1. The Chi-Square Test for Association
Math 137
Fall‘13
L. Burger

2. -The Chi Square Test
A statistical method used to determine goodness of fit
Goodness of fit refers to how close the observed data are to those predicted from a hypothesis
Note:
The chi square test does not prove that a hypothesis is correct
It evaluates to what extent the data and the hypothesis have a good fit

3. -Determine The Hypothesis: Whether There is an Association or Not
Ho : The two variables are independent (not matching up well!)
Ha : The two variables are associated (variables matching good --- called a ‘good fit!’)

4. Understanding the Chi-Square Distribution
You see that there is a range here: if the results were perfect you get a chi-square value of 0 (because obs = exp). This rarely happens: most experiments give a small chi-square value (the hump in the graph).
Note that all the values are greater than 0: that's because we squared the (obs - exp) term: squaring always gives a non-negative number.
Sometimes you get really wild results, with obs very different from exp: the long tail on the graph. Really odd things occasionally do happen by chance alone (for instance, you might win the lottery).

5. Critical Chi-Square
Critical values for chi-square are found on tables, sorted by degrees of freedom and probability levels. Be sure to use p = 0.05.
If your calculated chi-square value is greater than the critical value from the table, you “reject the null hypothesis”.
If your chi-square value is less than the critical value, you “fail to reject” the null hypothesis (that is, you accept that your theory about the expected ratio is correct).

6. -Calculating Test Statistics
Contrasts observed frequencies in each cell of a contingency table with expected frequencies.
The expected frequencies represent the number of cases that would be found in each cell if the null hypothesis were true ( i.e. the nominal variables are unrelated).
Expected frequency of two unrelated events is product of the row and column frequency divided by number of cases.
Fe= Fr Fc / N
Mean difference between pairs of values

7. -Calculating Test Statistics
Mean difference between pairs of values

8. 4. Calculating Test Statistics
Observed
frequencies
Expected frequency
Mean difference between pairs of values
Expected frequency

9. -Determine Degrees of Freedom
df = (R-1)(C-1)
Number of levels in column variable
Number of levels in row variable

10. -Compare computed test statistic against a tabled/critical value
The computed value of the Pearson chi- square statistic is compared with the critical value to determine if the computed value is improbable
The critical tabled values are based on sampling distributions of the Pearson chi-square statistic
If calculated 2 is greater than 2 table value, reject Ho

11. Example 1: One-dimensional
Suppose we want to know how people in a particular area will vote in general and go around asking them.
How will we go about seeing what’s really going on?
Republican
Democrat
Other 20 30 10

12. -Determine Hypotheses
Ho : There is no difference between what is observed and what is expected.
𝐻 𝑎 : There is an association between the observed and expected frequencies.
Solution: chi-square analysis to determine if our outcome is different from what would be expected if there was no preference

13. Plug in to formula 20 30 10 Observed Expected Republican Democrat
Other
Observed 20 30 10
Expected

14. Reject H0
The district will probably vote democratic, there is association between what is observed and what is expected.
However…

15. Conclusion of Example 1
Note that all we really can conclude is that our data is different from the expected outcome given a situation
Although it would appear that the district will vote democratic, really we can only conclude they were not responding by chance
Regardless of the position of the frequencies we’d have come up with the same result
In other words, it is a non-directional test regardless of the prediction

16. Example 2-Two Dimensional
Suppose a researcher is interested in voting preferences on gun control issues.
A questionnaire was developed and sent to a random sample of 90 voters.
The researcher also collects information about the political party membership of the sample of 90 respondents.

17. Bivariate Frequency Table or Contingency Table
Favor
Neutral
Oppose
f row
Democrat 10 30 50
Republican 15 40
f column 25
n = 90

18. Bivariate Frequency Table or Contingency Table
Favor
Neutral
Oppose
f row
Democrat 10 30 50
Republican 15 40
f column 25
n = 90
Observed
frequencies

19. Bivariate Frequency Table or Contingency Table
Row frequency
Bivariate Frequency Table or Contingency Table
Favor
Neutral
Oppose
f row
Democrat 10 30 50
Republican 15 40
f column 25
n = 90

20. Bivariate Frequency Table or Contingency Table
Favor
Neutral
Oppose
f row
Democrat 10 30 50
Republican 15 40
f column 25
n = 90
Column frequency

21. -Determine The Hypothesis
Ho : There is no difference between D & R in their opinion on gun control issue.
Ha : There is an association between responses to the gun control survey and the party membership in the population.

22. -Calculating Test Statistics
Favor
Neutral
Oppose
f row
Democrat
fo =10
fe =13.9
fo =30
fe=22.2 50
Republican
fo =15
fe =11.1
fe =17.8 40
f column 25
n = 90

23. -Calculating Test Statistics
Favor
Neutral
Oppose
f row
Democrat
fo =10
fe =13.9
fo =30
fe=22.2 50
Republican
fo =15
fe =11.1
fe =17.8 40
f column 25
n = 90
= 50*25/90

24. -Calculating Test Statistics
Favor
Neutral
Oppose
f row
Democrat
fo =10
fe =13.9
fo =30
fe=22.2 50
Republican
fo =15
fe =11.1
fe =17.8 40
f column 25
n = 90
= 40* 25/90

25. -Calculating Test Statistics=

26. -Determine Degrees of Freedom
df = (R-1)(C-1) = (2-1)(3-1) = 2

27. -Compare computed test statistic against a tabled/critical value
α = 0.05
df = 2
Critical tabled value = 5.991
Test statistic, 11.03, exceeds critical value
Null hypothesis is rejected

28. -Conclusion of Example 2
Ho : There is no difference between D & R in their opinion on gun control issue.
Ha : There is an association between responses to the gun control survey and the party membership in the population.
-Democrats & Republicans differ significantly in their opinions on gun control issues