Unformatted text preview: expected The expected in this case is a 1:1:1:1 ratio, or 250 for each class Chisquared = (280250 ) 2 + (270250 ) 2 + (220250 ) 2 + (230250 ) 2 250 250 250 250 3.6 + 1.6 + 3.6 + 1.6 = 10.4 5) Look up the Chisquared value in the table (Ch. 4, p.126 in 7 th ; Ch. 2, p.41 in 8 th edition). degrees of freedom (df) = number of classes  1. Reject the null hypothesis if the chisquare value you calculate is greater than the value given in the p = 0.05 column. For this problem, 3 df, the value of 10.2 is greater than the value for p(0.05) = 7.8 (and lies between the values for p(0.025) and p(0.01) This means that this great a deviation from the expected numbers for independent assortment is only seen 1 2.5% of the time, rarely. Therefore we reject the hypothesis of independent assortment. The data are thus consistent with linkage (but don't prove it)....
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This note was uploaded on 07/02/2010 for the course BIS 101 taught by Professor Simonchan during the Spring '08 term at UC Davis.
 Spring '08
 simonchan

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