chapter 9 analysis of two-way tables. two-way (i.e. contingency) tables: to classify & analyze...
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Chapter 9
Analysis of Two-Way Tables
Two-way (i.e. contingency) tables: to classify & analyze categorical data:
Binomial counts: ‘success’ vs. ‘failure’
Proportions: binomial count divided by total sample size
We’ll later see that inference via two-way tables is an alternative—with advantages & disadvantages—to the z-test for comparing two sample proportions:
. prtest hsci, by(white)
An advantage of two-way tables is that they can examine more than two variables.
A disadvantage of two-way tables is that they can only do two-sided hypothesis tests.
Here’s a two-way table:
. tab hsci white, cell Nonwhite White Total
not hsci 50 103 153
25.0% 51.5%76.5%
hsci 5 42 47
2.5% 21.0% 23.5%
Total 55 145 200
27.5% 72.5% 100.0%
Nonwhite White Total
not hsci 50 103 153
25.0% 51.5%76.5%
hsci 5 42 47
2.5% 21.0% 23.5%
Total 55 145 200
27.5% 72.5% 100.0%
The row variable: hsci vs. not hsci. The column variable: white vs. nonwhite.
Nonwhite White Total
not hsci 50 103 153
25.0% 51.5%76.5%
hsci 5 42 47
2.5% 21.0% 23.5%
Total 55 145 200
27.5% 72.5% 100.0%
Cells: each combination of values for the two variables (50, 103, 5, 42).
Nonwhite White Total
not hsci 50 103 153
25.0% 51.5% 76.5%
hsci 5 42 47
2.5% 21.0% 23.5%
Total 55 145 200
27.5% 72.5% 100.0%
Joint distributions: Each cell’s percentage of the total sample (50/200=.250; 103/200=.515; 5/200=.025; 42/200=.210).
Nonwhite White Total
not hsci 50 103 153
25.0% 51.5%76.5%
hsci 5 42 47
2.5% 21.0% 23.5%
Total 55 145200
27.5% 72.5% 100.0%
The marginal frequencies: the row totals (153, 47) & the column totals (55, 145).
Nonwhite White Total
not hsci 50 103 153
25.0% 51.5%76.5%
hsci 5 42 47
2.5% 21.0% 23.5%
Total 55 145 200
27.5% 72.5% 100.0%
The marginal distributions: each row total/sample total (76.5%, 23.5%). Each column total/sample total (27.5%, 72.5%).
Here are the same data displayed as column conditional probabilities:
. tab hsci white, col nofreq nonwhite white Total
no 90.91% 71.03% 76.5%
yes 9.09% 28.97% 23.5%
Total 100.0% 100.% 100.00%
The conditional distributions (i.e conditional probabilities): Column—divide each column cell count by its column total count.
Here are the same data displayed as row conditional probabilities:
. tab hsci white, row nofreq nonwhite white Total
not hsci 32.68% 67.32% 100.0%
hsci 10.64% 89.36% 100.0%
Total 27.5% 72.5% 100.00%
The conditional distributions (i.e conditional probabilities): Row—divide each row cell count by its row total count.
Tip: It’s usually best to compute conditional distributions (i.e. probabilities) across the categories of the explanatory variable.
E.g., tab hsci white, col: computes the conditional distributions across the categories of the explanatory variable race-ethnicity (i.e. white vs. nonwhite).
Alternatively, you may want to compare joint distributions (i.e. cell counts/total sample): tab hsci white, cell
We’ve discussed the following:
row variables
column variables
cells: each combination of values for the two variables.
joint distributions: each cell’s percentage of the total sample.
marginal frequencies
marginal distributions: each marginal frequency/total sample size
column conditional distributions: divide each column cell count by its column total count
row conditional distributions: divide each row cell count by its row total count.
And we’ve said that typically it’s best to compute the conditional distributions (i.e. probabilities) across the categories of the explanatory variable.
Or that it may be preferable to compare joint distributions (i.e. compare the cell probabilities).
Let’s next consider conceptual problems of two-way tables.
