Are fingerprint patterns genetic, or are they developmental? In 1892 Francis Galton compiled the following table on the relationship between the patterns on the same finger of 105 sibling pairs. Test the hypothesis that the patterns are independent for example, that knowing one sibling (A) has a Whorl on the finger does not help in predicting the pattern of the other (B).
A children
B children | Arches | Loops | Whorls | Totals |
Arches | 5 | 12 | 2 | 19 |
Loops | 4 | 42 | 15 | 61 |
Whorls | 1 | 14 | 10 | 24 |
Totals | 10 | 68 | 27 | 105 |
H0: Null Hypothesis: The patterns are independent
HA: Alternative Hypothesis: The patterns are dependent.
Under assumption of H0, the Expected Frequencies are obtained as follows:
Arches | Loops | Whorls | Total | |
Arches | 10X19/105=1.81 | 12.30 | 4.89 | 19 |
Loops | 5.81 | 39.50 | 15.69 | 61 |
Whorls | 2.38 | 16.19 | 6.43 | 25 |
Total | 10 | 68 | 27 | 105 |
From the above Table, the Table is calculated as follows:
O | E | (O - E)2/E |
5 | 1.81 | 5.63 |
12 | 12.30 | 0.01 |
2 | 4.89 | 1.70 |
4 | 5.81 | 0.56 |
42 | 39.50 | 0.16 |
15 | 15.69 | 0.03 |
1 | 2.38 | 0.80 |
14 | 16.19 | 0.30 |
10 | 6.43 | 1.98 |
Total = = | 11.17 |
= 11.17
ndf = ( r - 1) X (c - 1)
= (3 - 1) X (3 - 1) = 4
By Technology, p- value = 0.0247
Since p - value = 0.0247 is less than = 0.05, the difference is significant. Reject null hypothesis.
Conclusion:
The data do not support the claim that the patterns are
independent.
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