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Statistics Resources

This guide contains all of the ASC's statistics resources. If you do not see a topic, suggest it through the suggestion box on the Statistics home page.

Goodness-of-Fit

This Chi-Square test is used to determine if the distribution from a sample fits a known or hypothesized distribution. For example, determining if Skittle flavors are equally preferred.

Assumptions

  1. One categorical variable - can be dichotomous, nominal, or ordinal
  2. Independence of observations
  3. Groups are mutually exclusive - participants can only belong to one group
  4. At least 5 expected frequencies in each group

Running Chi-Square Goodness-of-Fit in SPSS

NOTE: these steps are for summated frequencies, not raw data

  1. Ensure you have two columns:
    • categorical variable with each category listed below it
    • frequency variable with the frequency for each category listed below it
  2. Data > Weight Cases
    • Select the "Weight cases by" radio button
    • Move the frequency variable into the "Frequency Variable" box
    • Click OK to go back to the data sheet
  3. Analyze > Nonparametric Tests > Legacy Dialogs > Chi-square
    • Move the categorical variable into the "Test Variable List" box
    • Adjust the "Expected Values" option if "All categories equal" is not the distribution you're hypothesizing
    • Click OK to run the test

*If your data is not summated, you'll want to use Analyze > Descriptive Statistics > Frequencies to first compute the frequencies for each group. Then, set up your data as indicated above.*

Interpreting the Output

  • [variable] (this box will be named the same as the test variable you entered)
    • Provides the observed and expected frequencies used for the calculation
  • Test Statistics
    • provides the results of the chi-square test
    • used to make a decision about the null hypothesis

Reporting Results in APA Style

A chi-square goodness-of-fit test was conducted to determine if the proportion of people suffering from depression differs based on marital status. There was not sufficient evidence to reject the hypothesis that the proportions were equal X^2(1, = 92) = 2.4, p = .35.

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