Simpson’s Paradox
An NSF study found that the median salary of newly graduated female engineers & scientists was just 73% of the median salary for males. Here are women’s median salaries in the 16 fields as a percentage of male salaries:
94% 96% 98% 95% 85% 85% 84% 100% 103% 100% 107% 93% 104% 93% 106% 100%
How can it be that, on average, the women earn just 73% of the median salary for males, since no listed % falls below 84%?
Because women are disproportionately located in the lower-paying fields of engineering & science.
That is, ‘field of science & engineering’ is a lurking variable (i.e. an unmeasured confounded variable) that influences the observed association between gender & salary.
Simpson’s Paradox: the reversal of a bivariate relationship due to the influence of a lurking variable.
Aggregating data has the effect of ignoring one or more lurking variables.
Another example: comparing hospital mortality rates.
Yet another: comparing airline on-time rates.
Conclusion from
Simpson’s Paradox
Always be on the lookout for lurking variables with aggregated data!!
A bivariate relationship may change direction when a third, control variable is introduced.
What’s a control variable?
Holding a variable constant makes it a control variable: doing so removes the part of the bivariate relationship that was caused by the control variable.
That is, controlling for a variable neutralizes its influence on the observed relationship.
E.g., controlling for field of science & engineering.
E.g., controlling for race/ethnicity.
To repeat, holding a variable constant removes its statistical effects from the bivariate association being examined.
Doing so ensures (more or less) that a bivariate relationship is assessed apart from the influence of the controlled variable: e.g., the relationship between a Montessori school program & student IQ scores, holding constant social class.
What’s better: statistical control or experimental control?
The answer returns us to the matter of observational study versus experimental study (see Moore/McCabe, chapter 3).
Good experimental design controls for all possible lurking variables. Why?
But statistical control cannot do so. Why not?
Moreover, statistical control is weakened by the imprecision of measurement of variables.
But we can’t experiment on everything.
Let’s consider the following variant on Simpson’s Paradox:
A bivariate association may not appear until a third, control variable is introduced.
The apparent absence of the bivariate relationship is called spurious non-association.
E.g., no association between years of education & level of income in post-WW II data, until controlling for age of respondents.
Conclusion from
Spurious Non-Association
Explore not just bivariate relationships but also multivariate relationships among all the variables of potential practical or theoretical relevance.
Here’s how to add a control variable to a two-way table in Stata:
bys female: tab hsci white, cell chi2
malenonwhite white Total
0 21 41 62
23.08% 45.05% 68.13%
1 2 27 29
2.20% 29.67% 31.87%
Total 23 68 91
25.27% 74.73% 100.00 %
Pearson chi2(1) = 7.6120 Pr = 0.006
femalenonwhite white Total
0 29 62 91
26.61% 56.88% 83.49%
1 3 15 18
2.75% 13.76% 16.51%
Total 32 77 109
29.36% 70.64% 100.00%
Pearson chi2(1) = 1.6744 Pr = 0.196
This example introduces a test of statistical significance for two-way tables.
The test is based on the Chi-square statistic.
male nonwhite white Total
0 21 41 62 23.08% 45.05% 68.13%
1 2 27 29 2.20% 29.67% 31.87%
Total 23 68 91 25.27% 74.73% 100.00 %Pearson chi2(1) = 7.6120 Pr = 0.006
femalenonwhite white Total
0 29 62 91 26.61% 56.88% 83.49%
1 3 15 18 2.75% 13.76% 16.51%
Total 32 77 109 29.36% 70.64% 100.00%
Pearson chi2(1) = 1.6744 Pr = 0.196
Note in the example that the two-way table for male tests insignificant.
How do two-way tables & their test of significance evaluate the data?
They do so by comparing expected & observed cell counts in terms of proportional distributions.
Back to the two-way table without the control variable::
. tab hsci white, cell
Nonwhite White Totalnot hsci 50 103 153
25.0% 51.5% 76.5%
hsci 5 42 47 2.5% 21.0% 23.5%
Total 55 145 200 27.5% 72.5% 100.0%
Describing Relations in Two-Way Tables
The original data must be counts.
Inference for two-way tables: compare the observed cell counts to the expected cell counts; then compute the Chi-square significance test.
We begin by computing the expected cell counts: row total times column total, divided by total sample size.
Premise: the null hypothesis of ‘statistical independence’ (i.e. no association between the variables) characterizes the data.
