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CHI-SQUARE GOODNESS OF FIT TEST u A nonparametric statistic u Nonparametric: u does not test a hypothesis about a population value (parameter) u requires.

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Presentation on theme: "CHI-SQUARE GOODNESS OF FIT TEST u A nonparametric statistic u Nonparametric: u does not test a hypothesis about a population value (parameter) u requires."— Presentation transcript:

1 CHI-SQUARE GOODNESS OF FIT TEST u A nonparametric statistic u Nonparametric: u does not test a hypothesis about a population value (parameter) u requires fewer assumptions and can be used to replace parametrics

2 Chi-Square Goodness of Fit u Purpose: Test whether an observed frequency distribution differs from a Null Hypothesis frequency distribution u Design: Individuals categorized into two or more groups

3 u Assumptions: u independent observations u mutually exclusive groups u expected frequencies at least 5 per cell

4 How it Works u Determine the frequencies you expect if the Ho is true. u Compare the observed frequencies to the Ho expected frequencies. u Large differences between observed and expected give a large value of chi- square, likely to be significant.

5 Formula for Chi-Square

6 Chi-Square Goodness of Fit Computation Example: Registered voters took a survey in which they indicated their political party preference. Determine whether there is a significant difference in the popularity of the parties.

7 Observed frequencies RepublicansDemocratsOthers 292422

8 STEP 1: Compute expected frequencies. Ho is equal frequencies, so divide total number of people by number of groups. fe = (29+24+22)/3 = 75/3 = 25 per group

9 STEP 2: For each group, compute (fo-fe) 2 and divide by fe.

10 Republicans Democrats Others

11 STEP 3: Add up the results across all the groups to get the chi-square.  2 =.64 +.04 +.36 = 1.04

12 STEP 4: Look up  2 -critical in table, using df = # of groups-1. df = 3-1= 2  2 crit = 5.99

13 STEP 5: Compare your x 2 to the critical value.  2 = 1.04  2 crit = 5.99 Not significant.

14 APA Format Sentence A chi-square goodness of fit test showed no significant difference among the three parties,  2 (2, N = 75) = 1.04, p >.05.


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