Plotting Lod Scores
Plot LOD score versus 'p' value for F2 data.
| class |
A_B_ |
A_bb |
aaB_ |
aabb |
Total |
| observed |
753 |
292 |
351 |
19 |
1415 |
The C2 for Ho: independence was
signigicant so we reject Ho.
| |
p/2
AB |
1-p/2
Ab |
1-p/2 |
p/2
ab |
| p/2
AB |
|
|
|
|
| 1-p/2
Ab |
|
|
|
|
| 1-p/2
aB |
|
|
|
|
| p/2
ab |
|
|
|
|
| class |
A_B_ |
A_bb |
aaB_ |
aabb |
Total |
| expected |
2+p2
4 |
(1-p)2
4 |
(1-p)2
4 |
p2
4 |
1415 |
We could use a Lod score test and vary 'p'.
z = Log10[
]
Log10[L(p)] = 753Log10(2+p2/4)
Log+ 292Log10[1-p2/4]
+ 351Log10[(1-p2/4)]
Log+ 19Log10[p2/4]
Log10(0.5) = 753Log10(9/16) + 292Log10[3/16]
Log+ 351Log10(3/16) + 19Log10[1/16]
= -678.4962
Let p =0.2, then Log10[L(p)]
= 753Log10(2.04/4) + 292Log10(0.96/4)
= + 351Log10(0.96/4) + 19Log10(0.04/4)
= 753(-0.2924) + 292(-0.6198) + 351(-0.6198)
= + 19(-2)
= -220.1772 - 180.9783 - 217.5498 - 38
= -656.7053
Let p = 0.25, then Log10[L(p)]
= 753Log10(0.51563) + 292Log10(0.2344)
= + 351Log10(0.2344) + 19Log10(0.01563)
= 753(-0.28766) + 292(-0630) + 351(-0.630)
= + 19(-1.806)
= -216.609 - 183.96 - 221.13 - 34.314
= -656.01335
Let p = 0.24, then Log10[L(p)]
= 753Log10(0.5144) + 292Log10(0.2356)
= + 351Log10(0.2356) + 19 Log10(0.0144)
= 753(-0.2887) + 292(-0.6278) + 35(-0.6278)
= + 19(-1.8416)
= -217.3911 - 183.3176 - 220.3578 - 34.9904
= -656.0569
Now the Lod scores are Log10[L(p)] - Log10[L(p)]
| p |
Log10[L(p)] |
Log10[L(0.5)] |
Lod |
| 0.2 |
-656.7053 |
-678.4962 |
21.7909 |
| 0.24 |
-656.0569 |
-678.4962 |
22.4395 |
| 0.25 |
-656.01335 |
-678.4962 |
22.4828 |
Previously, we found 
= 0.245, using the product method.