Expected cell counts: row total times column total, divided by total sample size.
nonwhite white Total
no 50 103 153
yes 5 42 47
Total 55 145 200
nonwhite white Total
no 50 103 153
yes 5 42 47
Total 55 145 200
. di (153*55)/200=42.075 . di (153*145)/200=110.925
. di (47*55)/200= 12.925 . di (47*145)/200= 34.075
How do the expected cell counts compare to the observed cell counts: Do the conditional probabilities appear to be equal for nonwhites & whites across no-hsci & yes hsci?
Expected count for each cell: its row total times its column total, divided by the total sample size.
Each expected cell count is based on the proportion of the total sample accounted for by its entire row & by its entire column.
The Chi-square test assumes independence (i.e. no association) between the conditional distributions of nonwhites & whites in honors science.
That is, each expected cell count reflects the null hypothesis of statistical independence (i.e. no association):
that the proportion of non-white honors science students is simply the proportion of non-white students in the population.
that the proportion of white honors science students is simply the proportion of white students in the population.
What’s the alternative hypothesis?
Chi-Square Test Assumptions
Random sample
Two categorical variables
Count data
At least 5 observations in 80% of the cells & no less than 1 observation in any cell (best if there’s at least 5 observations in all cells)
If the assumptions are fulfilled, use the Chi-square test:
tab hsci white, chi2
If the numbers of observations per cell don’t meet the assumptions, use ‘Fisher’s exact test’ (a non-parametric test, which may be very slow):
tab hsci white, exact
Chi-square statistic: measures how much the observed cell counts in a two-way table diverge from the expected cell counts.
It’s therefore a test of independence:
Ho: the variables are independent from each other
Ha: they are not independent from each other
Step 1: Chi-square = summation for all cells of (observed cell count – expected cell count)squared, divided by the cell’s expected count
Step 2: df = (# row vars –1) (# column vars – 1)
Step 3: Chi-square significance test=Chi-square/df
Chi-square/df statistic: positive values only
Has a distinct distribution for each degree of freedom (see Moore/McCabe)
Two-sided hypothesis test only
Chi-square Test: To Repeat…
Chi-square statistic: measures how much the observed cell counts in a two-way table diverge from the expected cell counts.
That is, it compares the sample distribution with a hypothesized distribution.
It’s a test of statistical independence (Ho: no association; Ha: association).
Step 1: Chi-square = summation for all cells of (observed cell count – expected cell count)squared, divided by the cell’s expected count
Step 2: df = (# row vars –1) (# column vars – 1)
Step 3: Chi-square significance test=Chi-square/df
Hypothesis Test
Ho: hsci whites = hsci nonwhites
Ha: hsci whites ~= hsci nonwhites (i.e. two-sided alternative)
Chi-square test: two-sided alternative hypothesis only.
. tab hsci white, cell chi2
nonwhite white Total
no 50 103 153
25.0% 51.5% 76.5%
yes 5 42 47
2.5% 21.0% 23.5%
Total 55 145 200
27.5% 72.5% 100.0%
Pearson chi2(1) = 8.7613 Pr = 0.003
Conclusion: Reject the null hypothesis.
Let’s repeat the earlier example to see what happens when we add a control variable to the two-way table:
bys female: tab hsci nonwhite, col chi2
female = malenonwhite white Total
0 21 41 62
23.08% 45.05% 68.13%
1 2 27 29
2.20% 29.67% 31.87%
Total 23 68 91
25.27% 74.73% 100.00%
Pearson chi2(1) = 7.6120 Pr = 0.006
female = femalenonwhite white Total
0 29 62 91
26.61% 56.88% 83.49%
1 3 15 18
2.75% 13.76% 16.51%
Total 32 77 109
29.36% 70.64% 100.00%
Pearson chi2(1) = 1.6744 Pr = 0.196
But when adding a control variable, beware of the consequences for sub-sample sizes.
If the sub-samples are too small, it may be hard to obtain statistical significance.
So always check the size of sub-samples.
Remember: cell counts should be >=5, & for at least 80% of the cells must be this large.
We can compare two population proportions either by the chi-square test or by the two-sample z-test—which give exactly the same result—because the chi-square test is equal to the square of the z-test.
