Lecture16 - BIOL 361 Final Exam Friday April 24th 12:30...

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BIOL 361 Final Exam Friday April 24 th 12:30 -3:00 pm in PAC 11,12 Duration = 2.5 hours
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Lecture 16: Categorical Data Analysis II (Two-Way Designs - Contingency Tables) Sources of Information Triola: Chapter 10 Motulsky: Chapters 26-29 Sokal & Rohlf: Chapter 17 Dytham: p. 60-64, 147-154 OUTLINE : 1. 1-Way GOF test - Intrinsic hypothesis 2. 2 x 2 Contingency Tables 3. R x C Contingency Tables
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• So far, for our examples the expected frequencies were determined on the basis of some extrinsic hypothesis – balanced sex ratio (based on chromosome sorting during meiosis) or phenotype frequencies followed expectations based on principles of Mendelian genetics, etc. • Sometimes, no a-priori basis for expected frequencies • Instead, parameters are estimated from the data themselves • Most common in GOF testing involving statistical distributions. 1-Way GOF Tests for Intrinsic Hypotheses
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Houseflies, when at rest, fold their wings over their abdomen. A student determined that this behavior was consistent in individual flies – they always had the left wing over the right wing, or vice versa. That is, flies exhibit “wingedness” similar to “handedness” in humans. Q: The student wanted to determine whether the distribution of wingedness in a laboratory colony was binomially distributed. Example (Intrinsic Hypothesis, Binomial data) :
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Left wing over right Frequency 0 out of 8 42 15 2 23 8 31 2 44 50 62 70 80 To fit a binomial distribution to these data, we need an estimate of p , the population proportion of left-winged flies. Since there is no a-priori biological rationale for specifying p , we estimate it from the data Æ intrinsic parameter estimates for the model She examined 150 random samples, each of 8 flies. Results :
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Can compute that p = 0.16 (= proportion who are left ‘winged’). Then, can use p = 0.16 and the binomial distribution to compute the expected frequencies. Doing this we get the following table. But, to avoid cells with unacceptably small frequencies (< 5), we pool adjacent cells. Summing the individual chi-square values for each cell yields χ 2 = 2.4048. The degrees of freedom are not k -1 (5-1 = 4), as you might expect, but rather k- 2 (5 -1-1 = 3). Left Winged Observed frequency Expected Frequency χ 2 0 42 37.1814 0.6246 1 52 56.6753 0.3828 2 38 37.7716 0.0014 3 12 14.3892 0.3967 4+ 6 4.006 0.9993
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Reason for the loss of an additional degree of freedom • we used the data to estimate p , the binomial parameter. • Every parameter that is estimated costs you a degree of freedom. • We don’t lose a degree of freedom for q (the proportion of flies that are ‘right winged’) because once p is known, q is automatically determined.
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Nominal Data: 2-way Designs (= Contingency Tables) So far, • considered one variable at a time Æ 1-way GOF tests • data could been presented in 1-way frequency tables Æ Observed data (values & categories) could be listed in a single row or column.
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Lecture16 - BIOL 361 Final Exam Friday April 24th 12:30...

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