Data Handling ZO4030 Lecture 5 Count data

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Data HandlingZO4030Lecture 5 Count data

• Andrew Jackson

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So far

• GLMs to study effects on continuous response data
• Fixed factors
• Linear covariates
• Normal errors(residuals)

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More types of data

• Count data
• E.g. the number of deaths resulting from IHD
  (Ischaemic Heart Disease) in an age group
• Positive integers
• Proportional data
• Number of males born as a proportion of females
• Binary data
• Species presence (True / False)
• Infection status
• Behaviour (ate / didnt eat)

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Proportional counts

• Flowering in 5 varieties of perennial plants
• 6 dose levels of a growth fertiliser applied
• Response is number flowered / number of flower
  buds

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What data type is our response variable?

• Number of successes (flowers) out of a number of
  attempts (flower buds)
• Remind you of anything?
• Number of heads (successes) out of repeated coin
  tosses (trials)

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The Binomial Distribution

• PDF Probability Density Function
• Y binom(p,N)
• p 0.5
• N 6

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For our flowers

• Flowers Binomial(p,Number)
• Now, the probability p is surely related to
  things like, dose and variety.
• So we have something likep b0 b1dose
  b2variety

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But

• We have a problem
• p is a probability and as such is strictly
  bounded by0 p 1

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Need to transform our relationship

• p b0 b1dose b2variety
• Aasdf
• This looks equationpretty terrible
• and it is!

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Can simplify though..

• p is probability of flowering
• 1-p is probability of not flowering
• Odds of flowering p/(1-p)
• Ln(p/(1-p)) logit(p)
• Logit(p) b0 b1dose b2variety
• And we have a straight line again!
• Perfect for a GLM (twiddles and all)

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Running Logistic Regression

• Now that its a simple GLM all we have to do is
  identify
• Response variable
• Fixed factors
• Covariates

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Running Logistic Regression in R

• In R this becomes
• Y lt- cbind(flowers,number-flowers)
• Model1 lt- glm(Ydosevariety, familybinomial)
• So this is an ANCOVA or a GLM with multiple
  slopes and intercepts

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Interpreting the output
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But what about the transformation?

• When we calculate the effect sizes (coefficients)
  they are on the transformed logit(p) scale
• So, variety D is 3.18 larger than variety A
• BUT.. This means the log(odds) are bigger

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Back-calculating

• To convert back to raw probabilities we use the
  equation
• For the intercept (i.e. variety A and dose0)p
  1 / (1exp(-(-4.6))) 0.001
• For variety Dp 1 / (1exp(-(-4.6 3.18)))
  0.19

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Visualising the output
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Binary Outcome Data
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Binary Dataset

• Fish infected with a parasite
• Binary response variable - Infected
• Gender as fixed factor
• Age and weight as covariates

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Our hypotheses

• That age affects parasite incidence
• That gender affects parasite incidence
• That weight affects parasite incidence

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Exactly as before

• Number of trials here 1 (just like a single
  coin toss)
• The probability of having the disease is modelled
  as a function of the explanatory variables

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Visualise the Data
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Run the model

• Coefficients
• Estimate Std. Error z
  value Pr(gtz)
• (Intercept) -0.109124 1.375388
  -0.079 0.937
• age 0.024128 0.020874
  1.156 0.248
• weight -0.074156 0.147678
  -0.502 0.616
• sexmale -5.969109 4.278066
  -1.395 0.163
• ageweight -0.001977 0.002006
  -0.985 0.325
• agesexmale 0.038086 0.041325 0.922
  0.357
• weightsexmale 0.213830 0.343265 0.623
  0.533
• ageweightsexmale 0.001651 0.003419 -0.483
  0.629

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What do these parameters mean?

• Again
• We have modelled the log(odds)
• And depending on what you want to compare or
  describe, you may need to back transform them
  using the earlier equation

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Summary

• Non-continuous data can be modelled using under
  the same GLM framework by picking the right
  family or distribution
• Binomial
• Proportion data
• Binary data
• Poisson
• Counts
• Things that happen at a given rate

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To do

• Revise your notes from last year on the
  chi-square test for analysing proportions
• R can do this test too
• Read up on the following distributions
• Binomial
• Poisson