7.2.Goodness-of-fit tests

Poisson distribution

Bárbara D. Bitarello

2025-11-25

Outline

•Goodness-of-fit tests

•The proportional model

•The \(\chi^2\) distribution

•Degrees of freedom

•The \(\chi^2\) test

•Using \(\chi^2\) to test if data are distributed according to a Poisson distribution

Recap: Goodness of Fit Tests

Does data come from a given distribution with specified parameters?

Using the \(\chi^2\) statistic to test if data are Poisson distributed !

Why Poisson? (1/2)

We often want to know if events independent in time/space. E.g.:

• Are some people lucky? ☘️
  
• Do baseball players go on streaks? ⚾
  
• Do cats stick together, or avoid one another? 🐯

Perhaps we can use real data and contrast it to expectations under a null model and test. What might that distribution look like?

Why Poisson? (2/2)

Assuming events are independent the Poisson distribution, the Poisson describes the expected probability of \(X\) of events (successes) in a block of time or space.

Poisson experiments (1/3)

Random Events in Time or Space

• The number of successes in the experiment can be counted (discrete).

• The mean number of events (successes) that occurs during a specific interval of time (or space) is known.

• Each outcome is independent.

• The probability that a success will occur is proportional to the size of the interval.

Poisson experiments (2/3)

Random Events in Time or Space

Example: Number of births per hour in a given hospital

• The number of successes in the experiment can be counted (discrete).

• The mean number of successes that occurs during a specific interval of time (or space) is known.

• Each outcome is independent.

• The probability that a success will occur is proportional to the size of the interval.

The Poisson distribution (1/)

Formally: a mathematical description of the probability of \(X\) successful outcomes when:

1. the number of attempts, \(n\), is high

2. the probability of success for each attempt, \(p\), is low and determined through a random process

3. the probability of each attempt, \(p\), is independent of prior success or failure

Remember: a discrete probability distribution (can only take values in the range 0,1,2,…

The Poisson distribution (2/)

The Poisson distribution describes the expected probability of \(X\) independent events in time/space.

The Poisson distribution (4/)

The Poisson distribution describes the expected probability of \(X\) independent events in time/space.

The Poisson Equation

Assuming events are random and independent, the probability of observing \(X\) events in a block of time or space equals:

\[\Huge{Pr[X] =\frac{e^{-\lambda} \lambda^{X}}{X!}}\]

• \(X\): Number of successes

• \(P[X]\): probability of occurrence of \(X\) successes in one trial

• \(\lambda\) (or \(\mu\)): the expected (mean) number of events in a block.

• \(e\) is the base of \(ln()\), aka Euler’s number (a constant, exp(1) in R).

Note: Many people use \(\mu\) in place of \(\lambda\)

Poisson: Fish Example (1/) 🐟

Parasites are a major force in human health, as well as evolution, ecology, and agronomy, animal husbandry.

Shaw et al. asked if the distribution of parasites or across individual Shad fish was random, or if some have an exceptional parasite burden? Here are their data:

num_parasites	num_fish
0	103
1	72
2	44
3	14
4	3
5	1
6	1
7	0
8	0
9	0

Shaw et al. 1998. Parasitology 117: 567 - 610.

Poisson: Fish Example (2/) 🐟

\(H_0\): Parasites are placed randomly on fish in a population i.e. parasite numbers follow a Poisson Distribution.

Poisson: Fish Example (3/) 🐟

\(H_A\): Parasites are not placed randomly on fish in a population i.e. parasite numbers do not follow a Poisson Distribution.

Steps in Hypothesis Testing

• State hypotheses

• Find expectations (via simulation or via model)

num_parasites	num_fish
0	103
1	72
2	44
3	14
4	3
5	1
6	1
7	0
8	0
9	0

total_fish	total_parasites
238	225

Option 1: A Lousy Simulation (1/) 🐟

We can put 225 on 238 fish many times to generate a null.

1. Take 225 parasites and:

   randomly place one parasite in one of 238 fish

   repeat for each parasite

   the same fish can be the target more than once

2. At the end, count the number of parasites in each fish

3. Repeat this many, many times

Option 1: A Lousy Simulation (2/) 🐟

Note: the observed data does not include X > 6 but they are possible and are thus shown here.

