A chi square test is often applied to two-way tables, like the one below.
This table represents a sample of 1,322 individuals. Of these individuals, 687 are male, and 635 are female. Also 143 are union members, 159 are represented by unions, and 1,020 are not affiliated with a union.
You might use a chi-square test if you want to learn something about the relationship of gender and union status. The question then might come up: should you use a test of independence, or a test of homogeneity?
Does it matter? Software doesn’t generally differentiate between the two, which leads to a final question: are they even different?
Well, yes and no. Read on!
Different: Independence versus Homogeneity
Independence and homogeneity do refer to different ideas. If union status and gender are independent, that means that union status and gender are unrelated. In other words, if you know someone’s union status, you won’t be able to make a better guess as to their gender.
If you know someone’s gender, you won’t be able to make a better guess as to their union status.
Homogeneity is different and refers to the concept of similarity. If you are familiar with linear regression, you might associate this with residuals. Residuals should be homogeneous, meaning they all come from the same distribution.
That idea applies to this two-way table as well. We may want to know if the distribution of union status is the same for men and women. In other words, does union status come from the same distribution for both men and women?
To test independence, we would not approach the question from the standpoint of gender or union status. We would take a sample of all employed individuals, and then break them down into the categories in the table.
To test homogeneity, we would approach it from the standpoint of gender. We would randomly sample individuals from within each gender, and then measure their union status.
Either approach would result in the table above.
Same: Chi-Square Statistics
Chi-square statistics for categorical data generally follow this formula:
For each of the six cells representing a combination of gender and union status, the number in the cell is the count we observe. “Expected” refers to what we would see in each cell under the null hypothesis. That means if gender and union status are independent (or if union status is homogeneous across the genders).
We calculate the difference, square it, and divide by the expected count for each cell. We then add these all together, and that is the chi-square test statistic.
Where do we get the expected counts for each cell?
Let’s examine the combination of male and union member under independence. If gender and union membership are independent, then how many male union members do we expect? Well,
– 10.81% of the sample are union members
– 51.96% are male
So, if they are independent, 10.81% x 51.96% is 5.62%, and 5.62% of 1,322 is 74.3. This is how many individuals we would expect to be male union members.
Now let’s consider male union members under homogeneity. Overall, 10.81% of the sample are union members. If this is the same for both males and females, then of the 687 males, we expect 74.3 to be union members.
Independence and homogeneity result in the same expected number of union members! It turns out this calculation is the same for every cell in the table. It follows that the chi-square statistic is also the same.
Does It Matter?
As it turns out, independence and homogeneity are two sides of the same coin. If gender and union status are independent, then union status is distributed the same way for males and females.
So which test should you say you are using, if they turn out the same?
Again, that comes back to how you have phrased your research question. Are you determining whether gender and union status are related. That is a test of independence. Are you looking for differences between males and females? That is a test of homogeneity.
If you analyze non-experimental data, is it helpful to understand experimental design principles?
Yes, absolutely! Understanding experimental design can help you recognize the questions you can and can’t answer with the data. It will also help you identify possible sources of bias that can lead to undesirable results. Finally, it will help you provide recommendations to make future studies more efficient. (more…)
Have you ever wondered why there are so many different types of experimental designs, and how a researcher would go about choosing among them to best address their research questions? (more…)
When learning about linear models —that is, regression, ANOVA, and similar techniques—we are taught to calculate an R2. The R2 has the following useful properties:
The range is limited to [0,1], so we can easily judge how relatively large it is.
It is standardized, meaning its value does not depend on the scale of the variables involved in the analysis.
The interpretation is pretty clear: It is the proportion of variability in the outcome that can be explained by the independent variables in the model.
The calculation of the R2 is also intuitive, once you understand the concepts of variance and prediction. (more…)
What are the best methods for checking a generalized linear mixed model (GLMM) for proper fit?
This question comes up frequently.
Unfortunately, it isn’t as straightforward as it is for a general linear model.
In linear models the requirements are easy to outline: linear in the parameters, normally distributed and independent residuals, and homogeneity of variance (that is, similar variance at all values of all predictors).
Generalized linear models—and generalized linear mixed models—are called generalized linear because they connect a model’s outcome to its predictors in a linear way. The function used to make this connection is called a link function. Link functions sounds like an exotic term, but they’re actually much simpler than they sound.
For example, Poisson regression (commonly used for outcomes that are counts) makes use of a natural log link function as follows:
Clearly, there is not a direct linear relationship of the x variables to the average count, but there is a “sort of linear” relationship happening: a function of the mean of y is related to a linear combination of x variables. In other words, the linear model has now been generalized to a bigger type of situation.
This can lead to confusion, though, because on the surface it looks very similar to what happens when we transform the dependent variable in a linear model, like a linear regression.
The key thing to understand is that the natural log link function is a function of the mean of y, not the y values themselves.
Transformations of Y
Below is a linear model equation where the original dependent variable, y, has been natural log transformed. That is, the natural log has been taken of each individual value of y, and that is being used as the dependent variable.
The linear model with the log transformation is providing an equation for an individual value of ln(y). We could also write it as follows, where we are modeling the mean of ln(y) (note the error term is no longer present):
This makes the difference a bit clearer. When we transform the data in a linear model, we are no longer claiming that y is normally distributed around a mean, given the x values — we are claiming that our new outcome variable, ln(yi), is normally distributed.
In fact, we often make this transformation specifically because the values of y do not appear to be normally distributed around their average.
In the case of the Poisson model, however, the link function does not change the distribution of the actual observations in some way to make them something other than Poisson distributed. Instead, the link function defines the relationship of the x variables directly to the mean of the Poisson distributed y. The individual observations then vary around this expected value accordingly.
The mean of the log is not the log of the mean
As you may know, if you have used this kind of data transformation in a linear model before, you cannot simply take the exponent of the mean of ln(y) to get the mean of y.
You might be surprised to know, though, that you can do this with a link function. If you have specific values of your x variables, you can calculate the predicted average count, μy based on those x values by inversing the natural log:
This ability to back-transform means (and regression coefficients) to a more intuitive scale is part of what makes generalized linear models so useful.
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