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Unformatted text preview: Obs minus expected Affected = 6156.9 = 4.1 Obs minus expected normal = 115119.1 = 4.1 (6156.9) 2 /56.9 = (4.1) 2 /56.9 = 0.295 (106119.1) 2 /119.1 = (4.1) 2 /119.1 = 0.141 X 2 = 0.436 1df p ~= 0.5 Conclusion: If the hypothesis is true (that alkaptonuria = simple recessive and we have correctly accounted for bias) then the difference between these data and the expected data would occur ~50% of the time. This is sufficiently likely for us to conclude that the "null hypothesis" is correct, namely that there is no significant difference between the observed and expected numbers....
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This note was uploaded on 10/27/2009 for the course BIO SCI 05580 taught by Professor Marsh during the Fall '09 term at UC Irvine.
 Fall '09
 MARSH

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