STP231 Brief Class Notes Ch10: Chi-square Tests, Instructor: Ela Jackiewicz

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1 Chi-square Tests for categorical variables Tests of independence: We will consider two situations: A) one variable with r categories in c independent samples from different populations or B) 2 variables with r and c categories. in one sample from one population In each case we obtain 2-way table and our hypotheses are: : No association between rows and columns (independence) : There is association between rows and columns (dependence) O= Observed counts are counts in each cell of the table from the sample(s) E=Expected counts (computed under :, E= R C, R=row total, C=column total, n Degrees of freedom= df=(r-1)(c-1), r=#of rows, c=#of columns Test statistics for each test : 2 = O E 2 E P-Value: To obtain a P-value (P) of a hypothesis test, we compute, assuming the null hypothesis is true, the probability of observing a value of the test statistic as extreme or more extreme than that observed. By extreme we mean far from what we would expect to observe if the null hypothesis were true. P-value =area right of the observed test statistics under Chi-square curve with df=(r-1)(c-1) Note: Our alternative is nondirectional, but in case of dichotomous variable(s) we can also have a directional hypothesis. We can test hypothesis, specifying that probability in one category of one of the variables is smaller/larger than in the other, as usual check directionality first. In that case p-value as computed above is divided by 2. Ex1. Sickle-cell anemia is a hereditary condition that can cause medical problems. Some researchers suggest that sickle-cell gene protects against malaria. In a study 543 African children were tested for the sickle- cell gene and for malaria (heavy infection). Results are presented in the table below. Do we have evidence of a relationship between sickle-cell gene and heavy malaria infection? Test by means of a Chi-square test, use =.05 Sickle_cell Gene Heavy Malaria Infection YES NO total YES NO total

2 This is a test of independence. We have a case of one sample and 2 categorical variables. : sickle-cell gene does not affect heavy malaria infections (variables are independent) : sickle-cell gene affects heavy malaria infections (variables are not independent) Expected counts: (188*136)/543=47.1 (355*136)/543=88.9 (188*407)/543=140.9 (355*407)/543=266.1 Expected counts are computed under assumption that null hypothesis is true. If that is the case, counts in each level of row variable have to be proportionally allocated in the corresponding levels of the column variable, for example: E 1,1 =fraction of malaria infected children*# of children with sickle-cell gene= (188/543)*136= (RC)/n E 1,2 =fraction of not infected children* # of children with sickle-cell gene= (355/543)* 136=(RC)/n and so on. Suppose we had the same number of infected and not infected children, then under null hypothesis we would expect all 136 cases with sickle-cell anemia to be split in half : 68 with malaria and 68 without malaria, and all 407 cases without the gene would also be split in half: infected and not infected (expected counts do not have to be integers) Our test statistics is χ 2 =5.33 df=1, p=0.021<.05. Reject null hypothesis, there is evidence of association, sickle cell gene seems to affect heavy malaria infections. We can check that % of infections is smaller among children with sickle -cell gene (26.5%), compared to 37.3% among children with no sickle-cell gene. Our alternative was nondirectional, we simply tested if sickle-cell gene affects malaria infections in any way. If we decide to test directional hypothesis that sickle-cell gene enhances survival, our p-value would be half of the one we received (0.0105). Ex2 Does pollster's gender have an effect on poll responses by men? A U.S News & World Report stated recently: On sensitive issues, people tend to give acceptable rather that honest responses; their answers may depend on the gender or race of interviewer. To support that claim, following data was provided for an Eagleton Institute poll in which surveyed men were asked if they agreed with a statement: Abortion is a private matter that should be left to the women to decide without government intervention. Two random samples of males were independently interviewed, one by male interviewers, the other by female interviewers. Data is shown below: Gender of Interviewer MALE FEMALE total AGREE Response DISAGREE total

3 Do we have evidence that proportions of AGREE/DISAGREE responses is different for male and female interviewers? Test using =.05. This is the case where we have two independent samples of males, one interviewed by female, another by male interviewers, variable is the type of response to the abortion question (agree or disagree). Hypotheses can be formulated as follows: : Proportions of agree/disagree responses are equal for male and female interviewers ( response independent of the sex of the interviewer) : Proportions of agree/disagree responses are different for male and female interviewers ( response not independent of the sex of the interviewer) Expected counts are: (rounded to nearest integer) (Examining expected counts we could see that they make sense. We have twice as many people interviewed by males than by females, so if null is true, expected counts in each agree and disagree category should be (and they are) twice as large for males then for females) χ 2 =6.53 df=1, p-value=0.011<0.05. Reject, we have evidence that proportion of agree/disagree responses depends on the gender of the interviewer. If we tested directional hypotheses that males tend to agree more with female than male interviewer, p- value for that test would be.5(.011)=.0055<.05, evidence in favor of alternative hypotheses. We could conclude that indeed there is more agreement with female than male interviewers (77% to 70%). Ex3. Table below presents results of investigation of a relationship between color of the helmet worn by the motorcycle drivers and whether or not they were killed or injured during the crash. Do you think there is an association between the helmet color and crash results? Use a Chi-square test of independence with =.10. Color of Helmet Black White Yellow/Orange Red Blue Crash Not injured result Injured or killed Test of independence, one samples (people that crushed on a motorcycle), 2 categorical variables: crush result and helmet color. : Crash result and helmet color are independent (not associated) : Crash result and helmet color are dependent (associated)

4 Expected counts are : (rounded to nearest integer) χ 2 =9.97 df=4, p=0.0409<0.05. Reject, we have evidence of association. Ex4. Following data was obtained to determine if person's occupation is independent of weather the cause of death of that person was a homicide or other cause Occupation Police Cashiers Taxi Drivers Guards Cause of Death Homicide Other Do we have evidence that cause of death and occupation are associated? Use 5% significance level. Here we have 4 independent samples from different occupations, one variable: cause of death. : Cause of death and occupation are not associated : Cause of death and occupation are associated Expected counts are: χ 2 =65.52 df=3, p-value=3.4*10-14 <0.05. Reject, there is overwhelming evidence that cause of death and occupation are associated. Steps in testing: P-VALUE APPROACH, α given 1. State and 2. Compute test statistic 3. determine the P-value 4. If P α reject ; otherwise, do not reject. 5. Interpret the result of the hypothesis test. Clearly answer question posed in the problem

5 Calculator (TI 83, 84) Chi-square test of independence: 1. Use MATRIX menu to put observed frequencies in a matrix 2. Use 2 Test, it is a nondirectional test, divide p-value by 2 for appropriate directional one Applicability of Chi-square methods of testing Conditions of validity: Design: Our contingency table must be appropriate to view in one of the two ways: (I) Few independent samples observed with respect to one categorical variable (II) One sample observed with respect to two categorical variables Observations within a sample(s) must be independent. Sample size: We must have a large enough sample(s), so that expected counts are all at least 5, some researchers would allow at most 20% of expected counts < 5. For large tables it is enough if average of all expected counts is 5 or more, even if some counts are less than 5. Scope of inference: As is the case for other hypotheses tests, i data arises from an experiment with random assignment to treatments, then for small p-value, we can draw a causal inference. If we draw experimental units randomly from the population, then we can extend our inference to the population. If we have observational study, then, for a small p-value, we can only infer that observed association, not simply due to chance, but we can't rule out other explanations.

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