1. Nonparametric Tests of Significance Statistics for Political Science Levin and Fox Chapter Nine Part One

2. What is a parametric test? When a test of significance requires: 1.Normality in the population (a normal distribution) or at least large samples so that the sampling distribution is normal. 2.An interval-level measure.

3. Nonparametric tests of significance have a list of requirements that do not include normality or the interval level of measurement. Nonparametric tests use a concept of power (power of a test): the probability of rejecting the null hypothesis when it is false and should be rejected. In other words, the probability of accurately claiming a statistically significant relationship does exist between two variables. Non-parametric Tests

4. Power varies from test to test: The more powerful the test, the more likely the null hypothesis is to be rejected when it is false and they have more difficult requirements to satisfy. Less powerful tests have less stringent requirements on the data and the null hypothesis may be retained when it should be rejected. Accuracy of different tests: The more powerful the test is, the more likely it is to accurately determine whether or not a statistically significant relationship does exist between variables. Non-parametric Tests

5. Nonparametric Tests: Two Nonparametric Tests: The Chi-Square Test: concerned with the distinction between expected frequencies and observed frequencies. The Median Test: A chi-square based test that also evaluates whether the scores fall above or below the median.

6. Some things to know about chi square: 1) It compares the distribution of one variable (DV) across the category of another variable (IV) 2) It makes comparisons across frequencies rather than mean scores. 3) It is a comparison of what we expect to what we observe. Null versus Research Hypotheses: The null hypotheses states that the populations do not differ with respect to the frequency of occurrence of a given characteristic, whereas a research hypothesis asserts that sample difference reflects population difference in terms of the relative frequency of a given characteristic. Nonparametric Tests:

7. Chi Square: Example: Political Orientation and Child Rearing Null Hypothesis: The relative frequency or percentage of liberals who are permissive IS the same as the relative frequency of conservatives who are permissive. Research Hypothesis: The relative frequency or percentage of liberals who are permissive is NOT the same as the relative frequency of conservatives who are permissive. Nonparametric Tests:

8. Chi Square: Example: Political Orientation and Child Rearing Expected and Observed Frequencies: The chi-square test of significance is defined by Expected and Observed Frequencies. Expected Frequencies (f e ) is the frequency we would expect to get if the hull hypothesis is true, that is there is no difference between the populations. Observed Frequencies (f o ) refers to results we actually obtain when conducting a study (may or may not vary between groups). Only if the difference between expected and observed frequencies is large enough do we reject the null hypothesis and decide that a population difference does exist. Nonparametric Tests:

9. Chi Square: Political Orientation and Child Rearing: Observed Frequencies 137 7 LiberalsConservatives Political Orientation Child-Rearing Methods Permissive Not Permissive Total 20 N = 40 Col. Marginal Row Marginal

10. Chi Square: Example: Political Orientation and Child Rearing Since the marginals are all equal, it is easy to calculate the expected frequencies: 10 in each cell. 10 LiberalsConservatives Political Orientation Child-Rearing Methods Permissive Not Permissive Total 20 N = 40 It is unusual for a study to produce row and column marginals that are evenly split.

11. Chi Square: Example: Political Orientation and Child Rearing Calculating expected frequencies when the marginals are not even: 1510 5 LiberalsConservatives Political Orientation Child-Rearing Methods Permissive Not Permissive Total 20 25 15 N = 40 To determine if these frequencies depart from what is expected (null) by chance alone, we have to calculate the expected frequencies. Row Marginal

12. If 25 of 40 respondents are permissive, than 62.5 % of them are permissive. To then determine the expected frequency, which asserts that Libs and Cons are the same (null) we have to calculate what would be 62.5% of 20 Libs and 20 Cons (the number of each that are in the study. 15 (12.5)10 (12.5) 5 (7.5)10 (7.5) LiberalsConservatives Political Orientation Child-Rearing Methods Permissive Not Permissive Total 20 25 (62.5%) 15 N = 40 The answer is 12.5 (62.5% of 20 or.625 x 20). We then know that the expected frequency for non permissive is 7.5 (20 – 12.5).

13. Calculating Expected Frequencies f e = (column marginal)(row marginal) N Example: fe = (20)(25) 40 = 500 40 = 12.5

14. 15 (12.5)10 (12.5) 5 (7.5)10 (7.5) LiberalsConservatives Political Orientation Child-Rearing Methods Permissive Not Permissive Total 20 25 (62.5%) 15 N = 40 The answer is 12.5 (62.5% of 20 or.625 x 20). We then know that the expected frequency for non permissive is 7.5 (20 – 12.5). Example: fe = (25)(20) 40 = 500 40 = 12.5

15. The Chi-Square Test Formula Where: f o = observed frequency in any cell f e = expected frequency in any cell Once we have the observed and expected frequencies we can use the following formula to calculate Chi-square.

16. Nonparametric Tests: Chi-Square Tests Observed Expected Subtract Square Divide by fe Sum After obtaining fo and fe, we subtract fe from fo, square the difference, divide by the fe and then add them up.

