1. Qualitative data – tests of association
The Chi-Square Distribution and Test for Independence
Hypothesis testing between two or more categorical variables
Sporiš Goran, PhD.

2. Chi-Square Distribution
The chi-square distribution results when independent variables with standard normal distributions are squared and summed.

3. Chi-square Degrees of freedom
df = (r-1) (c-1)
Where r = # of rows, c = # of columns
Thus, in any 2x2 contingency table, the degrees of freedom = 1.
As the degrees of freedom increase, the distribution shifts to the right and the critical values of chi-square become larger.

4. Chi-Square Test of Independence

5. Using the Chi-Square Test
Often used with contingency tables (i.e., crosstabulations)
E.g., gender x student
The chi-square test of independence tests whether the columns are contingent on the rows in the table.
In this case, the null hypothesis is that there is no relationship between row and column frequencies.
H0: The 2 variables are independent.

6. Requirements for Chi-Square test
Must be a random sample from population
Data must be in raw frequencies
Variables must be independent
Categories for each I.V. must be mutually exclusive and exhaustive

7. Example Crosstab: Gender x Student
Student
Not Student
Total
Males 46
(40.97) 71
(76.02)
117
Females 37
(42.03) 83
(77.97)
120
154
237
Observed
Expected

8. Special Cases Fisher’s Exact Test Strength of Association
When you have a 2 x 2 table with expected frequencies less than 5.
Strength of Association
Some use Cramer’s V (for any two nominal variables) or Phi (for 2 x 2 tables) to give a value of association between the variables.

9. Two chi square tests Goodness of fit Independence One variable
Determines how well the sample proportions match a pre-specified distribution
Independence
Two variables
Determines whether there is a relationship between two variables

10. Steps in hypothesis testing
State the hypotheses
null
research
Select an alpha level and determine the critical value
Compute the test statistic
Make a decision

11. Test for goodness of fit
Forms of the null hypothesis
No preference
There is no difference in proportions among the categories
Participants do not prefer one category over another
Example: Pepsi: 50%, Coke 50%
No difference from a comparison population
There is no difference between the sample distribution and a known (population) distribution
Example: ND: 20% Bl, 75% Br, 5% R
US: 20% Bl, 75% Br, 5% R

12. Test for goodness of fit
Null hypothesis
Specifies a distribution of proportions
Research hypothesis
Specifies that the distribution will be different than that indicated in the null hypothesis

13. Calculating the test statistic
Observed frequencies
the number of individuals from the sample who are classified in a particular category fo
Expected frequencies
the number of individuals from the sample who are expected to be classified in a particular category fe

14. Calculating the test statistic
Coin flip: What percentage of people will predict heads? tails?
Heads
Tails
Percentages
50%
Proportions .5

15. Calculating the test statistic
Expected frequency = fe = pn
n = 50 (sample size)
fe = .5 x 50 = 25
Expected
Heads
Tails
Proportions .5
Frequencies 25

16. Calculating the test statistic
Question: The last five flips were tails. What do you predict for the next flip?
Heads
Tails
Observed 35 15
Expected 25

17. Calculating the test statistic
Heads
Tails
Observed 35 15
Expected 25
x2 = ∑ (fo - fe)2 fe
Steps
find the difference between fo and fe for each category
square the difference
divide the squared difference by fe
sum the values from all categories

18. x2 = ∑ (fo - fe)2 = 4 + 4 = 8 Heads Tails Observed (fo) 35 15
Expected (fe) 25
fo - fe 10
-10
(fo - fe)2
100
(fo - fe)2/fe 4
x2 = ∑ (fo - fe)2 = = 8 fe

19. Chi square distribution
Critical range x2
Low chi square
Hi chi square

20. Critical values for chi square distribution
Table B.8

21. Test for goodness of fit
The greater the number of categories, the greater the likelihood of a large observed chi square value
Degrees of freedom (df)
The number of values that are free to vary
df = C – 1
C = the number of categories

22. Chi square distribution

23. Critical values for chi square distribution

24. Chi square distribution5%
3.84 x2
Critical value (df = 1,  = .05) = 3.84

25. Goodness of fit Make a decision
Critical value = 3.84 with df = 1 and  = .05.
Observed chi square = 8.0
8.0 > 3.84
Observed chi square is greater than critical value
We reject the null hypothesis
Conclude that category frequencies are different
People were more likely to predict heads than tails