Option 2: Derived by formula 🐟

\(P[X]=\frac{e^{-\lambda}\lambda^X}{X!}\)

\(e=2.718282\)

\(\lambda = \frac{tot.paras}{tot.fish}=225/238\)

Let’s try the first row:

\(X=0\)

\(P[0]=\frac{e^{-0.945}\times 0.945^0}{0!}\approx 0.388\)

\(\text{\# fish expected}=0.388\times 238=92.5\)

Repeat for each row…

Option 2: Derived by formula 🐟

num_parasites	num_fish	expect
0	103	92.471
1	72	87.420
2	44	41.322
3	14	13.022
4	3	3.078
5	1	0.582
6	1	0.092
7	0	0.012
8	0	0.001
9	0	0.000

total_fish	total_parasites
238	225

Proceed with testing?

• Find degrees of freedom
  df = # categories - 1 - # params estimated

Do we meet \(\chi^2\) assumptions?

• No expected values \(< 1\) &
  
• No more than 20% of expected values \(< 5\).

If yes, go on. If no, brainstorm.

num_parasites	num_fish	expect
0	103	92.471
1	72	87.420
2	44	41.322
3	14	13.022
4	3	3.078
5	1	0.582
6	1	0.092
7	0	0.012
8	0	0.001
9	0	0.000

total_fish	total_parasites
238	225

We do not meet assumptions, now what?

Options:

• Go on despite this: if test is robust to violations. (I don’t support this option!)

• Combine categories: if biologically sensible.

• Find a more appropriate test.

• Bypass traditional tests by using simulations (which we’ve done)

We can combine 4+ to meet assumptions

num_parasites	num_fish	tot.fish	tot.parasite	mu	expect	sq_dev
0	103	238	225	0.945	92.47	1.20
1	72	238	225	0.945	87.42	2.72
2	44	238	225	0.945	41.32	0.17
3	14	238	225	0.945	13.02	0.07
4+	5	238	225	0.945	3.77	0.40

\(\chi^ 2= 1.20 + 2.72 + 0.17 + 0.07 + 0.40 = 4.56\)

Hypothesis test

df = # categories - 1 - # params estimated = 5 - 1 - 1 = 3

new.expect |>
    summarise(chi2 = sum(sq_dev), p_value = pchisq(q = chi2, df = 3, lower.tail = FALSE))
## # A tibble: 1 × 2
##    chi2 p_value
##   <dbl>   <dbl>
## 1  4.56   0.207

Old fashioned moment

\(df=3; \chi^2 = 4.56\)

df | a	0.1	0.05	10^-2	10^-3	10^-4	10^-5	10^-6	10^-7
1	2.71	3.84	6.63	10.8	15.1	19.5	23.9	28.4
2	4.61	5.99	9.21	13.8	18.4	23.0	27.6	32.2
3	6.25	7.81	11.34	16.3	21.1	25.9	30.7	35.4

Assuming \(\alpha=0.05\), we fail to reject the NULL hypothesis (\(P>\alpha\)). We cannot exclude the idea that parasites are distributed at random across fish.

Are counts over or underdispersed

Over or underdispersed

• A particular feature of the Poisson distribution is that the variance equals the mean.

• Sample counts therefore vary more as the mean increases.

If the variance greatly exceeds the mean events are clumped

If the variance is much less than the mean events are dispersed.

Goodness of fit: Key points (1/)

• A test of \(\chi^2\) goodness of fit compares the frequency distribution of a discrete or categorical variable with the frequencies expected from a probability model.
• The test can be applied to many distributions
• More general than the binomial test because it can handle more than two categories.
• Goodness of fit is measured with the \(\chi^2\) test statistic. Its theoretical distribution is continuous. Probability is measured by the area under the curve.
• The \(\chi^2\) distribution predicts the deviation of count data from expectations under \(H_0\).

Goodness of fit: Key points (2/)

• Proportional probability model: events fall in different categories in proportion to the number of opportunities. • The Poisson distribution model: describes the frequency distribution of successes in blocks of time or space when successes happen independently and with equal probability over time or space.