Segment 16: The Towne Family - Again - 3/4/2012
Using Python, here are the following p-values we would get for the mutations as well as the code:
import scipy, scipy.stats array = [5, 23, 12] sum3 = 0.0 for j in range(len(array)): i = array[j] #print i sum3 = 0.0 while i <= 37: sum3 = sum3 + scipy.stats.binom.pmf(i,37,0.003) i += 1 print "Probability for mutations = " + str(array[j]) + ": " +str(sum3)
- Probability for mutations = 5: 9.77781793707e-08
- Probability for mutations = 23: 5.52227454159e-49
- Probability for mutations = 12: 9.18558952215e-22
To Think About 1
Can you think of a unified way to handle the Towne family problem (estimating r and deciding which family members are likely "non-paternal") without trimming the data? We'll show one such method in a later segment, but there is likely more than one possible good answer.