In ESunRoc/STT218: University of Rochester Statistics 218 Functions

knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)

library(URStat218)

All notes in the vignettes of this package are adapted from Dr. Joseph Ciminelli's fall 2019 section of Statistics 218. For a more complete version of the material herein contained, please reference the notes, attend the lectures, or ask Dr. Ciminelli himself.

Odds Ratio

Theory

We begin by taking a 2 $\times$ 2 contingency table on X and Y, where the rows are fixed:

|X (Explanatory)|Y(Response) | | | |:--: |:--: |:--: |:--: | | |1 |2 | Total | |1 | $n_{11}$ | $n_{12}$ | $n_{1+}$ | |2 | $n_{21}$ | $n_{22}$ | $n_{2+}$ |

For this 2x2 table, we have two independent binomial proportions, one for each level of X: $$n_{11} \sim \text{Binom}(n_{1+},p_{1|1})$$ $$n_{21} \sim \text{Binom}(n_{2+},p_{1|2})$$ Ultimately, we wish to test the significance of the difference in the two groups' proportions. If there is a significant difference, we conclude X and Y to be significantly associated, while the opposite leads to a correspondingly opposite conclusion.
As we are testing for independence, with unknown probabilities $p_{1|1}$ and $p_{1|2}$, we test against a null hypothesis of $H_0: p_{1|1} =p_{1|2}$; that is, that the proportions are independent and equal. Of the methods available to us, we note that R has built-in support for a difference-of-proportions test.
For any experiment, the success of a given trial is given simply as $\Omega=\frac{P(\text{success})}{P(\text{failure})}=\frac{P(\text{success})}{1-P(\text{success})}$. For any 2x2 table, we extend this to find the odds of a particular row. For row 1, we find the odds to be: $$\Omega_1=\frac{p_{1|1}}{p_{2|1}}=\frac{P(\text{success}|X=1)}{P(\text{failure}|X=1)}=\frac{P(\text{success}|X=1)}{1-P(\text{success}|X=1)}$$ And for row 2, we find them to be: $$\Omega_1=\frac{p_{1|2}}{p_{2|2}}=\frac{P(success|X=2)}{P(failure|X=2)}=\frac{P(success|X=2)}{1-P(success|X=2)}$$ We calculate the ratio of these odds (a quantity, expectedly, called the odds ratio) as: $$\theta=\frac{\Omega_1}{\Omega_2}=\frac{p_{1|1}/(1-p_{1|1})}{p_{1|2}/(1-p_{1|2})}=\frac{p_{1|1}/p_{2|1}}{p_{1|2}/p_{2|2}}.$$ Therefore, two categorical variables X and Y are independent if and only if $\theta=1$ (that is, $\Omega_1=\Omega_2$). $\theta$ will always be bound by 0 and $\infty$. In this range, if $\theta=1$, then we have independence between group membership and response (X and Y). If $\theta>0,$ there are higher odds favoring success in group 1 than in row 2 (category 1 is more likely to have a success). If $0<\theta<1,$ there are lower odds favoring success in row 1 than row 2 (category 1 is less likely to have a success). As $\theta$ increases past 1, the association between X and Y becomes stronger.
A luxury of the odds ratio is its ability to be written only in terms of joint probabilities: $$\theta=\frac{p_{11}/p_{12}}{p_{21}/p_{22}}=\frac{p_{11}\cdot p_{22}}{p_{12}\cdot p_{21}}$$ From this population odds ratio, we create our sample odds ratio simply as: $$\hat{\theta}=\frac{\hat{p}{11}\cdot \hat{p}{22}}{\hat{p}{12}\cdot \hat{p}{21}}=\frac{n_{11}n_{22}}{n_{12}n_{21}}$$ We note here that, like relative risk, there exists an improved sample estimator of the odds ratio achieved by adding $\frac{1}{2}$ to all cells. For the purposes of this class, our simpler estimator is sufficient.
Like the relative risk, it is easier for us to work on the log scale for inference purposes before back-transforming to our desired scale. When testing $H_0: \theta=1,$ we find the standard error of $\hat{\theta}$ to be: $$\hat{\sigma}((\hat{\theta}))=\sqrt{\frac{1}{n_{11}}+\frac{1}{n_{12}}+\frac{1}{n_{21}}+\frac{1}{n_{22}}}$$ We construct our interval, assuming large sample size for approximate normality, simply to be: $$(\hat{\theta})\pm z_{\alpha/2}\sqrt{\frac{1}{n_{11}}+\frac{1}{n_{12}}+\frac{1}{n_{21}}+\frac{1}{n_{22}}}$$ Exponentiating out the bounds of this interval yields the confidence interval at confidence level $1-\alpha$ of $$\left(e^{(\hat{\theta})-z_{\alpha/2}\sqrt{\frac{1}{n_{11}}+\frac{1}{n_{12}}+\frac{1}{n_{21}}+\frac{1}{n_{22}}}},e^{(\hat{\theta})+ z_{\alpha/2}\sqrt{\frac{1}{n_{11}}+\frac{1}{n_{12}}+\frac{1}{n_{21}}+\frac{1}{n_{22}}}}\right)$$ Again like relative risk, we reject $H_0$ and conclude association if 1 is within the bounds. Otherwise, we fail to reject $H_0,$ having found insufficient evidence for association.

Example

Consider the same table from the relative risk example. While the the idea behind the hypotheses remains the same, we reformulate them simply to be $H_0:\theta=1$ and $H_1:\theta\neq 1.$ $\hat{\theta}$ is found simply, using the formula above, to be $\theta=0.455,$ with its log being $(\hat{\theta})=-0.787$. Testing at the 95\% confidence level, we find our interval to be (0.389, 0.533). As 1 is not within this interval, we reject $H_0,$ having found statistically significant evidence to suggest that one's views on universal health care and their party affiliation are associated and we can be 95\% confident that this interval captures the true population odds ratio. Further, we are 95% confident that the odds a person is a Republican given they support universal health care are between 0.389 and 0.533 times those of that person being a Democrat.

In R

The odds functions identically to relrisk, save for the test statistic being calculated is $\hat{\theta}$ instead of r. This calculation is done using the formula above, or, in R: theta <- (n11*n22)/(n12*n21).

Example

Using the same example as above, and recalling the table created in the R example from relative risk, we simply apply the function:

tab <- matrix(c(491,740,813,558), nrow=2)
odds(tab)

As 1 is not within this interval, we reach the same conclusions as in the above examples and reject $H_0$.

ESunRoc/STT218 documentation built on Jan. 14, 2020, 2:39 a.m.