1. 1 Chapter 9 Nonparametric Tests of Significance

2. 2 Introduction  A test of significance, such as the t-test, is referred to as a parametric test when it requires: 1.Normality in the population 2.And an interval-level measure, it is referred to as a parametric test.  What about the social researcher who cannot employ a parametric test, that is, who either cannot honestly assume normality, who does not work with large numbers of cases, or whose data are not measured at the interval level?

3. 3 Power of a Test To understand the important position of nonparametric tests in social research, we must also understand the concept of the power of a test, the probability of rejecting the null hypothesis when it is actually false and should be rejected.

4. 4 Table 1: Cross-Tabulation of Seat Belt Use by Gender Use of Seat Belt MaleFemaleTotal Yes6171,4702,087 No1,3106391,949 Total1,9272,1094,036

5. 5 The Chi-Square Test  The chi-square is employed to make comparisons between frequencies rather than between mean scores –The null hypothesis for the chi-square test states that the populations do not differ with respect to the frequency of occurrence of a given characteristic –The research hypothesis says that sample differences reflect actual population differences regarding the relative frequency of a given characteristic  Example: Investigate the effect of political orientation and child-rearing permissiveness

6. 6 The Chi-Square Test Cont.  Concerned with the distinction between expected frequencies and observed frequencies –Expected frequencies (f e ): the terms of the null hypothesis, according to which the relative frequency (or proportion) is expected to be the same from one group to another. –Observed frequencies (f o ): the results that we actually obtain when conducting a study, and therefore, may or may not vary from one group to another  Only if the difference between expected and observed frequencies is large enough do we reject the null hypothesis and decide that a true population difference exists

7. 7 The Chi-Square Test Cont. Chi-Square = Subtract each expected frequency from its corresponding observed frequency, square the difference, divide by the expected frequency, and then add up these quotients for all the cells

8. 8 Degrees of Freedom Degrees of freedom = (r – 1)(c – 1)  Where r = the number of rows  Where c = the number of columns  Total column and total row do not count in this total Because the observed frequencies in Table 1 form two rows and two columns (2 x 2): df = (2-1)(2-1) = (1)(1)= 1

9. Critical Value of X 2 Once you have obtained your degrees of freedom, we need to determine our critical value of chi-square Use Table E in Appendix C of your book to determine the critical value Remember, if an alpha level is not specified, always use 0.05 Just like before, if your calculated chi-square value is greater than your critical chi-square value, we can reject the null hypothesis

10. Applicability of Chi Square Tests  One Way Chi-Square  Two Way Chi-Square  Comparing Several Groups  Median Test

11. 11 One Way Chi-Square: The five steps Step 1: Rearrange the data into a frequency distribution. Step 2: Obtain the expected frequency for each cell. Fe = N / k N = total number of observed frequencies K = total number of categories Step 3: Create a summary table. Step 4: Find the degrees of freedom. Step 5: Compare the obtained chi-square value with the appropriate chi-square value in the table in the back of the book.

12. Illustration Some politicians have been known to complain about the liberal press. To determine if in fact the press is dominated by left-wing writers, a researcher assesses the political leanings of a random sample of 60 journalists. He found that 15 were conservative, 18 were moderate, and 27 were liberal. Test the null hypothesis at the alpha level =.05 that all three political positions are equally represented in the print media.

13. fofefo-fe(fo-fe) 2 (fo-fe) 2 fe Conservative 15 Moderate 18 Liberal 27

14. fofefo-fe(fo-fe) 2 (fo-fe) 2 fe Conservative 15205251.25 Moderate 182024.20 Liberal 27207492.45 X 2 = 3.90 Df = k – 1 = 3 – 1 = 2 α=.05 X 2 Table = 5.991

15. Two Way Chi-Square Illustration A researcher is examining the child-rearing methods by political orientation. Test the null hypothesis at the alpha level =.05 that the relative frequency of liberals who are lenient is the same as the relative frequency of conservatives who are lenient. The researcher found that of the of the 20 liberals, 15 were permissive and 5 were not. The 20 conservatives were equally split: 10 permissive and 10 not permissive. Is there a statistical significance difference between political orientation on child- rearing practices?

16. 16 Two-Way Chi-Square: The five steps Step 1: Rearrange the data in the form of a 2x2 table. Step 2: Obtain the expected frequency for each cell. Step 3: Create a summary table. Step 4: Find the degrees of freedom. Step 5: Compare the obtained chi-square value with the appropriate chi-square value in the table in the back of the book.

