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Lecture 9

In this discussion, we'll focus on statistical tests for 2 X 2 tables.

Suppose you have two categorical variables--say, gender (male/female) and left-handedness. The chi-square test allows us to test whether there are statistically significant differences in one categorical variable (in this example, left-handedness) based on the other (in this example, gender).

Suppose, for instance, you wanted to examine whether women are more likely to be left-handed than men.

You can construct a 2 X 2 table to do that. Such a table is called a contingency table.

Suppose that 90 men and 110 women are each asked about his or her left-handedness. Of the 200 total individuals, 20 are left-handed, and 180 are right=handed.

Recall from previous lectures that the null hypothesis, H₀, is generally "no difference" or "no effect". In this case, H₀ would be that there is no association between gender and handedness. Suppose the researcher has some reason to expect that women are in fact more likely to be left-handed. In this case, then, H₁ (the alternative hypothesis) would be that women are more likely to be left-handed.

Let's talk first about the "null hypothesis".

I. The Null Hypothesis

If gender and left-handedness are not related, what counts would we see in the cells of a "contingency table"?

We can start by entering in the total number of men and women--which is given in the example. And, we can enter in the total number of individuals who should be left-handed, and the total number who should be right-handed--based on the information given in the sample.

So, we can set up a partial table with these totals:

Contingency Table for H₀ (No Effect of Gender on Handedness)

Then, the expected count can for each cell can be calculated as:

Row Total X Column Total

______________________                  = Cell Value

    Total n for table

For example, if there's no correlation between left handedness and gender, then (out of 110 women, if 10% of people are left handed) we would expect 11 women to be left handed (and 99 women to be right handed).

Given this, we can fill in the rest of the table--

	Left-Handed	Right-Handed	Total
Men	9 (10%)	81 (90%)	90
Women	11 (10%)	99 (90%)	110
Total	20 (10%)	180 (90%)	200

II. The Chi-Square Statistic

Of course, these are the expected counts--not the real counts.

How do we calculate a chi-square statistic?

• First, for each cell in the table, calculate: (observed count-expected count)²/(expected count)
• Then, total these quantiles over all the cells in the table.

So, in our example of handedness and gender, suppose that in reality, 15 women were left-handed, and 95 women were right-handed. 10 men are left-handed, and 80 men are right-handed.

We can then enter the (observed-expected)² / (expected) values into the table. Our actual contingency table looks like this:

	Left-Handed	Right-Handed	Total
Men	(10-9)2/(9)=.111	(80-81)2/(81)=.012	90
Women	(15-11)2/(11)=1.455	(95-99)2/(99)=.162	110
Total	20 (10%)	180 (90%)	200

And, we can add the values in the cells to get the chi-square:

X² = .111 + .012 + 1.455 + .162 = 1.740

III. Chi-Square Tables & P-Values

Recall that when we discussed t-tests, the question was often whether there was a significant difference between means (or whether the mean was significantly different from 0) in the population, based on what we saw in the sample. That is, we were using inferential statistics, because we were looking at samples and making generalizations to populations. In the case of means, we looked at t-tables to determine whether there were significant differences between our actual mean and our "null" mean. The t-tables provided a p-value, which was essentially the probability that we'd get as large or larger of a t if the null hypothesis was correct--that is, it can be seen as the probability of incorrectly rejecting the null hypothesis, or (put differently) accepting the "alternative hypothesis" even though the null hypothesis was true (in the population). In most cases, the p-value gives you the probability that you are incorrectly concluding that there is a significant effect or significant effect in the population--when, in fact, there is no such significant effect.

In this case, we're doing something very similar--we're looking at a chi-square table (instead of a t-table), but we'r estill getting a p-value and the p-value is still giving us the probability of incorrectly rejecting the null hypothesis.

IV. Additional Examples

(from Utts & Heckard)

A. Randomly Pick S or Q

92 college students were given a form that read "Randomly choose one of the letters S or Q". Of the 92 students, 66% (61/92) picked S. Another 98 students were given a form that read "Randomly choose oen of the letters Q or S." Of these students, 46% (45/98) picked S. Is this sufficient evidence to generalize that individuals are more likely to pick the first letter mentioned?

The null hypothesis is, as always, no effect.

The alternative hypothesis is that people are more likely to name an option (out of two) if it is mentioned first--that is, order has an effect on the answers people give.

So, we want to calculate out: (observed-expected)² / expected in each of the cells.

Our observed contingency table looks like:

	Q first in Question	S first in Question	Total
Respondent Named Q	53	31	84
Respondent Named S	45	61	106
Total	98	92	190



Remember that you can calculate out the expected values (what we'd expect if the null is correct) as (total row*total column)/total table. So, our expected (if the null is correct) contingency table looks like:



	Q first in Question	S first in Question	Total
Respondent Named Q	43.326	40.674	84
Respondent Named S	54.674	51.326	106
Total	98	92	190



So, to get the Chi-Square statistics, we need to calculate out (observed-expected)²/(expected) for each cell. The table with this calculations would be:



	Q first in Question	S first in Question	Total
Respondent Named Q	(53-43.326)2 / 43.326 = 2.160	(31-40.674)2 / 40.674 = 2.301	84
Respondent Named S	(45-54.674)2 / 54.674 = 1.712	(61-51.326)2 / 51.326 = 1.823	106
Total	98	92	190





And, the X² is therefore: 2.160 + 2.301 + 1.712 + 1.823 = 7.996

What is the degrees of freedom? For chi-squares, for 2 x 2 tables, df=1. In general, degrees of freedom for chi-square statistic = (# rows - 1) X (# columns - 1)

According to a chi-square table, for df=1, and X²=7.996, the p-value is .005--which means there is only a very, very small chance (5 in 1000) that, if the "null hypothesis" was correct, there would be a chi-square of this magnitude. So, we can "reject the null hypothesis" and "accept the alternative hypothesis that question ordering influences respondents' answers."



