Testing association between exposure and disease

In a particular example from the book Epidemiologic Research by Kleinbaum [example 15.1], I have three problems.

Consider the data in table 01. These data pertain to a follow-up study concerning the possible association between obesity and mortality(from all causes) among white women aged 60–75 from a midwestern urban population.

$$\text{table 01. Density type of data from an Obesity Study}$$

$$\begin{array}{l|cc|l} & \text{Obese}(E) & \text{Nonobese}(\bar E) & \\ \hline \text{Deaths during 1960-1967}&30&36&m_1=66\\ \text{Person-Years of follow-up,1960-1967}&699&399&L=2098\\ \hline \end{array}$$

Here, $a=30 , b=36, L_1=699, L_0=1399$, so that incidence density ratio, $$\hat {IDR}=\frac{\hat {ID_1}}{\hat {ID_0}}=\frac{30/699}{36/1399}=1.67$$

Our hypothesis is:

$$H_0:IDR=1\quad\text{versus}\quad H_A:IDR>1$$

Since $\hat {ID_1}$ and $\hat {ID_0}$ are not binomial proportions, standard hypothesis testing methods cannot be used here.

• My first question is: why are $\hat {ID_1}$ and $\hat {ID_0}$ not binomial proportions ? (Though I don't know, why would $\hat {ID_1}$ and $\hat {ID_0}$ be binomial proportions?) If they aren't binomial proportions, why can't standard hypothesis testing methods be used here?

Returning to our particular example, the next step is described as:

One method for testing $H_0$ versus $H_A$ is to assume that each of the $m_1$ observed cases represents an independent Bernoulli trial, with "success" and "failure" defined as being exposed and unexposed categories, respectively. The probabilities of "success" and "failure" under $H_0$ are $$p_0=\frac{L_1}{L}\quad\text{and}\quad q_0=\frac{L_0}{L}$$ , respectively.

• My second question is: Why are the probabilities defined under null hypothesis? What would be happened if the probabilities are not defined under null hypothesis?

Again, Returning to our example, in the next step they have computed the $p$-value and interpreted based on these $p$-value.The procedure to find these $p$-value is first to calculate the $Z$ statistic as we have large sample

$$Z=\frac{(A-m_1p_0)}{\sqrt{m_1p_0q_0}}=\frac{30-(66)(.333)}{\sqrt{(66)(.333)(.667)}}=2.10$$ so that $$pr(A\ge 30 |H_0)=pr(Z>2.10|H0)=.0179$$

Since $p<0.05$, based on the sample we conclude that there is strong evidence of a real exposure-disease association.

• My last question is: If our conclusion is based on the $p$-value, where is the necessity to compute the estimate $\hat {IDR}=1.67$ at the very beginning of the example? What does the estimate $\hat {IDR}=1.67$ tell us ?

1. $ID_{1}$ and $ID_{0}$ aren't binomial proportions because they aren't proportions at all. ID stands for incidence density and is a rate. In a proportion, the numerator is a subset of the denominator, so the value is bounded between 0 and 1. Here, the denominator is person-time, and the numerator is people (deaths). Imagine if one person dies after 6 months of follow-up - if this is your whole study, you have ID = (1 death)/(0.5 person-years) = 2. So the rate is bounded by 0 and infinity. As for 'standard hypothesis testing', I think what the author means is that you cannot use the same approach as for the risk ratio (cumulative incidence ratio) or odds ratio - both of which are comparing proportions. However, you are actually using the standard approach of calculating a Z-score and obtaining a p-value. You just have to use a different formula for the IDR than for the other measures.