There are too many errors in the cited example. We don't know animal age upon arrival. We also don't know how long the young dolphins lived after the study was terminated. We also know that juvenile mortality will cause average mortality to be non-normally distributed. The sample size is too small to be meaningful in a larger context, and it is highly questionable that it has meaning for conditions at "the facility."

My guess was that you were dealing with a binomial variable and a small sample size (n=7). So t statistic at 0.975 would be 2.365. Multiply this by the square root of (1/7 * p * (1-p) where p=ASR. The 95% CI is roughly 0.88 to 1. ... which is not the answer given in the cited text. I tried using z=1.96 and I got a CI of 0.9 to 1 ... which is also not the given answer. I tried replacing ASR with DSR in the formulae, but when I raise the lower 95% CI for DSR to the 365.25th power, I get a lower bound for ASR of 0.136. I acknowledge that this approach assumes a large enough sample size that the central limit theorem holds -- and I agree with all those who criticize this assumption. It was the simplest method. I also tried thinking of this as a disease model where dead is used in place of "infected." I tried using the total sample days which was 30887, and the live animal days which was 27321, using a program called pooledinfrate, and no luck. I also tried it with 27321 live days, 1 death, and a total of 27322 days, and no luck.  

Hopefully my failures will assist others.

Jenny,

Just be aware that there is an enormous literature on age based life table analysis. It started with trying to understand human populations.

Kurt,

The use of animal days for n is flawed (and I know that you are just describing what was done, and don't necessarily agree with the method). I would suspect that the probability that animal 1 is alive tomorrow is not independent of the probability that it was alive on all previous days. What you have are seven sets of 5551 conditional probabilities (or fewer for the young animals). The described method uses an inflated n, and this results in confidence intervals that are too narrow.

This is a great opportunity to make a method confront reality. Use the random number generator in your computer (research a bit on flaws in computer random number generators). You can then create "animals" that are have a probability of dying each day. You can make hundreds or thousands of them. They can even "reproduce". You can then select five of these at random, and place them in a "facility." You can repeat this several thousand times to see how things work out. The advantage of this approach is that you know what the true answer must be, and you can compare it to the outcome of the random sample.

So to start, the DSR is 0.9999634. I can generate random numbers and count the number of times I have to call the random number generator before I get a value greater than 0.9999634. I used a uniform random number generator. I did this 56 times, and found that on average a DSR of 0.9999634 will result in an organism that lives 76 years. Is this reasonable? The maximum life span was 438 years.

You cannot blindly put numbers into formulae -- well, you can but you get meaningless junk out. ASR and related values have a maximum of exactly 1. If you are using a formula of -(1/ln(ASR)) then ASR must be less than 1 because the expression is undefined at 1. The 95% CI for life expectancy with these numbers is 25.269 years to immortal.

So lets say that all the animals at the facility are treated the same. They get roughly the same quantity of food, the water conditions are the same, and the vet care is the same. One might suspect that the living conditions for animal 1 are not independent of the living conditions for the other animals. We can ask what is the probability that an animal will die in any given year. There are 15 years in this study, but not all animals were present for all years. So in 1985 I have 5 animals.

(note: the condition of dead versus alive is akin (mathematically) to the condition of infected versus healthy because both are binary variables. So dead = infected. Testing for infection can be expensive, so often samples are pooled: you test 100 individuals, but each sample is a combined output from five individuals. This sacrifices the quality of your estimated infection rate for savings in the number of samples processed. In this case we will pool all animals in a given year in order to reduce the effect of the correlation between animals on our answer)

So this gives me one observation. The pool size is 5. The number of dead = 0, and the number of pools = 1. I find the equivalent values for the other years. The end result is that for this facility the expected yearly survival rate is 0.9877 with a 95% CI of 0.9411 to 0.9993. This is still not a great answer because the program PooledInfRate thinks that you have 81 animals when you really only have 7. PooledInfRate is at http://www.cdc.gov/westnile/resourcepages/mosqSurvSoft.html.

You could say that there were 7 animals at this facility and one died in the 15 year period of observation. This gives a survival rate of 0.857 with a 95% CI of 0.485 to 0.992. However, this assumes that all seven animals were present for all years, and that is also not true. So what you could do is argue that this is the pessimistic answer. The answer in the preceding paragraph is the optimistic answer. The truth lies somewhere between.

It might be useful to look at the infant mortality rate. It will be high, but is it different between different facilities? You could also look at health problems. This would avoid a practice of the facility's veterinarian finding that an animal's health is declining and moving it to a new facility before it has a chance to die. Of course these data have their own set of problems compounded by the over all problem of small sample sizes.