17. Nonparametric Tests: Chi-Square Tests df = (r-1)(c-1) Where r = the number of rows of observed frequencies c = the number of columns of observed frequencies Formula for Finding the Degrees of Freedom

18. Since there are two rows and two columns of observed frequencies in our 2 x 2 table df = (r-1)(c-1) df = (2-1)(2-1) = (1)(1) = 1 Next Step, Table E, where we will find a list of chi-square scores that are significant at.05 and.01 levels. Table E (.05, df = 1): 3.84 Obtained X = 2.66 Retain null 2

19. Step by Step Chi-Square Test: 1) Subtract each expected frequency from its corresponding observed frequency 2) Square the difference 3) Divide by the expected frequency, and then 4) Add up these quotients for all the cells to obtain the chi-square value

20. Comparing Several Groups When comparing more than two groups, you use essentially the same process as when comparing 2 x 2 tables. Nonparametric Tests:

21. Chi Square: Example: Political Orientation and Child Rearing: 2 x 2: Null Hypothesis: The relative frequency of permissive, moderate, and authoritarian child-rearing methods IS the same for Protestants, Catholics, and Jews. Research Hypothesis: The relative frequency of permissive, moderate, and authoritarian child-rearing methods is NOT the same for Protestants, Catholics, and Jews. Nonparametric Tests:

22. Chi Square: Example: Political Orientation and Child Rearing Calculating expected frequencies when the marginals are not even: 15 (12.5)10 (12.5) 5 (7.5)10 (7.5) LiberalsConservatives Political Orientation Child-Rearing Methods Permissive Not Permissive Total 20 25 (62.5%) 15 N = 40 Comparing Several Groups

23. Chi Square: Example: Political Orientation and Child Rearing 15 (12.5)10 (12.5) 5 (7.5)10 (7.5) LiberalsConservatives Child-Rearing Methods Permissive Not Permissive Total 20 25 (62.5%) 15 N = 40 Comparing Several Groups

24. Correcting for Small Frequencies Generally, chi square should be used with great care whenever some of the frequencies are below Five (5). Though, this is not a hard and fast rule.

25. Yate’s Correction HOWEVER, when working with a 2x2 table where any expected frequency is less than 10 but greater than 5, use Yate’s correction which reduces the difference between the expected and observed frequencies. The vertical indicate that we must reduce the absolute value (ignoring minus signs) of each fo – fe by.5

26. Yate’s Correction Smoking Status Nationality AmericanCanadian Nonsmokers Smokers 15 (11.67) 6 (9.33) 5 (8.33) 10 (6.67) 20 16 N = 36 Total2115 Observed Expected Subtract Subtract.5 Square Divide by fe Sum

27. Requirements for the use of Chi-Square Requirements for the use of Chi-Square: 1.A comparison between two or more samples. 2.Nominal data must be used. 3.Samples should have been randomly selected. 4.The expected cell frequencies should not be too small.

28. Median Test Median Test: Is a simply nonparametric test for determining the likelihood that two or more random samples have been taken from populations with the same median. It involves conducting a Chi-Square test where one of the dimensions is whether the scores fall above or below the median of the two groups combined.

29. Median Test Median Test: Example: Gender and Embarrassment Experiment asked 15 men and 12 women with average singing ability to sing in front of expert judges (sound familiar?) Men (N = 15)Women (N = 12) 4 6 9 11 12 15 16 18 19 21 23 24 25 26 27 1 2 3 5 7 8 10 13 14 17 20 22

30. Median Test Median Test: Example: Gender and Embarrassment Experiment asked 15 men and 12 women with average singing ability to sing in front of expert judges (sound familiar?) Step 1: Find Median of the two samples Mdn = (N + 1) ÷ 2 = 27 + 1 ÷ 2 Mdn = 14

31. Median Test Median Test: Example: Gender and Embarrassment Experiment asked 15 men and 12 women with average singing ability to sing in front of expert judges (sound familiar?) Step 1: Find Median of the two samples Mdn = (N + 1) ÷ 2 = 27 + 1 ÷ 2 Mdn = 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 1 2 3 4 5 6 7 8 9 10 11 12 13 14

32. Median Test Median Test: Example: Gender and Embarrassment Experiment asked 15 men and 12 women with average singing ability to sing in front of expert judges (sound familiar?) Step 2: Count the Number in each sample falling above and below the median Median Gender MenWomen Above Below 10 5 3939

33. Median Test Median Test: Example: Gender and Embarrassment Mdn = (N + 1) ÷ 2 = 27 + 1 ÷ 2 Mdn = 14 Men: Below Mdn = 5 Above Mdn = 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Women: Below Mdn = 9 Above Mdn = 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 1 2 3 4 5 6 7 8 9 10 11 12 13 14

34. Median Test Median Test: Example: Gender and Embarrassment Mdn= 14 minutes MenWomen 4 6 9 11 12 15 16 18 19 21 23 24 25 26 27 1 2 3 5 7 8 10 13 14 17 20 22 14: 10 above, 5 below 14: 3 above, 9 below

35. Step 2: Count the Number in each sample falling above and below the median 13 14 N = 27 15 12 Median Gender MenWomen Above Below 10 (7.22) 5 (7.78) 3 (5.78) 9 (6.22) Total Observed Expected Subtract Square Subtract.5 Divide by fe Sum

36. Step 3: Calculate the Degrees of Freedom df = (r-1)(c-1) = (2-1)(2-1) = (1)(1) = 1 We then go to Table E, which tells us that at.05 and a df = 1 chi-square must exceed 3.84 to be considered statistically significant. Obtained X = 3.13 Table E = 3.84 Retain Null hypothesis There is insufficient that men and women differ in terms of embarrassment. 2