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Probability Models & Frequency Data
Goodness of Fit Proportional Model Chi-square Statistic Example R Distribution Assumptions Poisson Distribution Example R 2
Goodness of Fit
Goodness of fit tests are used to compare any observed frequency distribution against an expected frequency distribution.
We previously did specialized examples of this for a probability distribution (the 50:50 expected right- vs. left-hand toad example) and binomial distribution (sperm genes on X chromosome of mice).
The binomial test we did is a specialized form for categorical variables with only two outcomes. Here we will introduce a more generalized form.
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Proportional Model
The proportional model is one of the simplest
probability model. The frequency of occurrence of
events is proportional to the number of
opportunities (e.g., X chromosome example).
What would we do, however, if we had multiple
proportions? A more generalized form of this test
is the chi-square (χ
2) goodness-of-fit-test
.
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Example 8.1: Under the proportional model, one would expect babies born in the U.S. to be born in equal proportions across the days of the week (i.e., 14.28% per day). Is this true? Shown are a random sample of 350 births from across the U.S. During the year 1999.
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63
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Goodness-of-Fit Test
The χ2 goodness-of-fit test use the chi-square statistic (based upon the chis-square distribution) to compare frequency data to a model stated by the null
hypothesis.
Continuing with our example:
H0: The probability of birth is the same every day of the week. HA: The probability of birth is not the same every day of the week.
Again, H0 and HA are statements about the population from which the sample is obtained.
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frequencies under the null model. In examining the calender for 1999, we see that there are not an even number of each day (52) in the year (there was an additional Friday), so we need to adjust for this.
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The calculation of the expected frequencies is straight forward.
Expected = 350 ⋅ (52/365) = 49.863 NB: the sum of the expected frequencies must sum to the total observed (350).
Once you have a full set of observed and expected frequencies, one can then determine a chi-square statistic and associated probability.
Goodness-of-Fit Test
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Chi-square Statistic
The chi-square statistic measures the discrepancy between between observed and expected
frequncies (make sure to always use the absolute frquencies [counts] not relative frequencies [proportions]).
Chi-square for each element can be calculated as:
2=
∑
Observed−Expected 2 Expected = 33−49.863 49.863 =5.70 9Chi-square Statistic
χΤηε 2 statistic is additive across all levels, so: χ2 = 5.70 + 1.58 + 3.46 + 3.46 + 0.16 + 0.53 + 0.16 = 15.05 We now have a calculated test statistic and as usual need to compare it to a table value at a particular degree of freedom to make our decision. In other words, is 15.05 large enough to be significantly different?
df = (number of categories) -1 = 7-1 = 6
From Statistical Table A in your text, we see that at df = 6, the critical value for χ2 is 12.59. Therefore, we reject the null hypothesis and conclude that there are unequal proportions of births among days.
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Chi-square Statistic
This type of problem can most easily be solved using a table format:
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Assuming equal probabilities this can be very easily done in R using chisq.test:
> births<-c(33,41,63,63,47,56,47) > chisq.test(births)
Chi-squared test for given probabilities data: births
X-squared = 15.24, df = 6, p-value = 0.01847 How can we do this with the unequal probabilities that we have? This is a bit more complicated, but still straightforward:
Chi-square Statistic
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> obsbirths<-births > days<-c(52,52,52,52,52,53,52) > expbirths<-350*(days/365) > expbirths [1] 49.86301 49.86301 49.86301 49.86301 49.86301 50.82192 [7] 49.86301 > chi<-sum((obsbirths-expbirths)^2/expbirths) > chi [1] 15.05676 > ?pchisq > pchisq(chi,df=6) [1] 0.9801802 > pchisq(chi,df=6,lower.tail=FALSE) [1] 0.01981982 What's going on here?13
Chi-square Distribution
The chi-square distribution is a theoretical probability distribution (analogous to normal, binomial, poisson, etc.).
Note that the distribution is not symmetrical and is highly skewed.
When df = 1 then asymptotic to both axes!
