Simple measures of mortality
Before we look at examples of cohort and static life tables, we will first give a worked example showing how the different measures of mortality are calculated.
We will use a similar example to that used when looking at measures of disease frequency, except that now we have a number of deaths occurring as a result of the infections. Eight pigs are observed over twelve weeks.
| 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 ||10||11||12
|1|| || || || || || ||infected||died||9
|2|| || || || || || || || || || || || ||12
|4|| || || || || || || || || || || || ||12
|5|| || || ||infected||12
|6|| || || || || || || || || || || || ||12
|7|| || || || || || || || || || || || ||12
|8|| || || || ||infected||died||7
|Total time at risk:||80
- Number of deaths
There were 3 deaths over the 12 week period.
The cumulative mortality for the 12 week period was 3/8 which is 0.375.
There are 4 cases of which 3 died, so the case fatality was 3/4 = 0.75. But note that the animal that was infected, but remained alive, may have died shortly afterwards. The case fatality is very dependent on the duration of observations.
Using the exact method, we add up the number of weeks at risk to give a total of 80 pig weeks. The mortality rate is then given by 3/80 which is 0.037 per pig week.
Using the approximate method, we estimate the average number at risk over the period by adding the number at risk at the start and end of the period, and dividing by two. In this case we have 8 present at the start, 5 at the end, so the average number present is 6.5. The crude mortality rate is then given by 3/(6.5×12) = 0.038 per pig week.
We can also estimate the instantaneous mortality rate per week from the cumulative mortality (=1 - cumulative survival) over 12 weeks as [- ln (1 - cumulative survival)]/12 = [-ln(.625)]/12 = 0.039.
Standard cohort life table
Cohort life table with censoring
There is one obvious way you can make the interval shorter, and also make more efficient use of the data. That is to deal with each event/death individually and use a variable interval length set by the time between individual deaths. This is known as the Kaplan-Meier (or product-limit) approach.
In this approach the event times themselves define the length of the intervals at which the cumulative survival probability (S) is calculated. This ensures that the interval times used are the shortest possible and we use all the available data (usually a good idea!).
Kaplan-Meier life table
Worked example of a fertility table
For a fertility table we start the table with a column giving the proportion of females surviving to the mid-point of the interval. The next column gives the number of female offspring new column added, the mi column, which details the age-specific fertility. We then multiply these two columns together to give the total number of females born in each age category. The sum of this column is the net reproductive 'rate' (Ro):
|Survivorship and fertility table
of females born
in each interval
Vi and time
of interval (x)
||Σ = 3.139 = Ro
||Σ = 81.845
The net reproductive rate is not a true rate, but a multiplication factor - in fact the multiplication factor per generation. It will equal the number of females in generation (n+1) divided by the number in generation n.
But how long is a generation?
The cohort generation time (Tc) is defined as the mean length of time elapsing between the birth of the parents, and the birth of the offspring. It is estimated by dividing the sum of the Vix column (81.45) by the net reproductive rate (3.139) to give 26.1 weeks.
In summary, the population should increase by a factor of 3.139 times every 26.1 weeks. We look at how the net reproductive rate is related to the population rate of increase, per unit time, in the Related Topic on population change.
Static life table
A static life table is derived from the age structure of a single sample of a population, at a particular time. The age structure is taken to reflect the fate of a cohort of animals born at time 0. As we have pointed out above, this is only true if there is a stable age distribution and the population is stationary. Providing these conditions are met, mortality rates can be calculated in the same way as with cohort life tables.
The data represent a random sample taken from an insect population. The age distribution of the sample is therefore assumed to be representative of that of the population.
|Static Life Table
||ni − ni+1
The mortality rate per day appears to be rather high initially (0.081/indiv/day), then varies between 0.0404 and 0.0667/indiv/day, before increasing in later life to 0.15/indiv/day.
Note that, because we do not to know the exact ages, the estimates of cumulative mortality and mortality rate are a little biased in this type of life table. Instead we only know the number within rather wide age ranges. However it seems this error of age classification does not affect the estimate too much - provided the mortality does not change too abruptly from one age class to the next.
In the example above, survival appears to vary between age groups. But this may just be a result of sampling error. In some insect populations it may be valid to assume that survival (or at least adult survival) is constant over the different age categories. In this case there are various ways to obtain a pooled estimate of the mortality rate. A weighted mean is one possibility - in this case giving an average mortality rate of 0.070. But another method is more widely used, especially in medical entomology.
Providing there is a stable age distribution, and the population is stationary, the age structure will reflect the fate of a cohort of animals born at time 0. Hence we can represent the total sample in the following way:
|Total in sample|| = ||No + NoS + NoS2+ NoS3....+ NoS∞
- No = the number in the youngest age group,
- S is the cumulative survival per age interval.
It is then straightforward to show that S is given by the proportion that age groups older than No make up of the population.
Using the above example, there are 80 individuals in the youngest age class out of a total of 165 individuals. Hence S is given by 80/165 = 0.485
The daily cumulative survival is the bth root of S, where b is the length of the age interval. This equals 0.930 in this example.
The daily instantaneous mortality rate = − ln (daily cumulative survival) = 0.072
The big problem with this approach is that a stable age distribution is very rarely achieved in natural populations, and a stationary population is even rarer! When a population is increasing, survivorship will be underestimated, and mortality rates will be overestimated. The opposite bias is produced when the population is decreasing.
It is sometimes argued that, despite the biases, a static life table is better than no life table, especially since they are relatively easy to construct. But this is not good advice. The methods given here should only be used when:
- A stable age distribution has been demonstrated (by examining several sequential age distributions).
- A stationary population has been demonstrated (by examining several sequential population estimates).
In fact it is now possible to correct an estimate, depending on whether the population is increasing or decreasing. There are also other methods which require neither a stable age distribution nor a stationary population. References for these methods are given below.