Chi-square test advantage: can compare more than two populations (e.g., SES by race-ethnicity in hsb2.dta); but the original data must be counts.
z-test advantage: can test either one-sided or two-sided alternatives; the original data may be counts or proportions.
. prtest hsci, by(white)
Two-sample test of proportion nonwhite: Number of obs = 55
white: Number of obs = 145
------------------------------------------------------------------------------
Variable | Mean Std. Err. z P>|z| [95% Conf. Interval]
---------+--------------------------------------------------------------------
nonwhite | .0909091 .0387638 2.34521 0.0190 .0149335 .1668847
white | .2896552 .0376696 7.68936 0.0000 .2158241 .3634863
---------+--------------------------------------------------------------------
diff | -.1987461 .0540521 -.3046863 -.0928059
| under Ho: .0671451 -2.95995 0.0031
------------------------------------------------------------------------------
Ho: proportion(nonwhite) - proportion(white) = diff = 0
Ha: diff < 0 Ha: diff ~= 0 Ha: diff > 0
z = -2.960 z = -2.960 z = -2.960
P < z = 0.0015 P > |z| = 0.0031 P > z = 0.9985
. tab hsci white, cell chi2
nonwhite white Total
no 50 103 153
25.0% 51.50% 76.50%
yes 5 42 47
2.50% 21.0% 23.50%
Total 55 145 200
27.50% 72.50% 100.0%
Pearson chi2(1) = 8.7613 Pr = 0.003
pr=.003 for Chi-square test & for z-test.
Other Useful Stata Commands
findit tabchi: displays observed & expected frequencies, various types of residuals (raw, pearson, adjusted), & various tests of significance.
findit tabout: to make publication-style contingency & other tables
See the following class documents: ‘Making contingency tables in Stata’; ‘Making working & publication-style tables in Stata’.
For greater depth concerning contingency tables & their various significance tests, see Agresti & Finlay, Statistical Methods for the Social Sciences, chap. 8.
Summary: Two-Way Tables
Two-way tables: categorical data—binomial counts (‘success’ vs. ‘failure’) or proportions (binomial counts divided by the total sample size); but the data must be counts
Row variables? Column variables? Cells?
Marginal frequencies? Marginal distributions?
Joint distributions?
Row & column conditional distributions?
How to compute expected cell frequencies? What do they represent?
Null hypothesis? Alternative hypothesis?
How to compute the Chi-square test?
How to compute its degrees of freedom?
Chi-square assumptions?
Advantages/disadvantages of inference via two-ways tables versus inference via z-test for two sample proportions?
Chi-square test of significance: equals the square of the z-test for comparing sample proportions, but the Chi-square test requires the original data to be counts.
Simpson’s Paradox: aggregating the data ignores lurking variables.
Moral of the story: beware of the relations portrayed in aggregated data (i.e. look out for lurking variables)!
Spurious non-association: a bivariate association appears only when a third, control variable is introduced.
Moral of this story: the same as for Simpson’s Paradox.
Finally, when should we use contingency tables & the Chi-square test?
As part of bivariate exploratory data analysis, including in preparation for regression analysis
Or when we don’t have enough observations to do regression analysis (i.e. perhaps categorize the data and do cross-tabs).
Here’s how to do the tables for Moore/McCabe, problem 9.1:
. tabulate educ age [freq=years], chi2
a1 a2 a3 Total
e1 5,325 9,152 16,035 30,512
e2 14,061 24,070 18,320 56,451
e3 11,659 19,926 9,662 41,247
e4 10,342 19,878 8,005 38,225
Total 41,387 73,026 52,022 166,435
Pearson chi2(6) = 9.6e+03 Pr = 0.000
. tabulate educ age [freq=years], cell chi2
a1 a2 a3 Total
e1 3.20 5.50 9.63 18.33
e2 8.45 14.46 11.01 33.92
e3 7.01 11.97 5.81 24.78
e4 6.21 11.94 4.81 22.97
Total 24.87 43.88 31.26 100.00
Pearson chi2(6) = 9.6e+03 Pr = 0.000
The ‘row’ or ‘col’ options may be preferable to use.
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