How to resolve the algorithm Verify distribution uniformity/Chi-squared test step by step in the 11l programming language

Published on 12 May 2024 09:40 PM
#11l

How to resolve the algorithm Verify distribution uniformity/Chi-squared test step by step in the 11l programming language

Table of Contents

Problem Statement

Write a function to determine whether a given set of frequency counts could plausibly have come from a uniform distribution by using the

χ

2

{\displaystyle \chi ^{2}}

test with a significance level of 5%.
The function should return a boolean that is true if and only if the distribution is one that a uniform distribution (with appropriate number of degrees of freedom) may be expected to produce. Note: normally a two-tailed test would be used for this kind of problem.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Verify distribution uniformity/Chi-squared test step by step in the 11l programming language

Source code in the 11l programming language

V a = 12
V k1_factrl = 1.0
[Float] c
c.append(sqrt(2.0 * math:pi))
L(k) 1 .< a
   c.append(exp(a - k) * (a - k) ^ (k - 0.5) / k1_factrl)
   k1_factrl *= -k

F gamma_spounge(z)
   V accm = :c[0]
   L(k) 1 .< :a
      accm += :c[k] / (z + k)
   accm *= exp(-(z + :a)) * (z + :a) ^ (z + 0.5)
   R accm / z

F GammaInc_Q(a, x)
   V a1 = a - 1
   V a2 = a - 2
   F f0(t)
      R t ^ @a1 * exp(-t)

   F df0(t)
      R (@a1 - t) * t ^ @a2 * exp(-t)

   V y = a1
   L f0(y) * (x - y) > 2.0e-8 & y < x
      y += 0.3
   I y > x
      y = x

   V h = 3.0e-4
   V n = Int(y / h)
   h = y / n
   V hh = 0.5 * h
   V gamax = h * sum(((n - 1 .< -1).step(-1).map(j -> @h * j)).map(t -> @f0(t) + @hh * @df0(t)))

   R gamax / gamma_spounge(a)

F chi2UniformDistance(dataSet)
   V expected = sum(dataSet) * 1.0 / dataSet.len
   V cntrd = (dataSet.map(d -> d - @expected))
   R sum(cntrd.map(x -> x * x)) / expected

F chi2Probability(dof, distance)
   R 1.0 - GammaInc_Q(0.5 * dof, 0.5 * distance)

F chi2IsUniform(dataSet, significance)
   V dof = dataSet.len - 1
   V dist = chi2UniformDistance(dataSet)
   R chi2Probability(dof, dist) > significance

V dset1 = [199809, 200665, 199607, 200270, 199649]
V dset2 = [522573, 244456, 139979,  71531,  21461]

L(ds) (dset1, dset2)
   print(‘Data set: ’ds)
   V dof = ds.len - 1
   V distance = chi2UniformDistance(ds)
   print(‘dof: #. distance: #.4’.format(dof, distance), end' ‘ ’)
   V prob = chi2Probability(dof, distance)
   print(‘probability: #.4’.format(prob), end' ‘ ’)
   print(‘uniform?  ’(I chi2IsUniform(ds, 0.05) {‘Yes’} E ‘No’))

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