17. Child-Rearing Methods LiberalsConservatives Permissive1510 Not Permissive 510

18. 18 Calculating the Expected Frequencies In order to calculate the expected frequency (fe) use the following formula: For the top left value (permissive liberals = 15), Fe = (25)(20) 40 = 12.5

19. Finding Fe Child- Rearing Methods LiberalsConservativesTotal Permissive151025 Not Permissive 51015 Total20 N = 40 Upper Left [15] = (25)(20) / 40 = 12.5 Upper Right [10] = (25)(20) / 40= 12.5 Bottom Left [5] = (20)(15) / 40 = 7.5 Bottom Right [10]= (15)(20) / 40 = 7.5

20. The Chi-Square Test Cont. CellFoFeFo – Fe(Fo - Fe) 2 (Fo - Fe) 2 Fe Upper Left 15 Upper Right 10 Lower Left 5 Lower Right 10

21. 21 The Chi-Square Test Cont. CellFrequency Observed Frequency Expected Fo – Fe(Fo - Fe) 2 (Fo - Fe) 2 Fe Upper Left 1512.5 Upper Right 1012.5 Lower Left 57.5 Lower Right 107.5

22. 22 The Chi-Square Test Cont. CellFrequency Observed Frequency Expected Fo – Fe(Fo - Fe) 2 (Fo - Fe) 2 Fe Upper Left 1512.52.5 Upper Right 1012.5-2.5 Lower Left 57.5-2.5 Lower Right 107.52.5

23. 23 The Chi-Square Test Cont. CellFrequency Observed Frequency Expected Fo – Fe(Fo - Fe) 2 (Fo - Fe) 2 Fe Upper Left 1512.52.56.25 Upper Right 1012.5-2.56.25 Lower Left 57.5-2.56.25 Lower Right 107.52.56.25

24. 24 The Chi-Square Test Cont. CellFrequency Observed Frequency Expected Fo – Fe(Fo - Fe) 2 (Fo - Fe) 2 Fe Upper Left 1512.52.56.25.50 Upper Right 1012.5-2.56.25.50 Lower Left 57.5-2.56.25.83 Lower Right 107.52.56.25.83 X 2 = 2.66

25. 25 Yates’s Correction Yate’s correction reduces the size of the chi-square statistic. For use only with 2 x 2 tables. The virtual lines indicate absolute value. Disregard the sign when subtracting. |3 – 4| = |-1| = 1

26. Example Some recent studies have suggested that the chance of a baby being born prematurely are increased when the mother suffers from chronic oral infections such as periodontal disease. An interested researcher collected the following data. Applying Yates’s correction, conduct a chi-square analysis to test the null hypothesis that pregnant women who suffer from chronic oral infections are no more likely to give birth prematurely than women who do not suffer from chronic oral infections.

27. 27 Illustration: Yates's Correction 1.Some recent studies have suggested that the chance of a baby being born prematurely are increased when the mother suffers from chronic oral infections. An interested researcher collected the following data. 2.Applying Yate’s correction, conduct a chi-square analysis to test the null hypothesis that pregnant women who suffer from chronic oral infections are no more likely than women who do not. Suffers from Chronic Oral Infections Baby Born Premature YesNo Yes448 No657

28. foFe|fo-fe| -.5(|fo-fe| -.5) 2 (|fo-fe| -.5) 2 fe Upper Left 44 Upper Right 8 Bottom Left 6 Bottom Right 57

29. fofe|fo-fe| -.5(|fo-fe| -.5) 2 (|fo-fe| -.5) 2 fe Upper Left 4422.6120.89436.3919.30 Upper Right 829.3920.89436.3914.84 Bottom Left 627.3920.89436.3915.93 Bottom Right 5735.6120.89436.3912.25 X 2 = 62.31 Df = 1, a =.05, X 2 table = 3.841

30. 30 Comparing Several Groups Imagine that we are investigating the relationship between a technician’s skill in repairing computers and if they were trained in the new and improved course or the customary one. 1.We will be drawing information from 100 trainees. 2.We were asked to categorize the trainees into their repair skill based on the final examination and which course they took. Null: The relative frequency of above average, average, and below average repair skills is the same for those who took the customary class and the new and improved class. Research: The relative frequency of above average, average, and below average repair skills is not the same for those who took the customary class and the new and improved class.

31. 31 Chi-Square Test for Several Groups Step 1: Rearrange the data in the form of a table. Step 2: Obtain the expected frequency for each cell. Step 3: Construct a summary table. Step 4: Find the number of degrees of freedom. Step 5: Compare the obtained chi-square with the appropriate chi-square value in Table E.

