In vjcitn/CSHstats: discussions and exercises on classical statistics for CSHL 2022

The "null" situation

Here are two variables that are independent. Let's think of x as an "exposure" indicator, and y as an "outcome".

set.seed(234)
x = rbinom(100,1,.3)
y = rbinom(100,1,.4)

The 2x2 table relating exposure and outcome is:

library(gmodels)
CrossTable(x,y)

Notice that the marginal frequencies (column and row proportions for 0 and 1) are very similar for x and y. This arises by construction, because the two variables here were simulated separately.

Exercise: What is the interpretation of the numbers 0.389 and 0.393 in this table?

A test for independence of x and y was produced by R.A. Fisher.

fisher.test(table(x,y))

Exercise: What is the interpretation of this p-value?

How to produce "correlated binomial" responses?

A very simple way of producing dependent binary responses is via a logistic regression relationship. We produce x, which can be random or deterministic, and then let the probability of y=1 depend on the value of x. R's vectorized computations make this very simple.

x = rbinom(100,1,.3)
y = rbinom(100,1, plogis(.2+x))
# summary(glm(y~x, fam=binomial))

Now the cross-tab is

CrossTable(x,y)

and Fisher's test is

fisher.test(table(x,y))

Exercise: What is the interpretation of the numbers 0.56 and 0.812 in this table? What is the interpretation of the Fisher's test p-value? What features of the simulation could be changed to make the p-value smaller?

Note: the odds ratio reported in the Fisher test is recovered (some discrepancy owing to rounding):

(.812/(1-.812))/(.56/(1-.56))

Connection to the hypergeometric distribution

We'll start with a very simple problem. An urn contains 10 balls, some red, some blue. We pick three balls without replacement and note their color. The table below shows that two balls were red, one blue.

color = rep(c("red", "blue"), each=5)
picked = c(1,1,0,0,0,1,0,0,0,0)
ta = table(color,picked)
ta

The hypergeometric distribution is used to model such a situation, in which a draw of $n$ balls from an urn holding $N$ balls yields $r$ balls in a given condition. For this table, the probability of this draw is given by

$\frac{{5 \choose 2} \cdot {5 \choose 1}}{10 \choose 3}$. In this case, Fisher's test indicates no association between color and presence in the draw.

fisher.test(ta)

We can use a 1-sided test:

fisher.test(ta, alt="less")

Use the hypergeometric probability function to produce this p-value:

sum(dhyper(0:2,5,5,3))

Note that if the proportions are preserved but the frequencies are multiplied by 10, we have a different two-sided p-value, but similar odds ratio.

ta2 = 10*ta
fisher.test(ta2)

Exercise: A reformulation of the analysis of the larger 2x2 table is:

g = glm(formula = ta2[, 2:1] ~ factor(rownames(ta2)), family = binomial)
summary(g)

Show that the log of the cross product ratio of the elements of ta2 is given by the coefficient of 'red' in the binomial model above.

vjcitn/CSHstats documentation built on July 31, 2023, 2:31 p.m.