Two-way tables and odds ratios

• Last semester, we discussed contingency tables. Contingency tables give the frequency of observations that fall into a cross-classification of two or more categorical variables. When two variables are involved we call it a two-way table. When three categories are involved, we speak of a three-way table, and so on.
• For most of this module we will focus on two-way tables, but I will also consider the case of more complicated tables at the conclusions.
• The simplest sort of contingency table is one made up of two dichotomous variables. In this case, we get a 2x2 table. Powers and Xie give an example of a 2x2 table in which one category is education ("more than high school") and the other category is expressed approval of premarital sex:

  	Attitude
  Education	Disapprove	Approve	Total
  High School or less	873	1190	2063
  College and above	533	1208	1741
  Total	1406	2398	3804

  (discuss column and row marginals, conditional probabilities)

• There is a general syntax for these types of tables.
  • $f_{ij}$ gives the actual frequency for the $i$th row and $j$th column (give example).
  • $F_{ij}$ gives the expected frequency under some model.
  • Column and row marginals are written as $f_{i+}$ and $f_{+j}$, respectively.
  • The grand total is written as $f_{++}$.
  • $I$ and $J$ are the number of categories in the row and column variables, respectively.

• Given this table we can ask whether there is an association between education and attitudes towards premarital sex. We did this last semester with the $\chi^{2}$ test.
  • If there is no association, then the probability of being in $f_{ij}$ is simply the product of the probability of being in marginal category $i$ and category $j$:
    $$F_{ij}=(\frac{f_{i+}}{f_{++}}\frac{f_{+j}}{f_{++}})/f_{++}=\frac{f_{i+}f_{+j}}{f_{++}}$$
    Use the first cell as an example:
    $$f_{1+}/f_{++}=2063/3804=0.5423$$
    $$f_{+1}/f_{++}=1406/3804=0.3696$$
    $$F_{11}=(0.5423)(0.3696)(3804))=762.5$$

  • We can compare the expected count under independence to the actual count using the Pearson $\chi^{2}$ statistic:
    $$X^{2}=\sum_{i}\sum_{j}(F_{ij}-f_{ij})^{2}/F_{ij}$$

  • Under the null hypothesis of independence, this statistic is distributed as a $\chi^{2}$ with (I-1)(J-1) degrees of freedom. In the case of the 2x2 table, we have one degree of freedom. In the table above, we get a statistic of 55.5 with 1 degree of freedom. The probability of this value under the null hypothesis is exceedingly small, so we reject the null hypothesis that there is no association. Clearly, education and attitudes towards premarital sex are related.
• But how are they related? We would like some way of measuring the relationship. Last semester, we used conditional distributions to examine this relationship, but it would be useful to have something more mathematically concise. For that task we turn to odds ratios, again.
  • First, let's get the distribution of attitudes conditional on education.

    	Attitude
    Education	Disapprove	Approve	Total
    High School or less	0.423	0.577	1
    College and above	0.306	0.694	1
    Total	0.370	0.630	1

  • Now let's calculate the odds of approval for each education group:
    $$.577/.423=1.364$$
    $$.694/.306=2.268$$

  • Now calculate the odds ratio between the educational groups:
    $$2.268/1.364=1.66$$

  • So college-educated individual's odds of approval are about two-thirds greater than non-college-educated individual's odds.
  • What if we reversed this and calculated the odds of having attended college conditional on attitude?

    	Attitude
    Education	Disapprove	Approve	Total
    High School or less	0.621	0.496	0.542
    College and above	0.379	0.504	0.458
    Total	1	1	1

    $$0.379/0.621=0.610$$
    $$0.504/0.496=1.016$$
    $$1.016/.610=1.66$$

  • The odds ratios are identical! The odds ratio is like the correlation coefficient - it does not distinguish between the independent and dependent variable. It just provides a measure of association.
  • Odds ratios are also invariant to changes in sample size and marginal distribution.
  • We didn't actually need to calculate the conditional distributions to get this odds ratio. Let's say we want to get the odds of approval for each educational group. These odds are given directly by the counts as:
    $$f_{12}/f_{11}=1190/873=1.364$$
    $$f_{22}/f_{21}=1208/533=2.266$$
    The odds ratio is then given by:
    $$\frac{f_{22}/f_{21}}{f_{12}/f_{11}}=\frac{f_{22}f_{11}}{f_{21}f_{12}}=\frac{873*1208}{533*1190}=1.66$$

  • Thus for any 2x2 table, the odds ratio is given by:
    $$OR=\frac{f_{22}f_{11}}{f_{21}f_{12}}$$

  • This is sometimes referred to as the cross-product because one divides the product of the diagonal elements by the product of the off-diagonal elements.
  • As before, we will often see these odds-ratios in logged form.
  • For a 2x2 table, there is only one odds-ratio, but in larger tables, one can identify more than one possible odds ratio.
  • Let's expand the table Powers and Xie are using. In actuality, the two categories they used for education and attitudes were collapsed from four original categories for both, giving the followinf 4x4 table:

    	Is pre-marital sex wrong?
    		Almost	Some-
    Education	Always	always	times	never
    Less than HS	332	99	141	311
    HS	313	129	258	480
    Some College	199	87	218	423
    College+	176	71	208	359

