07 - Tests of significance The 2 test of "goodness of...

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MCB140 2-6-08 1 Tests of significance The χ 2 test of “goodness of fit” (Karl Pearson)
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MCB140 2-6-08 2 Classical problem “No one can tell which way a penny will fall, but we expect the proportions of heads and tails after a large number of spins to be nearly equal. An experiment to demonstrate this point was performed by Kerrich while he was interned in Denmark during the last war. He tossed a coin 10,000 times and obtained altogether 5,067 heads and 4,933 tails.” MG Bulmer Principles of Statistics
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MCB140 2-6-08 3 Hypothesis vs. observation Hypothesis: the probability of getting a tail is 0.5. Observation: 4,933 out of 10,000. Well?!! How can we meaningfully – quantitatively – construct a test that would tell us, whether the hypothesis is, most likely, correct, and the deviation is due to chance – or (alternatively) – the hypothesis is incorrect, and the coin dislikes showing its “head” side for some mysterious reason? Sampling errors are inevitable, and deviations from perfection are observed all the time. The goodness of fit test has been devised to tell us, how often the deviation we have observed could have taken place solely due to chance.
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MCB140 2-6-08 4 χ 2 2 = - ( ) O E E
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MCB140 2-6-08 5 The procedure Come up with an explanation for the data (“the null hypothesis”). Ask yourself – if that explanation were correct, what should the data have been? E.g., if the hypothesis is that the probability of getting “tails” is 50%, then there should have been 5,000 tails and 5,000 heads. This set of numbers forms the “expected data.” Take the actual – observed – data ( critical point : take the primary numbers, not the frequencies or percentages – this is because the “goodness of fit” is a function of the absolute values under study). Plug them into the following formula: χ 2 2 = - ( ) O E E
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MCB140 2-6-08 6 Calculate p value. If it’s .05 or below, the hypothesis is incorrect – the deviation you see in the data is unlikely to be due to chance. If it’s above .05, the hypothesis stands.
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MCB140 2-6-08 7 SMI? Take a pure-breeding agouti mouse and cross it to a pure-breeding white mouse. Get 16 children: all agouti (8 males, 8 females). Cross each male with one female (randomly). Get 240 children in F2: 175 agouti and 65 white (ratio: 2.692).
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MCB140 2-6-08 8 Calculating the chi square value Let’s  hypothesize  that  we  are  dealing  with  simple  Mendelian  inheritance  (the  null  hypothesis ).  If  this  were  true,  then  we  would  expect  that the 240 children would have split: 180 agouti : 60 white.  For agouti mice:  (175-180) 2 / 180=0.139  For white mice:  (65-60) 2 / 60=0.417  sum ( ) of agouti and white = 0.139 + 0.417 = 0.556 
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MCB140 2-6-08 9 Evaluating the null hypothesis There are only two classes here, so we must use the “1 degree of  freedom” line in the table. For  χ 2 =0.556, the  lies between 0.1 and 
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This note was uploaded on 08/01/2008 for the course MCB 140 taught by Professor Urnov during the Spring '08 term at University of California, Berkeley.

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07 - Tests of significance The 2 test of "goodness of...

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