Download Contingency Tables

Sociology 601
Lecture 11: October 6, 2009
 No office hours Oct. 15, but available all day Oct. 16
 Homework
 Contingency Tables for Categorical Variables (8.1)
 some useful probabilities and hypothesis tests based on contingency tables
 independence redefined.
 The Chi-Squared Test (8.2) [Thursday]
 When to use Chi-squared tests (8.3) [Thursday]
 chi-squared residuals
Homework
 Stata ttests: means and proportions – using categorical,
dummy, interval/continuous variables
 P values with the T table: t=3, n=9, what is P?
 # 30 – industrial plant – part C
 # 52 – random number generator
 Small sample significance test
 # 54 – e is incorrect
3
Definitions for a 2X2 contingency table
 Let X and Y denote two categorical variables
 Variable X (Explanatory/Independent variable)
can have one of two values: X = 1 or X = 2
 Variable Y (Response/Dependent variable)
can have one of two values:Y = 1 or Y = 2
 nij denotes the count of responses in a cell in a table
Structure for a 2X2 contingency table
 Values for X and Y variables are arrayed as follows:
Value of Y:
Value
of X:
1
2
1
n11
n12
total X=1
2
n21
n22
total X=2
total Y=1 total Y=2
(grand
total)
Some useful definitions
 The unconditional probability P(Y = 1):
= (n11 + n21 )/ (n11 + n12 + n21 + n22 )
= the marginal probability that Y equals 1
 The conditional probability P(Y = 1, given X = 1): = n11 / (n11 + n12)
= P ((Y = 1) | (X = 1))
 The joint probability P(Y = 1 and X = 1):
= n11 / (n11 + n12 + n21 + n22 )
= P ((Y = 1)  (X = 1))
= the cell probability for cell (1,1)
Example:
Support health
care spending?
Yes
No
Tot
Support Law Enforcement?
Yes
No
292
25
14
9
306
34
Tot
317
23
340
 What is the unconditional probability of favoring increased spending on law
enforcement?
 What is the conditional probability of favoring increased spending on law
enforcement for respondents who opposed increased spending on health?
 What is the joint probability of favoring increased spending on law
enforcement and opposing increased spending on health?
Hypothesis tests based on
contingency tables:
 Usually we ask: is the distribution of Y when X=1 different than the distribution ofY
when X=2?
 Null Hypothesis: the conditional distributions of Y, given X, are equal.
Ho: P ((Y = 1) | (X = 1)) – P((Y = 1) | (X = 2)) = 0
alternatively, Ho: Y|X=1 - Y|X=2 = 0
 This type of question often comes up because of its causal implications.
 For example: “Are childless adults more likely to vote for school funding than parents?”
A confusing new definition for independence
 Previously we used the term independence to refer to groups of observations.
 “White and hispanic respondents were sampled independently.”
 In this chapter, we use independence to refer to a property of variables, not
observations.
 “Political orientation is independently distributed with respect to ethnicity”
 Two categorical variables are independent if the conditional distributions of one variable are
identical at each category of the other variable.
Democrat
Independent
Republican
Total
white
440
140
420
1000
black
hispanic
Total
44
110
594
14
35
189
42
105
567
100
250
1350
Contingency tables in STATA
 The 1991 General Social Survey Contains data on Party
Identification and Gender for 980 respondents.
 See Table 8.1, page 250 in A&F
 Here is a program for inputting the data into STATA interactively:
input str10 gender str12 party number
female
democrat
279
male
democrat
165
female
independent
73
male
independent
47
female
republican
225
male
republican
191
end
Contingency tables in STATA
 Here is a command to create a contingency table, and its output
. tabulate gender party [freq=number]
|
party
gender | democrat independe republica |
Total
-----------+---------------------------------+---------female |
279
73
225 |
577
male |
165
47
191 |
403
-----------+---------------------------------+---------Total |
444
120
416 |
980
 The following slide adds row, column, and cell %
. tabulate gender party [freq=number], row column cell
+-------------------+
| Key
|
|-------------------|
|
frequency
|
| row percentage
|
| column percentage |
| cell percentage |
+-------------------+
|
party
gender | democrat independe republica |
Total
-----------+---------------------------------+---------female |
279
73
225 |
577
|
48.35
12.65
38.99 |
100.00
|
62.84
60.83
54.09 |
58.88
|
28.47
7.45
22.96 |
58.88
-----------+---------------------------------+---------male |
165
47
191 |
403
|
40.94
11.66
47.39 |
100.00
|
37.16
39.17
45.91 |
41.12
|
16.84
4.80
19.49 |
41.12
-----------+---------------------------------+---------Total |
444
120
416 |
980
|
45.31
12.24
42.45 |
100.00
|
100.00
100.00
100.00 |
100.00
|
45.31
12.24
42.45 |
100.00
8.2 Developing a new statistical significance test
for contingency tables.
support
environment?
Yes
No
Tot
support tax reform?
Yes
No
150
100
200
50
350
150
Tot
250
250
500
 “Is the level of support for the environment dependent on the level of support
for tax reform.”
 If so, these two measures are likely to have some causal link worth investigating.
With a 2x2 table, we can use a t-test for
independent-sample proportions.
. prtesti 250 .6 250 .8
Two-sample test of proportion
x: Number of obs =
250
y: Number of obs =
250
-----------------------------------------------------------------------------Variable |
Mean
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------x |
.6
.0309839
.5392727
.6607273
y |
.8
.0252982
.7504164
.8495836
-------------+---------------------------------------------------------------diff |
-.2
.04
-.2783986
-.1216014
| under Ho:
.0409878
-4.88
0.000
-----------------------------------------------------------------------------diff = prop(x) - prop(y)
z = -4.8795
Ho: diff = 0
Ha: diff < 0
Pr(Z < z) = 0.0000
Ha: diff != 0
Pr(|Z| < |z|) = 0.0000
Ha: diff > 0
Pr(Z > z) = 1.0000
Moving beyond 2x2 tables:
 Comparing conditional probabilities is fine when there are only two comparisons and two
possible outcomes for each comparison.
 The Chi-Square (2) test is a new technique for making comparisons more flexible.

2 is like a null hypothesis that every cell should have the frequency you would expect if the
variables were independently distributed.
 fe is the expected count for each cell.
 fe = total N * unconditional row probability * unconditional column probability
 A test for the whole table will combine tests for fe for every cell.