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Explanations
Poisson was a French mathematician, and amongst the many contributions he made, proposed the Poisson distribution, using the example of the number of soldiers accidentally injured or killed from kicks by horses. This distribution became useful as it models events, particularly uncommon events.
Javascript Program
Counts of events, based on the Poisson distribution, is a frequently encountered model in medical research. Examples of this are number of falls, asthma attacks, number of cells, and so on. The Poisson parameter Lambda (λ) is the total number of events (k) divided by the number of observation units (n) in the data (λ = k/n). The unit forms the basis or denominator for calculation of the average, and need not be individual cases or research subjects. For example, the number of asthma attacks may be based on the number of child months, or the number of pregnancies based on the number of women years in using a particular contraceptive. Poisson is different to the Binomial parameter of proportion or risk where proportion is the number of individuals classified as positive (p) divided by the total number of individuals in the data (r = p/n). Proportion or risk must always be a number between 0 and 1, while λ may be any positive number. For examples, if we have 100 people, and only 90 of them go shopping in a week then the binomial risk of shopping is 90/100 = 0.9. However, some of the people will go shopping more than once in the week, and the total number of shopping trips between the 100 people may be 160, and the Poisson Lambda is 160/100 = 1.6 per 100 person week Large Lambda (λ=k/n) values, say over 200, assumes an approximately normal or geometric distribution, and the count (or sqrt(count)) can be used as a Parametric measurement. If the events occur very few times per individual, so that individuals can be classified as positive or negative cases, then the Binomial distribution can be assumed and statistics related to proportions used. In between, or when events are infrequent, the Poisson distribution is used. Some clarification of nomenclature may be useful. - Counts of events (e.g. number of asthma attacks recorded) are represented by k. This count must be in terms of how many events over a defined period or environment (e.g. in 100 attacks in 300 children over 6 months, or 10 cells seen in 5 microlitres of fluid),
- The mean count, or count rate (k/n) is represented by λ. e.g. 100 attacks(k) in 1800 children months (n) produces λ=100/1800 = 0.06 attacks per child month (λ)
**Example 1:**Looking back, the number of patient complaints per month in a hospital ward was 2,5,4, and 3 in the last 4 months, averaged to 3.5 per month (λ). Since a new ward manager arrived the number of complaint last month was 6 (k). We want to know what was the probability that this count was no different to 3.5.Using the Poisson Test, we found that the probability of 6 or more complaints being the same as λ=3.5 is 0.14. If we use p< 0.05 as a decision criteria, then we cannot confidently conclude that 6 is greater than the mean of 3.5. **Example 2:**Following the events in Example 1, in the same ward, the total number of complains over the following 6 months was 30 (averaged to 5 per month). We now want to know whether this should be considered as higher than the previous average of 3.5.If the mean (λ) is 3.5 per month, then over 6 months, the number of complaints expected is 6 x 3.5 = 21. Using the Poisson Test, the probability of having 30 or more complaints when the expected was 21 is 0.0374. If we use p<0.05 as the decision criteria, then we can conclude that 30 complaints over 6 months is significantly greater than the expected average of 3.5 per month. **Example 3:**We wish to establish some standards for detecting blood in the urine by microscopic examination. If the urine is normal, we expect, on average, 3 or less red blood cells per high power field under the microscope. We wish to establish the maximum total red cell count over 5 high power fields acceptable for normality.If the average (λ) is 3/hpf, then the expected total count over 5 fields is 15. Using a sequence of Poisson Probability Tests, the probabilities and cumulative probabilities of observing counts from 0 to 25 are shown in the following table. Expected λ Observed k Probability <=k Probability >=k 15 0 <0.0001 >0.9999 15 1 <0.0001 >0.9999 15 2 <0.0001 >0.9999 15 3 0.0002 >0.9999 15 4 0.0009 0.9998 15 5 0.0028 0.9991 15 6 0.0076 0.9972 15 7 0.018 0.9924 15 8 0.0374 0.982 15 9 0.0699 0.9626 15 10 0.1185 0.9301 15 11 0.1848 0.8815 15 12 0.2676 0.8152 15 13 0.3632 0.7324 15 14 0.4657 0.6368 15 15 0.5681 0.5343 15 16 0.6641 0.4319 15 17 0.7489 0.3359 15 18 0.8195 0.2511 15 19 0.8752 0.1805 15 20 0.917 0.1248 15 21 0.9469 0.083 15 22 0.9673 **0.0531**15 23 0.9805 **0.0327**15 24 0.9888 0.0195 15 25 0.9938 0.0112 From column 4, the probability of observing 22 or more cells is 0.0531, and 23 or more cells 0.0327. If we use p<0.05 as unlikely, we can conclude that we are unlikely to see a total of 23 or more cells in 5 high power fields if we expect an average of 3 per high power field. We can therefore use observing a total of 23 or more red blood cells in 5 high power fields as the diagnostic criteria for detecting blood in the urine.
## ReferencesPoisson Distribution by WikipediaSteel RGD, Torrie JH Dickey DA (1997) Principles and Procedures of Statistics. A Biomedical Approach. The McGraw-Hill Companies, Inc New York. p. 558
R Codes for Poisson Test
The probabilities of obstaining a count >=k or <=k, given the average count (lambda)
# function PoissonP <- function(lambda,k) # lambda is expected count, k is observed count { return (exp(log(exp(-lambda)) + log(lambda) * k - log(factorial(k)))) } The main sequence of the program
myDat = (" Lambda K 2.2 4 6.7 5 10.9 15 ") df <- read.table(textConnection(myDat),header=TRUE) # conversion to data frame df # display input data (lambda and k) PLtEq <- vector() # probability of a count <= k PGtEq <- vector() # probability of a count >= k for(i in 1:nrow(df)) { c = 0 for(j in 0:df$K[i]) { p = PoissonP(df$Lambda[i],j) c = c + p } PLtEq = append(PLtEq, c) PGtEq = append(PGtEq, 1 - c + p) } df$PLtEq <- PLtEq df$PGtEq <- PGtEq df # display data frame including Probabilities of a count <= k and >=KThe results are > df # display input data (lambda and k) Lambda K 1 2.2 4 2 6.7 5 3 10.9 15 > df # display data frame including Probabilities of a count <= k and >=K Lambda K PLtEq PGtEq 1 2.2 4 0.9275037 0.1806476 2 6.7 5 0.3406494 0.7978410 3 10.9 15 0.9126344 0.1387804 |