4.4  An example

Suppose the event of interest is the probability of females working full-time (W). This is coded (1) for those who are in full-time work and 0 if not working full-time (NW). For simplicity we postulate that whether a female is working depends solely on her age (A) and a continuous measure of health status (H), and that these variables are independent. The first step is to fit the logit model. Suppose this resulted in the following equation:

(8)

Case A:

Let the means of the independent variables be A = 50 and H = 50 so that evaluating Z at the means yields:

Z = 0.03 – 0.015(50) + 0.006(50 )= -0.42

so that Pr (W) = 1/(1+e^-z) = 1/(1+e^0.42 ) = 0.40

and Pr (NW) = 1 – 0.40 = 0.60 and

= or the odds of an average 50-year-old female working are 0.66 to 1.

Log odds = ln(0.66) = –0.42, where ln is the natural logarithm.

Case B:

Now consider the change when we allow the age to increase by one unit (a year in this case) from 50 to 51 while holding H at its original level of 50; ie, we are interested in estimating the effect of an increase in age on the probability of working holding constant a female's health score.

Z = 0.03 – 0.015(51) + 0.006(50) = -0.435

so that Pr (W) = 1/(1+e^-z ) = 1/(1+e^0.435 ) = 0.39

and Pr (NW) = 1 – 0.39 = 0.61 and

or the odds of an average 51-year-old female working are 0.65 to 1.

So that the odds ratio =

which we can verify as = e^-0.015 = 0.98.

This implies that a one year increase in age decreases the odds of working by a factor of 0.98, or similarly decreases odds by 2%. In Case A, the odds of a 50-year-old female working full-time were 0.66 to 1. For a 51-year-old female the odds are (0.66*0.98) to 1, or 0.65 to 1. Note this is not the same as the probability of a female working full-time which, for a one year increase in age, fell from 0.40 in Case A to 0.39 in Case B. The difference in the latter probabilities is the marginal effect which we discuss in the following section.

Case C:

In some instances we might be interested in a non-marginal change. Consider the case of a 65-year-old female, again isolating the effect of a change in age by holding the health status constant at its initial mean level. In this case the predicted value of Z is given by:

Z = 0.03 – 0.015(65) + 0.006(50) = -0.645

so that Pr (W) = 1/(1+e^-z ) = 1/(1+e^0.645) = 0.34

and Pr (NW) = 1 – 0.34 = 0.66 and

or the odds of an average 65-year-old female working are 0.52 to 1.

Log odds = ln(0.52) = –0.645,

and the odds ratio =

We can now compute the change in the log odds corresponding to a change in age from 50 to 65:

(Log Odds Case C) – (Log Odds Case A) = -0.645 – (-0.420) = -0.225.

Note that this result is simply 15 times the value of β₁; ie, 15 * -0.015 = -0.225.