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Chapter 9

Nonparametric Tests of Significance

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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.

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Table 1: Cross-Tabulation of Seat Belt Use by Gender

Use of Seat Belt

Male Female Total

Yes 617 1,470 2,087

No 1,310 639 1,949

Total 1,927 2,109 4,036

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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

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The Chi-Square Test Cont.

e

eo

f

ffX

22

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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 X2

• 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

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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.

fo fe fo-fe (fo-fe)2 (fo-fe)2

fe

Conservative

Moderate

Liberal

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

Liberals Conservatives

Permissive 21 11

Not Permissive

14 13

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Calculating the Expected FrequenciesIn order to calculate the expected frequency (fe)

use the following formula:

For the top left value (permissive liberals = 21),

Fe = (32)(35) 59

= 19

Nfe

)totalmarginalcolumn)(totalmarginal(row

The Chi-Square Test Cont.

Cell Fo Fe Fo – Fe (Fo - Fe)2 (Fo - Fe)2 Fe

Upper Left

Upper Right

Lower Left

Lower Right

On Your Own

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

Gender

Male Female

Safe 100 81

Unsafe 33 82

End Day 1

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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

e

eo

f

ffX

2

2 5.

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.

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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

Yes No

Yes 34 9

No 14 25

fo fe |fo-fe| - .5 (|fo-fe| - .5)2 (|fo-fe| - .5)2

fe

Upper Left

Upper Right

Bottom Left

Bottom Right

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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.

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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

Skill Customary New and Improved

Above average 15 19

Average 25 21

Below average 10 10

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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.

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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.

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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