biol139-lecture20-2011

Class number observed number expected2 number

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Unformatted text preview: 44 142 yyR_ yyR_ 124 142 yyrr 146 142 (1 2 ( ( 1 5 4 −1 4 2 ) 2 4 4 −1 4 2 ) ( 1 2 4 −1 4 2 ) 2 1 4 6 −1 4 2 ) 2 + 142 + + 142 142 142 χ = 1 .0 1 + 0 .0 2 8 + 2 .2 8 + 0 .1 1 3 = 3 .4 3 5. Use the chi-square value and number of degrees of freedom to determine a p value : the probability that a deviation from the predicted numbers at least as large as that observed in the experiment will occur by chance (Table 5.1 in text) chance Remember: Χ 2 = 3.43 and df = 3 Remember: 3.43 p value ≈ 0.3 null hypothesis (no linkage) is accepted value is If observed values deviate from values predicted by our hypothesis, is this significant (reflecting a real difference between observation + theory) or insignificant insignificant (reflecting random sampling error or chance)? P value lets us know this…. 6. Evaluate the significance of the p value. The 6. convention is that a 0.05 p value is the boundary convention between accepting and rejecting the null hypothesis. P value – the probability that a deviation from the predicted numbers at least value as large as that observed in the experiment has occurred due solely to chance Very small p values indicate a high degree of significant difference significant between observed and expected data i.e. between 0.05 = 5% chance that the observation is due to chance alone (in other words, a 95% probability that the observation is not due to chance and so is a REAL difference) REAL If very low, deviation of observed from expected results becomes significant, very significant, unlikely hypothesis being tested explains data (i.e. if p value = 0.01, only 1 in 100 times will that observation be due to chance alone) If high, likely hypothesis being tested is correct, deviation from expected is high insignificant (i.e. p value = 0.90 -90 out of 100 times the observation would occur by chance if the null hypothesis is correct) occur Are genes A and B linked in testcross AB/ab x ab/ab ? Are AB/ab ab/ab ab ab ab ab χ 2= (18-12.5)2 + (15-12.5)2 12.5 12.5 χ 2= 2.42 + 0.5 + 2.42 + 0.5 χ 2= 5.84 Df =...
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