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Association tests


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Association Tests

The statistical tests we have covered up to now leave out a substantial portion of life science projects. Specifically, we are referring to data that is binary, categorical and ordinal. To give a very specific example, consider genetic data where you have two groups of genotypes (AA/Aa or aa) for cases and controls for a given disease. The statistical question is if genotype and disease are associated. As in the examples we have been studying previously, we have two populations (AA/Aa and aa) and then numeric data for each, where disease status can be coded as 0 or 1. So why can’t we perform a t-test? Note that the data is either 0 (control) or 1 (cases). It is pretty clear that this data is not normally distributed so the t-distribution approximation is certainly out of the question. We could use CLT if the sample size is large enough; otherwise, we can use association tests.

Lady Tasting Tea

One of the most famous examples of hypothesis testing was performed by R.A. Fisher. An acquaintance of Fisher’s claimed that she could tell if milk was added before or after tea was poured. Fisher gave her four pairs of cups of tea: one with milk poured first, the other after. The order was randomized. Say she picked 3 out of 4 correctly, do we believe she has a special ability? Hypothesis testing helps answer this question by quantifying what happens by chance. This example is called the “Lady Tasting Tea” experiment (and, as it turns out, Fisher’s friend was a scientist herself, Muriel Bristol).

The basic question we ask is: if the tester is actually guessing, what are the chances that she gets 3 or more correct? Just as we have done before, we can compute a probability under the null hypothesis that she is guessing 4 of each. If we assume this null hypothesis, we can think of this particular example as picking 4 balls out of an urn with 4 green (correct answer) and 4 red (incorrect answer) balls.

Under the null hypothesis that she is simply guessing, each ball has the same chance of being picked. We can then use combinatorics to figure out each probability. The probability of picking 3 is ${4 \choose 3} {4 \choose 1} / {8 \choose 4} = 16/70$. The probability of picking all 4 correct is ${4 \choose 4} {4 \choose 0}/{8 \choose 4}= 1/70$. Thus, the chance of observing a 3 or something more extreme, under the null hypothesis, is $\approx 0.24$. This is the p-value. The procedure that produced this p-value is called Fisher’s exact test and it uses the hypergeometric distribution.

Two By Two Tables

The data from the experiment above can be summarized by a two by two table:

tab <- matrix(c(3,1,1,3),2,2)
rownames(tab)<-c("Poured Before","Poured After")
colnames(tab)<-c("Guessed before","Guessed after")
              Guessed before Guessed after
Poured Before              3             1
Poured After               1             3

The function fisher.test performs the calculations above and can be obtained like this:


	Fisher's Exact Test for Count Data

data:  tab
p-value = 0.2429
alternative hypothesis: true odds ratio is greater than 1
95 percent confidence interval:
 0.3135693       Inf
sample estimates:
odds ratio 

Chi-square Test

Genome-wide association studies (GWAS) have become ubiquitous in biology. One of the main statistical summaries used in these studies are Manhattan plots. The y-axis of a Manhattan plot typically represents the negative of log (base 10) of the p-values obtained for association tests applied at millions of single nucleotide polymorphisms (SNP). The x-axis is typically organized by chromosome (chromosome 1 to 22, X, Y, etc.). These p-values are obtained in a similar way to the test performed on the tea taster. However, in that example the number of green and red balls is experimentally fixed and the number of answers given for each category is also fixed. Another way to say this is that the sum of the rows and the sum of the columns are fixed. This defines constraints on the possible ways we can fill the two by two table and also permits us to use the hypergeometric distribution. In general, this is not the case. Nonetheless, there is another approach, the Chi-squared test, which is described below.

Imagine we have 250 individuals, where some of them have a given disease and the rest do not. We observe that 20% of the individuals that are homozygous for the minor allele (aa) have the disease compared to 10% of the rest. Would we see this again if we picked another 250 individuals?

Let’s create a dataset with these percentages:

dat <- data.frame(disease, genotype)
dat <- dat[sample(nrow(dat)),] #shuffle them up
    disease genotype
249   cases       aa
68  control    AA/Aa
167 control    AA/Aa
129 control    AA/Aa
162 control    AA/Aa
215 control       aa

To create the appropriate two by two table, we will use the function table. This function tabulates the frequency of each level in a factor. For example:

AA/Aa    aa 
  200    50 
control   cases 
    220      30 

If you provide the function with two factors, it will tabulate all possible pairs and thus create the two by two table:

tab <- table(genotype,disease)
genotype control cases
   AA/Aa     180    20
   aa         40    10

Note that you can feed table $n$ factors and it will tabulate all $n$-tables.