  • We could look at a lot of different odds ratios in this table, each involving four cells. However, it turns out that for an $I$x$J$ table there are only $(I-1)(J-1)$ non-redundant odds ratios. By non-redundant, I mean that we can derive all other possible odds ratios in the table if we have the non-redundant ones. Therefore, all the information about association in the table is contained in the non-redundant odds ratios.
  • For convenience, we usually take the local odds ratios:
    $$\theta_{ij}=\frac{f_{ij}f_{(i+1)(j+1)}}{f_{(i+1)j}f_{i(j+1)}}$$

    	Is pre-marital sex wrong?
    Education	Always vs. almost	almost vs. sometimes	sometimes vs. never
    LHS vs. HS	1.382	1.404	0.843
    HS vs. Some Coll.	1.061	1.253	1.043
    Some Coll. vs. College	0.923	1.169	0.890

  • Any other odds ratio can be calculated from these nine local odds ratios.

    Let's take the odds ratio between less than high school and some college and between always and sometimes wrong. Calculated directly this odds ratio is:

    $$\frac{218*332}{199*141}=2.579$$
    Compared to individuals with less than a high school education, individuals with some college education are 2.579 times as likely to say that pre-marital sex is sometimes wrong vs. always wrong.

    We could have derived this from our local odds ratios by multiplication:

    $$(1.382)(1.404)(1.061)(1.253)=2.579$$
    In essence, we have to ``traverse" the horizontal and vertical distance between points on the table to derive any particular odds ratio.
  • Interpreting local odds ratios
    • Lets look at the lower corner of the table. The odds ratio here turns out to be 0.89. What does this mean? College-educated individuals are 0.89 times as likely as some-college educated individuals to say that pre-marital sex is never vs. sometimes wrong.

    • How about the 1.253 in the middle of the table? Some-college educated individuals are 1.253 times as likely as high school graduates to say that pre-maritial sex is sometimes vs. almost always wrong.

    • Because both of our categories are ordinal, we can interpret all of the odds ratios as indicating how tolerance for pre-marital sex changes as respondents move up in education. Odds ratios greater than one indicate that tolerance goes up across that boundary as education goes up across that boundary. Odds ratios less than one indicate the opposite.

    • Thus although most of the local odds ratios are greater than one, indicating increasingly liberal attitude with eduation, There are some interesting deviations as we move across the some college vs. college boundary where tolerance actually appears to decline at the extremes. In general the greatest shift in attitudes for all educational comparisons happens at the almost vs. sometimes boundary.

    • Remember that last semester we looked at a similar question on abortion attitudes by treating both education and attitudes as quantitative variables. The results here show the the pitfall of that approach. There are some clear non-linearities in how education affects attitudes that we would miss if we took the former approach.
  • Log-odds ratios
    • As before, we often work with the log of the odds ratios rather than the odds ratios themselves. For our local odds ratios, this gives:

      	Is pre-marital sex wrong?
      Education	Always vs. almost	almost vs. sometimes	sometimes vs. never
      LHS vs. HS	0.324	0.340	-0.170
      HS vs. Some Coll.	0.059	0.225	0.042
      Some Coll. vs. College	-0.080	0.156	-0.117

    • Whereas before, 1 marked the boundary between a positive and negative relationship, now 0 marks that boundary.
    • When we work with the log of the odds ratios, all of our multiplicative relations from before become additive.

      So, the log of the local odds ratio is given by:

      $$\log(\theta_{ij})=\log(f_{ij})+\log(f_{(i+1)(j+1)})-\log(f_{(i+1)j})-\log(f_{i(j+1)})$$
      Before, to get from the local odds ratios to any other odds ratio, we multiplied. Now we add:
      $$\log(1.382)+\log(1.404)+\log(1.061)+\log(1.253)=\log(2.579)$$
      $$0.324+0.340+0.059+0.225=0.948$$

    • One of the main advantages of the log-odds ratio is that it is easier to derive inferences for than the odds ratio directly. In large samples, the sampling distribution of the log-odds ratio is roughly normal with a standard deviation of
      $$\sqrt{\frac{1}{f_{ij}}+\frac{1}{f_{(i+1)j}}+\frac{1}{f_{i(j+1)}}+\frac{1}{f_{(i+1)(j+1)}}}$$
      Let's use this to calculate a 95% confidence interval for the local odds ratio between LHS and HS respondents for saying almost always vs. always wrong.

      First, lets calculate the standard deviation of the sampling distribution for the log-odds ratio:

      $$\sqrt{\frac{1}{332}+\frac{1}{99}+\frac{1}{313}+\frac{1}{129}}=0.155$$
      Now we use this value to get the confidence interval for our log-odds ratio:
      $$0.324 \pm 0.155*1.96=(0.020,0.628)$$
      Now we exponentiate this to get the confidence interval for our odds ratio:
      $$(e^{0.020},e^{0.628})=(1.02,1.87)$$
      We are 95% confident that the range of values for the odds ratio in the population does not include one, so we would reject the hypothesis that there was no assocation at this level of the two variables.