I want to test the association between 2 variables : - one is an ordered factor from quantiles of a numeric value ("Q1", "Q2", "Q3", "Q4") - the other is an unordered factor from groups of a character value ("group A", "group B", "group C").
If I make a usual chi square test, I get the association between those variables.
Is there a way I can't take into account the ordering of my first variable ? If yes, which test is appropriate ?
As illustrated by the R example below, given that the assumptions of the ordinal regression model are satisfied, you gain statistical power by testing for an effect of the factor group in an ordinal regression model (it's one line of code) as compared to testing for any kind of deviation from independence between your response (y below) and group via a chi-square test.
> # Simulated data
> n <- 30
> group <- factor(rep(c("A","B","C"),each=n))
> set.seed(1)
> z <- rnorm(3*n, mean=c(0,.5,1)[group],sd=1.5) # unobserved latent variable
> y <- rep(0,3*n)
> y[z>0] <- 1
> y[z>1] <- 2
> y[z>2] <- 3
> table(y,group)
group
y A B C
0 13 9 6
1 9 7 6
2 6 10 12
3 2 4 6
>
> # Fit the model
> y <- ordered(y)
> library(MASS)
> anova(polr(y ~ group), polr(y ~ 1))
Likelihood ratio tests of ordinal regression models
Response: y
Model Resid. df Resid. Dev Test Df LR stat. Pr(Chi)
1 1 87 241.1152
2 group 85 234.1001 1 vs 2 2 7.015068 0.02997073
>
> # Compare to a chi-square test
> chisq.test(table(y, group))
Pearson's Chi-squared test
data: table(y, group)
X-squared = 7.2792, df = 6, p-value = 0.2958
ordinal~groups to the model ordinal~1. But how does this take account of the order ? Also, how do you see the actual gain of power in your output ? And to be fully annoying (sorry), is there any way to get this concept using base R models (like lm, glm, aov etc) ?
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Commented
Feb 15, 2018 at 10:05
y as an ordered factor (with the ordered function). The increased power is only indicated by the $p$-values under the two approaches. To truly see the gain in power you would need to simulate a large number of data sets under $H_1$ and then estimate the probability of rejecting $H_0$ under each approach. But in general, you are more likely to reject $H_0$ if your alternative $H_1$ is more specific. In the above case, $H_1$ has $3+2=5$ parameters in the ordinal regression whereas $H_1$ under the chi-square test has $3\times 4 -1=11$ parameters.
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Commented
Feb 15, 2018 at 10:18