Sociology 601 Lecture 11: October 6, 2009

No office hours Oct. 15, but available all day Oct. 16Homework

Contingency Tables for Categorical Variables (8.1)some useful probabilities and hypothesis tests based on

contingency tablesindependence redefined.

The Chi-Squared Test (8.2) [Thursday]

When to use Chi-squared tests (8.3) [Thursday]chi-squared residuals

HomeworkStata 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 variablesVariable X (Explanatory/Independent variable)

can have one of two values: X = 1 or X = 2Variable 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 tableValues for X and Y variables are arrayed as follows:

Value of Y:

1 2

Value

of X:

1 n11 n12 total X=1

2 n21 n22 total X=2

total Y=1 total Y=2 (grand total)

Some useful definitionsThe 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 Law Enforcement? Yes No TotSupport health Yes 292 25 317care spending? No 14 9 23

Tot 306 34 340

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 of Y 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 independencePreviously 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 44 14 42 100

hispanic 110 35 105 250

Total 594 189 567 1350

Contingency tables in STATAThe 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 numberfemale democrat 279male democrat 165female independent 73male independent 47female republican 225male republican 191end

Contingency tables in STATAHere 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 tax reform? Yes No Totsupport Yes 150 100 250environment? No 200 50

250Tot 350 150 500

“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 Ha: diff != 0 Ha: diff > 0

Pr(Z < z) = 0.0000 Pr(|Z| < |z|) = 0.0000 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.