B. Heart Attacks & Aspirin



In 1988, the Steering Committee of the Physicians' Health Study Research Group released the results of a five-year randomized experiment conducted using 22,071 male physicians between the ages of 40 and 84. The purpose of the experiment was to determine if taking aspirin reduces the risk of a heart attack. The physicians had been randomly assigned to one of the two treatment groups. One group took an ordinary aspirin tablet every other day, while the other group took a placebo. None of the physicians knew whether he was taking the actual aspirin or the placebo.

The actual observed results are as follows:



Observed Results
Treatment	Heart Attacks	No Heart Attacks	Total
Aspirin	104	10,933	11,037
Placebo	189	10,845	11,034
Total	293	21,778	22,071



Recall that to get the "contingency table" with the expected counts, for each cell, we just multiply the "total" for the row by the "total" for the column--and then divide by the total number of observations in the entire table (i.e., 22,071).

So, our expected table (that is, the results we'd expect if there was no association between aspirin and heart attacks) looks something like this:




Expected Results
Treatment	Heart Attacks	No Heart Attacks	Total
Aspirin	146 . 520	10,890 . 480	11,037
Placebo	146 . 480	10,887 . 520	11,034
Total	293	21,778	22,071


And, to get the Chi-Square, recall that in each cell, we subtract the expected value from the observed value--square that--and then divide by the expected value. That is, we calculate:



(observed-expected)² / (expected)




Chi-Square Calculations
Treatment	Heart Attacks	No Heart Attacks	Total
Aspirin	12.339	0.166	11,037
Placebo	12.343	0.166	11,034
Total	293	21,778	22,071


Chi-Sq = 12.339 + 12.343 + .166 + .166 = 25.014

With a df of 1, that chi-square is significant at the .0001 level--which means that at most standard levels of confidence (i.e., 95%, or .05), you can reject the null hypothesis and conclude that there is a relationship between taking aspirin and heart attacks.

V. Comments on Causation

Statistical significance does not necessarily mean that two variables are causally related--that is, that one variable causes the other. Generally, in order to draw conclusions about causation, you need a good theory, and often contnrols for other factors that might influence your dependent variable. So, for instance, in the case of left-handedness, it's not clear that being female "causes" left-handedness--just that there's an association between being female and being left handed. It may be that there's a third variable -- some sort of genetic predisposition, perhaps, or socialization -- that is associated with being female, and that causes "left handedness".

That said, the experiments on heart attacks and aspirin might tell us something about causation, for two reasons: first, medical researchers may have a theory about what it is about aspirin that would cause fewer heart attacks, and second, in an experiment, other possible factors influencing heart disease would presumably randomize out across these two very large, randomly selected groups. But the association between heart attacks and aspirin itself, in the absence of other information, doesn't tell us anything about causation, just correlation (or association, or a relationship).

V. Comments on Non-Significant Results (Very Important!!!)

When a result is not significant -- that is, the p-value is greaster than the conventionally used threshold of .05, and you therefore can't reject the null hypothesis of "no effect" or "no relationship"--what does that tell us?

It does not tell us that there's no association between the two variables.

It merely tells us that there isn't enough evidence to conclude that such a relationship exists in the population.

In other words, even if we found no significant association between left-handedness and gender, it might still be possible that one exists in the population-- we just don't have the evidence to draw that conclusion.

VI. Extensions

While all the examples that we've used focus on 2 X 2 tables, one can calculate a chi-square statistic for larger tables. The process is the same:

• First, calculate out expected frequencies by multiplying the total for that row by the total for that column, and divide by the total # of observations
• Then, calculate out -- for each cell -- (observed - expected)² / (expected)
• Then, add up those values across cells.
• Then (at a pre-determined confidence level, such as 95% (.05)), consult a chi-square table to see if the statistic is significant

So, for example: the following table presents data regarding three types of external occipital protuberance, based on gender.

Type	Women	Men	Totals
Type I	427	89	516
Type II	52	94	146
Type III	21	317	338
Totals	500	500	1,000

The following table represents the expected counts:

Type	Women	Men
Type I	258	258
Type II	73	73
Type III	169	169

And, then you can calculate out for each cell the value of (observed-expected)² / (expected). You can then sum up all of those values of (O-E)² / E across all cells of the table (in this case, 6 values would be added up), and perform a chi-square test. The degrees of freedom = (# columns - 1) * (# rows - 1).

It appears that gender is significantly associated with type of occipital protuberance.

VII. Questions

1. In a retrospective observational study, researchers asked women who were pregnant with planned pregnancies how long it took them to get pregnant. Length of time to pregnancy was measured according to the number of cycles between topping birth control and getting pregnant. Women were also categorized according to whether or not they smoked, with smoking defined as having at least one cigarette per day for at least the first cycle during which they were trying to get pregnant.

The observed counts are as follows:


Observed Results
First Cycle	Two or More Cycles	Total
Smoker	29	71	100
Non-Smoker	198	288	486
Total	227	359	586




• Calculate out the expected counts in a table.
• Calculate out the chi-square
• Is there an association between smoking and length of time for pregnancy (at the .05 level)?
  Click here for a chi-square table.
• If so, why can't we conclude that smoking causes a longer time for pregnancy?