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If χ2 is a random variable with a chi-square distribution:
χ2 is a positive real number
The density function depends only on n (df) The expected value of χ2 = n
The variance of χ2 = 2 n
The graph of f (χ2) is not symmetrical
The graph of f (χ2) approaches symmetry as ν ∞=
Chi-square Distribution
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16 We can explore the properties of the chi-square distribution through the use of R functions and graphics:
> par(mfrow=c(2,2),mar=c(3,4,3,3)) > layout.show(4) > plot(dchisq(1,df=1:30)) > plot(dchisq(5,df=1:30)) > plot(dchisq(10,df=1:30)) > plot(dchisq(15,df=1:30)) 17 18
Chi-square Assumptions
The sampling distribution of the chi-square
statistic only approximately follows the chi-square
distribution (but pretty closely).
Two assumptions apply:
1) None of the categories should have an
expected frequency less than one.
2) No more than 25% of the categories should
have expected frequencies less than five.
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Goodness-of-Fit Test
Two Proportions
-The chi-square goodness of fit test is a very general one and can be used in a variety of situations.
It can also be used when there are only two proportions, a replacement for the binomial test, but at a cost...it is much less powerful in this situation. So, use the binomial test whenever appropriate.
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Poisson Distribution
The poisson distribution describes the number of successes in blocks of time or space, when successes happen independently of each other and occur with equal probability at every point in time or space.
The poisson is often useful in biological studies because it is a starting place for evaluating whether or not an observed pattern is random or not.
If the null model is rejected, the distribution may be either clumped or dispersed.
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Poisson Distribution
A clumped distribution arises when the presence of one success is increases the probability of success for adjacent observations (e.g., occurrences of a contagious disease). A dispersed distribution is the opposite: the presence of one success decreases the probability of success for adjacent observations (e.g., animals with well defended territories).
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Poisson Distribution
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Poisson Distribution
The poisson distribution is constructed using the probability of X successes occurring in any given block of time or space:
Pr[ X successes]=e
−x
X !
Where mu is the mean number of independent successes in time or space (expressed as a unit count) and e is the base of the natural log.
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Poisson Distribution
Example
-Example 8.6 provides the example of an assessment of the fossil record. They ask, do extinctions occur randomly through the fossil record or are their periods where extinction rates are unusually high (mass extinctions) compared to background rates? Fossil marine invertebrates are an ideal taxa to test this question as they
preserve well. The data are the number of recorded extinctions in 76 contiguous blocks of time.
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The hypotheses are:
H0: The number of extinctions per time interval has a Poisson distribution. HA: The number of extinctions per time interval does not have a P distr.
We need to begin by estimating μ, the mean number of extinctions per time interval. As usual, μ, can be estimated by x-bar (= 4.21, n = 76).
We need to use the same protocol and generate expected values to compare to our observed values, so return to the formula for calculation of the poisson distribution.
Poisson Distribution
Example 27Poisson Distribution
Example -Pr[3 extinctions]=e −4.21 4.213 3! For example, for 3 extinctions:Expected[3 extinctions] = 76 x 0.1846 = 14.03
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Poisson Distribution
Example
-We now have a chi-square test statistic calculated. We need to determine the degrees of freedom. In the broadest sense, df normally is n – 1. However, in a variety of circumstances, we need to also subtract the number of parameters being estimated from the data. So, df = 8-1-1=6.
The critical value for χ2 of 12.59 at P = 0.05 and df = 6 is 12.59. Thus, we reject the null hypothesis and conclude extinctions are non-random.
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> extinctions<-c(0,13,15,16,7,10,4,2,1,2,6) > ?dpois
> dpois(extinctions, 4.21)
[1] 1.484637e-02 3.111768e-04 2.626347e-05 6.910575e-06 [5] 6.905011e-02 7.156129e-03 1.943289e-01 1.315693e-01 [9] 6.250321e-02 1.315693e-01 1.148102e-01
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> extinctions2<-c(13,15,16,7,10,4,2,9) > chisq.test(extinctions2)
Chi-squared test for given probabilities data: extinctions2
X-squared = 18.7368, df = 7, p-value = 0.009053
Poisson Distribution
Example
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We can explore the properties of the chi-square distribution through the use of R functions and graphics:
