have two categorical variables measured on a single population. Homogeneity test, we have a single categorical variable measured independently on two or more populations.

44 Null and Alternative Hypothesis The null hypothesis is that the row variable is independent (not related to) of the column variable. The alternative hypothesis is that the row variable is dependent (related to) of the column variable.

45 Assumptions and Conditions Counted Data Condition Randomization Condition and < 10% All expected counts are at least 5

46 Example Chi-Square Test of Independence We have counts of 626 individuals categorized according to their tattoo status and their hepatitis status. Hepatitis C No Hepatitis C Total Tattoo, parlor Tattoo, elsewhere None Total Is the chance of having hepatitis C related to whether people had tattoos and where they got their tattoos?

47 Example Chi-Square Test of Independence Step 1 Identify population of interest. State the hypothesis in words and symbols. H o : Tattoo status and hepatitis status are independent. H a : Tattoo status and hepatitis status are not independent

48 Example Chi-Square Test of Independence Step 2 Choose the appropriate inference procedure and verify conditions for it use. Conditions: 1. Counted Data We have counts of the number of individuals in categories of two categorical variables. 2. Randomization Condition These data are from a retrospective study of patients being treated for something unrelated to hepatitis. Although they are not SRS, they were selected to avoid biases and should be representative of the general population.

49 Example Chi-Square Test of Independence 3. Expected cell frequency condition The expected value are (shown below) are all at least 5. Hepatitis C No Hepatitis C Total Tattoo, parlor 17/ / Tattoo, elsewhere 8/ / None 22/ / Total (52) (513)

50 Example Chi-Square Test of Independence Step 3 Carry out the procedure. 2 ( Obs Exp) x Exp nd Matrix Observed 2 nd Matrix p value Expected

51 Example Chi-Square Test of Independence Step 4 Interpret your results in context of the problem. There is sufficient evidence to reject the H o since p-value = < = 0.05, and conclude that the hepatitis status is not independent of tattoo status.

52 Chi-Square and Causation Chi-square tests are common. Test for independence are especially widespread. Unfortunately, many people interpret a small p- value as proof of causation. Just as correlation between quantitative variables does not demonstrate causation, failure of independence between two categorical variables does not show a causeand-effect relationship between them.

53 Chi-Square and Causation The chi-square test for independence treats the two variables symmetrically. There is no way to differentiate the direction of any possible causation from variable to the other. In our example it is unlikely that have hepatitis causes one to crave a tattoo. Also there s never any way to eliminate the possibility that a lurking variable is responsible for the observed lack of independence. For example It might be people who have body piercings or those who inject drugs are both more likely to get tattooed and more likely to contract hepatitis C.

54 Types of Chi-Square Distributions Goodness-of-fit A test of whether the distribution counts in one categorical variable matches the distribution predicted by a model. Homogeneity A test comparing the distribution of counts for two or more groups on the same categorical variable Independence A test of whether two categorical variables are independent examines the distribution of counts for one group of individuals classified according to both variables.