1 Chapter 9 Nonparametric Tests of Significance

2 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.

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

4 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

5 The Chi-Square Test Cont.

6 Degrees of Freedom Degrees of freedom = (r – 1)(c – 1)  R = the number of rows  C = the number of columns Because the observed frequencies in Table 1 form two rows and two columns (2 x 2): df = (2-1)(2-1) = (1)(1)= 1

Critical Value of X 2 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

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

9 One Way Chi-Square: 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 alpha =.05 that all three political positions are equally represented in the print media. fofefo-fe(fo-fe) 2 (fo-fe) 2 fe Conservative Moderate Liberal

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

Two Way Chi-Square Illustration A researcher is examining the child-rearing methods by political orientation. The researcher found that of the of the 35 liberals, 21 were permissive and 14 were not. The 24 conservatives: 11 permissive and 13 not permissive. Is there a statistical significance difference between political orientation on child-rearing practices? 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.

Child-Rearing Methods LiberalsConservatives Permissive 2111 Not Permissive 1413

Calculating the Expected Frequencies In order to calculate the expected frequency (fe) use the following formula: For the top left value (permissive liberals = 21), Fe = (32)(35) 59 = 19

Finding Fe Child- Rearing Methods LiberalsConservativesTotal Permissive Not Permissive Total3524N = 59 Upper Left [21] = (32)(35) / 59 = 19 Upper Right [11] = (32)(24) / 59 = 13 Bottom Left [14] = (27)(35) / 59 = 16 Bottom Right [13]= (27)(24) / 59 = 11

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

16 The Chi-Square Test Cont. fofefo-fe(fo-fe) 2 fe Upper Left Upper Right Bottom Left Bottom Right

On Your Own A researcher has collected information to find out if fear of crime is related to gender. Your task is to determine if there is a relationship between fear of crime and gender. Using chi square, test the null hypothesis at the alpha level =.05. Gender MaleFemale Safe10081 Unsafe3382

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19 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

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.

21 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 Yes349 No1425

fofe|fo-fe| -.5(|fo-fe| -.5) 2 (|fo-fe| -.5) 2 fe Upper Left Upper Right Bottom Left Bottom Right

fofe|fo-fe| -.5(|fo-fe| -.5) 2 (|fo-fe| -.5) 2 fe Upper Left Upper Right Bottom Left Bottom Right X 2 = Df = 1, a =.05, X 2 table = 3.841

24 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.

25 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.

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

Course Taken SkillCustomaryNew and Improved Total Above average Average Below Average10 20 Total50 N = 100

Obtain the frequencies expected Course Taken SkillCustomaryNew and Improved Total Above average Average Below Average 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

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

Create a summary table fofefo-fe(fo-fe) 2 fe Upper Left Upper Right Middle Left Middle Right Bottom Left Bottom Right10 00 X 2 =.82

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?

32 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.

33 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.

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

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

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.

37 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.

Initial Results, Finding the Median Man Woman Position of Median = N + 1 / 2 = / 2 = 16.5 Find the median

Table, Expected Frequencies WomanManTotal Above Median4913 Not above Median 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

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 Bottom Right 79.5

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

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?