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Number of Progeny Required for Minimum Confidence Interval
Source: Pg. 192 Statistical Genomics: Linkage, mapping and QTL analysis. B.H. Liu.
Let c = the maximum range of the 95% confidence interval for our estimate of p; i = I(p) is the expected information content per individual as p is varied; N is the number of individuals required to obtain the desired confidence interval. For the case of a 95% confidence interval, we know that Za = 1.96. 2(1.96) = 3.92. The formula is as follows:
Example:
For the testcross family with repulsion linkage and p = 0.4, we have shown that
I(p) = i = |
1 |
= |
1 |
= 4.167 = i. |
p(1-p) |
0.4(0.6) |
Now let us find the number of individuals required for a 95% confidence interval for p = 0.4 in the case of testcross family when we want the size of the confidence interval to be c = 0.02
and using the formula
N > |
(3.92)2 |
= |
(3.92)2 |
= 904 individuals. |
c2I(p) |
(0.02)2(4.167) |
Example:
Let us find the number of individuals needed in our family when we estimate p for the testcross when p = 0.05 and we want the 95% confidence interval to be c = 0.02.
N > |
(3.92)2 |
= 1830. |
(0.02)2(21.05) |
We knew that I(p) = 21.05 because I(p) = 1/ p(1-p)
= 1/ 0.05(1 - 0.05) = 21.05 = i.
Example:
Let us find N when we have p = 0.4 for the F2 family in repulsion-phase linkage.
I(p) = |
2(1 + 2p2) |
= 1.46 then N> |
(3.92)2 |
= 26500. |
(1 - p2)(2 + p2) |
(0.02)2(1.46) |
We can use the above formulas to plan experiments to have a pre-determined level of precision.
Copyright
2000©, Ted Helms
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