1. Chapter 11 Other Chi-Squared Tests

2. Chi-Square Statistic
Measures how far the observed values are from the expected values
Take sum over all cells in table
When is large, there is evidence that H0 is false.

3. Goodness of Fit (GOF) Tests
We can use a chi-squared test to see if a frequency distribution fits a pattern.
The hypotheses to these tests are written a little different than we have seen in the past because they are usually written in words.

4. GOF Hypothesis Test
A researcher wishes to see of the number of adults who do not have health insurance is equally distributed among three categories:
Category
Less than 12 years
12 years
More than 12 years
Frequency (observed) 29 20 11

5. GOF Hypothesis Test Step 1
Ho: The number of people who do not have health insurance is equally distributed over the three categories.
Ha: The number of people who do not have health insurance is not equally distributed over the three categories.

6. GOF Hypothesis Test
Step 2
α = 0.05
Step 3

7. GOF Hypothesis Test Step 4 Test Statistic = 8.1
Put Observed in L1 and Expected in L2
L3 = (L1-L2)2 / L2
Sum L3
χ2 cdf (test stat, E99, df) = p-value = 0.017
Category
Less than 12 years
12 years
More than 12 years
Frequency (observed) 29 20 11
Expected

8. GOF Hypothesis Test Step 5 Step 6 Reject Ho
There is enough evidence to suggest that the number of people who do not have health insurance is not equally distributed over the three categories.

9. Two-Way Tables
Summarizes the relationship between two categorical variables
Row = values for one categorical variable
Column = values for other categorical variable
Table entries = number in row by column class

10. Example of Two-Way Table
Age Group
Educ
25 to 34
35 to 54
55+
Total
No HS
5,325
9,152
16,035
30,512 HS
14,061
24,070
18,320
56,451
C. 1-3
11,659
19,926
9,662
41,247
C. 4+
10,342
19,878
8,005
38,225
41,388
73,028
52,022
166,438

11. Independence of Categorical Variables.
Is there a relationship between two categorical variables. Are two variables related to each other?
Unrelated = independent
Related = dependent

12. Example
A 1992 poll conducted by the University of Montana classified respondents by sex and political party.
Sex: Male and Female
Party: Democrat, Republican, Independent
Is there evidence of an association between gender and party affiliation?

13. Hypothesis Test for Independence
HO: Sex and political party affiliation are independent (have no relationship).
HA: Sex and political party affiliation are dependent (are related to each other.)

14. Two-Way Tables Describe table with # of rows (r) and # of columns (c)
r x c table
Each number in table is called a cell
r times c cells in table
We will use the two-way table to test our hypotheses.

15. Data Democrat Republican Independent Total Male 36 45 24 105 Female 4833 16 97 84 78 40
202

16. Expected Counts
If HO is true, we would expect to get a certain number of counts in each cell.
Expected cell count = row total * column total
table total

17. Example of Expected Counts
Cell – Male and Democrat
Expected Count =
Cell – Male and Republican
Cell – Male and Independent

18. Two-Way Table of Expected Counts
Democrat
Republican
Independent
Total
obs
exp 1 36
43.66 45
40.54 24
20.79
105 2 48
40.34 33
37.46 16
19.21 97 84 78 40
202

19. Expected Counts Expected cell count is close to observed cell count
Evidence Ho is true
Expected cell count is far from observed cell count
Evidence Ho is false

20. Chi-Squared Test for Independence
To test these hypotheses, we will use a Chi-Squared Test for Independence if the assumptions hold.
Assumptions:
Expected Cell Counts are all > 5

21. Chi-Square Statistic
Measures how far the observed values are from the expected values
Take sum over all cells in table
When is large, there is evidence that H0 is false.
Your calculator will do this for you.

22. Chi-Square Statistic Cell – Male and Democrat
Observed Count = 36
Expected Count = 43.66
Cell – Male and Republican
Observed Count = 45
Expected Count = 40.54

23. Chi-Square Statistic Repeat this process for all 6 cells χ2 = 4.85
As long as the assumptions are met
χ2 will have a χ2 distribution with d.f. (r-1)(c-1)
Sex has 2 categories (so r = 2); party has 3 categories (so c = 3)
We have (2-1)(3-1) = 2 degrees of freedom

24. Hypothesis Test P-value = P(χ2 > 4.85)= 0.09
Decision: Since p-value > α = 0.05, we will Do Not Reject HO.
Conclusion: There is no evidence of a relationship between sex and political party affiliation.

25. Homogeneity of Proportions
Samples are selected from different populations and a researcher wants to determine whether the proportion of elements are common for each population.
The hypotheses are:
Ho: p1 = p2 = p3 = p4
Ha: At least one is different

26. Homogeneity of Proportions
An advertising firm has decided to ask 92 customers at each of three local shopping malls if they are willing to take part in a market research survey. According to previous studies, 38% of Americans refuse to take part in such surveys. At α = 0.01, test the claim that the proportions are equal.

27. Homogeneity of Proportions
Step 1
Ho: p1 = p2 = p3
Ha: At least one
is different
Step 2
α = 0.01
Step 3
Mall A B
Mall C
Total
Will Participate 52 45 36
133
Will not participate 40 47 56
143 92
276

28. Homogeneity of Proportions
Step 4
Put into your calculator
Observed in matrix A
Expected in matrix B
Test statistic = 5.602
P-value = 0.06

29. Homogeneity of Proportions
Step 5
Do Not Reject Ho
Step 6
There is not sufficient evidence to suggest that at least one is different.