Chi-Squared Test

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Chi-Squared Test
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Chi-Square Test

• Compare observed data with data you would expect
  to obtain according to a specific hypothesis
• How much deviation can occur before you conclude
  something other than chance at work?
• ?2 S(o-e)2 / e
• Tests the null hypothesis
• States that there is no significant difference
  between the expected and observed result

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Lizard Example

• Male phenologies in a population of side-blotched
  lizards
• Three male phenotypes
• Very aggressive orange-throat males
• Moderately aggressive blue-throat males
• Sneaker yellow-throat males
• The three phenotypes vary in their frequencies
  over time

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Scientific Hypothesis

• Scientific hypothesis
• The three male phenotypes are present in equal
  frequencies in the population
• Research question
• Are there differences in the frequencies of male
  phenotypes in the population of lizards
• Chi-square (goodness of fit)
• How well does our observed frequency distribution
  match the expected frequency distribution

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Step 1

• Determine number of observed and expected in each
  category
• Obtained a sample of 51 males from the population
• 12 orange-throat males
• 30 blue-throat males
• 9 yellow-throat males
• Expected frequency is 111
• 51/3 17
• We expect 17 of each individuals

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Step 2

• Calculate X2

Orange Blue Yellow
Observed (o) 12 30 9
Expected (e) 17 17 17
Deviation (o-e) -5 13 -8
Deviation2 (d2) 25 169 64
d2/e 1.47 9.94 3.76
X2 Sd2/e 15.17 X2 Sd2/e 15.17
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Step 3

• Determine degrees of freedom (df)
• Number of values we can pick freely without being
  constrained by other values
• Number of categories in the problem minus 1
• Three categories (orange, blue, yellow)
• 2 df

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Step 4

• Determine a relative standard to serve as the
  basis for rejecting the hypothesis
• Commonly used in biological research plt0.05
• P value probability of rejecting the null
  hypothesis (no difference between observed and
  expected) when it is actually true
• Probability of making an error of falsely stating
  that there is a significant difference between e
  and o when there is not a significant difference.
• p lt 0.05
• Stating that there is a less than 5 chance of
  error of stating there is a difference when there
  actually isnt a significant difference

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Step 5

• Refer to chi-square distribution table
• Locate appropriate df (2)
• Locate the value corresponding to the p value of
  0.05

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Conclusions

• If your X2 gt distribution table
• Conclude observed numbers significantly different
  from the expected
• X2 (15.17) is greater than 5.99 (df2, p0.05)
• Reject the hypothesis that the three male
  phenotypes are present in equal frequencies
• Significant difference in frequencies among the
  three phenotypes

Degrees of Freedom (df) Probability (p) Probability (p) Probability (p) Probability (p) Probability (p) Probability (p) Probability (p) Probability (p) Probability (p) Probability (p) Probability (p)
Degrees of Freedom (df) 0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001
1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83
2 0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82
3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27
4 0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.28 18.47
5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52
Nonsignificant Nonsignificant Nonsignificant Nonsignificant Nonsignificant Nonsignificant Nonsignificant Nonsignificant Significant Significant Significant
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Plant and Environment Interaction

• Galapagos Tree Fern
• Present in 4 life zones on the islands
• Miconia Zone
• Brown Zone
• Scalesia Zone
• Transition Zone

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• Hypothesis
• The cactus tree fern is present in equal
  frequencies across the life zones
• Observed data
• Miconia Zone - 13
• Brown Zone - 15
• Scalesia Zone - 22
• Transition Zone - 18