32. Create the table Course Taken SkillCustomaryNew and Improved Above average1519 Average2521 Below average10

33. Course Taken SkillCustomaryNew and Improved Total Above average151934 Average252146 Below Average10 20 Total50 N = 100

34. Obtain the frequencies expected Course Taken SkillCustomaryNew and Improved Total Above average 151934 Average252146 Below Average 10 20 Total50 N = 100 UL = (34)(50) / 100 = 17 UR = (34)(50) / 100 = 17 ML = (46)(50) / 100 = 23 MR = (46)(50) / 100 = 23 BL = (20)(50) / 100 = 10 BR = (20)(50) / 100 = 10

35. Create a summary table fofefo-fe(fo-fe) 2 fe Upper Left15 Upper Right19 Middle Left25 Middle Right21 Bottom Left10 Bottom Right10

36. Create a summary table fofefo-fe(fo-fe) 2 fe Upper Left1517-24.24 Upper Right191724.24 Middle Left252324.17 Middle Right2123-24.17 Bottom Left10 000 Bottom Right10 000 X 2 =.82

37. Obtain the degrees of freedom / Make a decision Df= (r-1)(c-1)Computed X 2 =.82 = (3-1)(2-1)Critical X 2 = 5.99 = 2 a =.05 X 2 = 5.99 Reject or retain the null hypothesis?

38. 38 Requirements for the Use of Chi- Square A comparison between two or more samples. Nominal data. Random sampling. The expected cell frequencies should not be too small.

39. 39 The Median Test 1.For ordinal data, the median test is a simple nonparametric procedure for determining the likelihood that two or more random samples have been taken from populations with the same median. 2.The median test involves performing a chi- square test of significance on a cross-tab in which one of the dimensions is whether the scores fall above or below the median of the two groups combined.

40. 40 Requirements for the Use of a Median Test A comparison between two or more median. Ordinal data. Random sampling.

41. Summary -Parametric requirements cannot always be met -Non-parametric procedures are less powerful but still quite useful -The most popular include: chi-square and the median test -Chi-square can be calculated for nominal level data with two or more categories -Median test is used with ordinal data -Both non-parametric tests must meet certain requirements

42. The Median Test For ordinal data, use the median test. Recall the formula for the position of the median: Count the number above and below the median these will function similar to observed frequencies. Use Yate’s correction if necessary and calculate chi-square statistic.

43. 43 Illustration: Median Test Suppose an investigator wanted to examine gender perceptions. Two samples were obtained with one half told the author was a woman and the other half a man. The students were told to evaluate the story from 1-8; the higher the score, the better the story. Step 1: Find the median of the two samples combined. Step 2: Count the number in each sample falling above the median and not above the median. Step 3: Perform a 2x2 Chi- square test of significance.

44. Initial Results, Finding the Median Man6882563868226843 Woman6511343655135663 Position of Median = N + 1 / 2 = 32 + 1 / 2 = 16.5 Find the median

45. Table, Expected Frequencies WomanManTotal Above Median4913 Not above Median 12719 Total16 N = 32 Expected Frequencies: UL = (13)(16) / 32 = 6.5BL = (19)(16)( / 32 = 9.5 UR = (13)(16) / 32 = 6.5BR = (19)(16)( / 32 = 9.5 Observed Frequencies

46. Summary Table fofe|fo-fe||fo-fe| –.5(|fo-fe| –.5) 2 (|fo-fe| –.5) 2 fe Upper Left 46.5 Upper Right 96.5 Bottom Left 129.5 Bottom Right 79.5

47. Summary Table Completed fofe|fo-fe||fo-fe| –.5 (|fo-fe| –.5) 2 (|fo-fe| –.5) 2 fe Upper Left 46.52.524.62 Upper Right 96.52.524.62 Bottom Left 129.52.524.42 Bottom Right 79.52.524.42 X 2 = 2.07

48. Obtain the degrees of freedom / Make a decision Df= (r-1)(c-1)Computed X 2 = 2.07 = (2-1)(2-1)Table X 2 = 3.84 = 1 a =.05 X 2 = 3.84 Reject or retain the null hypothesis?

49. 49 Requirements for the Use of a Median Test A comparison between two or more median. Ordinal data. Random sampling.

50. Summary -Parametric requirements cannot always be met -Non-parametric procedures are less powerful but still quite useful -The most popular include: chi-square and the median test -Chi-square can be calculated for nominal level data with two or more categories -Median test is used with ordinal data -Both non-parametric tests must meet certain requirements