The typical statistics we use to summarize these results is the odds ratio (OR). We compute the odds of having the disease if you are an “aa”: 10/40, the odds of having the disease if you are an “AA/Aa”: 20/180, and take the ratio: $(10/40) / (20/180)$

(tab[2,2]/tab[2,1]) / (tab[1,2]/tab[1,1])
[1] 2.25

To compute a p-value, we don’t use the OR directly. We instead assume that there is no association between genotype and disease, and then compute what we expect to see in each cell of the table (note: this use of the word “cell” refers to elements in a matrix or table and has nothing to do with biological cells). Under the null hypothesis, the group with 200 individuals and the group with 50 individuals were each randomly assigned the disease with the same probability. If this is the case, then the probability of disease is:

[1] 0.12

The expected table is therefore:

expected <- rbind(c(1-p,p)*sum(genotype=="AA/Aa"),
genotype control cases
   AA/Aa     176    24
   aa         44     6

The Chi-square test uses an asymptotic result (similar to the CLT) related to the sums of independent binary outcomes. Using this approximation, we can compute the probability of seeing a deviation from the expected table as big as the one we saw. The p-value for this table is:

[1] 0.08857435

Large Samples, Small p-values

As mentioned earlier, reporting only p-values is not an appropriate way to report the results of your experiment. Many genetic association studies seem to overemphasize p-values. They have large sample sizes and report impressively small p-values. Yet when one looks closely at the results, we realize odds ratios are quite modest: barely bigger than 1. In this case the difference of having genotype AA/Aa or aa might not change an individual’s risk for a disease in an amount which is practically significant, in that one might not change one’s behavior based on the small increase in risk.

There is not a one-to-one relationship between the odds ratio and the p-value. To demonstrate, we recalculate the p-value keeping all the proportions identical, but increasing the sample size by 10, which reduces the p-value substantially (as we saw with the t-test under the alternative hypothesis):

[1] 1.219624e-09

Confidence Intervals for the Odds Ratio

Computing confidence intervals for the OR is not mathematically straightforward. Unlike other statistics, for which we can derive useful approximations of their distributions, the OR is not only a ratio, but a ratio of ratios. Therefore, there is no simple way of using, for example, the CLT.

One approach is to use the theory of generalized linear models which provides estimates of the log odds ratio, rather than the OR itself, that can be shown to be asymptotically normal. Here we provide R code without presenting the theoretical details (for further details please see a reference on generalized linear models such as Wikipedia or McCullagh and Nelder, 1989):

fit <- glm(disease~genotype,family="binomial",data=dat)
coeftab<- summary(fit)$coef
              Estimate Std. Error   z value     Pr(>|z|)
(Intercept) -2.1972246  0.2356828 -9.322803 1.133070e-20
genotypeaa   0.8109302  0.4249074  1.908487 5.632834e-02

The second row of the table shown above gives you the estimate and SE of the log odds ratio. Mathematical theory tells us that this estimate is approximately normally distributed. We can therefore form a confidence interval and then exponentiate to provide a confidence interval for the OR.

ci <- coeftab[2,1] + c(-2,2)*coeftab[2,2]
[1] 0.9618616 5.2632310

The confidence includes 1, which is consistent with the p-value being bigger than 0.05. Note that the p-value shown here is based on a different approximation to the one used by the Chi-square test, which is why they differ.

ExercisesWe showed how to calculate a Chi-square test from a table. Here we will show how to generate thetable from data which is in the form of a dataframe, so that you can then perform an associationtest to see if two columns have an enrichment (or depletion) of shared occurrences.Download thefile into your Rworking directory, and then read it into R:d = read.csv(“assoctest.csv”)1.This dataframe reflects the allele status (either AA/Aa or aa) and the case/control statusfor 72 individuals. Compute the Chi-square test for the association of genotype withcase/control status (using thetablefunction and thechisq.testfunction). Examine thetable to see if there appears to be an association. What is the X-squared statistic?2.Compute Fisher’s exact testfisher.testfor the same table. What is the p-value?

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