Key points:

• Chi square is a nonparametric test. That means it has no assumptions. Thus, it has low power.
• Chi square goodness of fit tells you whether a set of proportions of one variable (e.g. toss of dice) distributes as one would expect from theory.
• Chi square test of association is a measure of relationship between an ordinal D.V. and nominal I.V. for samples taken from the same population. To do the chi square test of association, you set up a contingency table. The variables are
  • f sub o = observed frequency of occurrence in each cell
  • f sub e = expected frequency of occurrence in each cell
  • f sub e = (row total)*(column total)/(overall total)
  • degrees of freedom = (rows - 1)*(columns - 1)
  • n = sample size
  • alpha = the error probability you are willing to accept (.01, .05, or .10 - usually choose .05) - you need this for the table look-up.
• Chi_square = sum over all cells [(observed frequency - expected frequency)**2/(expected frequency)]
• If the calculated value exceeds the critical value in the table, then reject the null hypothesis.
• SPSS will print out the contingency coefficient (c), which is used when each variable has two or more levels. C tells us about the strength of the relationship between the two variables in the table.

Median test

• This is a special type of chi square test where you compare the medians of two groups (so it is ordinal) instead of the mean (which is interval/ratio), so you throw out information (or didn't have the right information). It has low power.
• If you rejected H zero, then by how much?? You need more information.
• If you have only a 2 x 2 table, use r sub phi = sqrt[chi_square/n]. This tells you about the strength of the relationship between the variables.
• If you have more rows/columns than 2 x 2, then you use the contingency coefficient to get the strength of the relationship:
  c = sqrt[chi_square/(n + chi_square)]
  
• A small relationship is .1 - .3, medium is .4 - .7, and high is .8 or higher.
• With r, you can square the correlation and get the % of the variance predicted by the relationship, but you cannot do this here. These are not very powerful statistics.

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