Trail+R

Wood Grass Grass Grass Grass Wood

1 Black Green Orange Yellow

6 1 6 7 2 Black Green Orange Red Yellow

5 2 1 9 3 3 Black Green Orange Red Yellow

5 2 4 4 5 4 Black Green Orange Red Yellow

7 1 3 3 6 5 Black Green Orange Red Yellow

5 3 4 3 5 6 Black Green Orange Yellow

6 3 6 5


 * Calculating Chi2**

__**Independent replicate 1 (at one end of the field)**__

Woodland-intermediate-grassland
 * Population || Black || Green || Orange || Red || Yellow || Total (row) ||
 * 1 || 6 || 1 || 6 || 0 || 7 || 20 ||
 * 2 || 5 || 2 || 1 || 9 || 3 || 20 ||
 * 3 || 5 || 2 || 4 || 4 || 5 || 20 ||
 * Total (column) || 16 || 5 || 11 || 13 || 15 || Grand total= 60 ||


 * Expected Values || 5.3333 || 1.6666 || 3.6666 || 4.3333 || 5 ||

EV for green is less than 3 (1.6666). Therefore, we will combine the green category with yellow.


 * Population || Black || Orange || Red || Yellow + Green || Total (row) ||
 * 1 || 6 || 6 || 0 || 8 || 20 ||
 * 2 || 5 || 1 || 9 || 5 || 20 ||
 * 3 || 5 || 4 || 4 || 7 || 20 ||
 * Total (column) || 16 || 11 || 13 || 20 || Grand total= 60 ||
 * Expected Values || 5.3333 || 3.6666 || 4.33333 || 6.6667 ||  ||


 * Population || Black || Orange || Red || Yellow + Green || Total (row) ||
 * //X2// 1 || 0.083 || 1.485 || 4.333 || 0.26667 || 6.168 ||
 * //X2// 2 || 0.021 || 0.267 || 5.023 || 0.417 || 5.728 ||
 * //X2// 3 || 0.021 || 3.267 || 0.026 || 0.017 || 3.331 ||
 * Total (column) || 0.125 || 5.016 || 9.382 || 0.700 || **15.2** ||

Degrees of freedom= 6
 * The chi squared value of 15.2 is slightly greater than 15.086 (P=0.01).
 * Values more than or equal to 15.086 are expected to occur 1% of the time if the null hypothesis was true.
 * Therefore it is very rare for deviations between observed and expected values to be explained by random sampling error.
 * To reject the null hypothesis (that t he difference in allele frequencies is not significant between populations and therefore the polymorphism of snails is attributed to selection) the chi square would have to be greater than 12.592 and **it is** indeed.
 * This means that polymorphisms between the populations is more likely to be significantly different with a chi square value this big (frequencies are significantly different; phenotypes are not in the same frequencies).

__**Independent replicate 2 (at the other end of the field)**__
 * Population || Brown || Green || Orange || Red || Yellow || Total (row) ||
 * 4 || 5 || 2 || 4 || 4 || 5 || 20 ||
 * 5 || 7 || 1 || 3 || 3 || 6 || 20 ||
 * 6 || 5 || 3 || 4 || 3 || 5 || 20 ||
 * Total (column) || 17 || 6 || 11 || 10 || 16 || Grand total= 60 ||


 * Expected Values || 5.6666 || 2 || 3.6666 || 3.333 || 5.333 ||


 * Population || Brown || Orange || Red || Green+ Yellow || Total (row) ||
 * 4 || 5 || 4 || 4 || 7 || 20 ||
 * 5 || 7 || 3 || 3 || 7 || 20 ||
 * 6 || 5 || 4 || 3 || 8 || 20 ||
 * Total (column) || 17 || 11 || 10 || 22 || Grand total= 60 ||


 * Expected Values || 5.6666 || 3.6666 || 3.3333 || 7.333 ||


 * Population || Brown || Orange || Red || Yellow + Green || Total (row) ||
 * //X2// 1 || 0.078 || 0.03 || 0.133 || 0.015 || 0.256 ||
 * //X2// 2 || 0.667 || 0.12 || 0.333 || 0.015 || 1.135 ||
 * //X2// 3 || 0.078 || 0.03 || 0.333 || 0.061 || 0.502 ||
 * Total (column) || 0.815 || 0.18 || 0.799 || 0.091 || 1.89 ||

Degrees of freedom= 6
 * Chi square is slightly greater than 1.635 (p=0.95). Values equal to or greater than 1.635 are expected to occur 95% of the time based on random chance alone if the null hypothesis was true.
 * Therefore 95% of the time it is probable that the deviations between observed and expected valuescan be explained by random sampling error.
 * Our chi square is less than 12.592 and therefore it is more likely for the null hypothesis (that t he difference in allele frequencies is not significant between populations and therefore the polymorphism of snails is attributed to selection) to be accepted.

The replicative independent sampling gave very different results. At the first end of the field, genetic drift had a significance on the chi squared test as opposed to the other end of the field where selection had a significance on the chi squared test.
 * Comments:**

Comparing frequencies of same habitat populations

__**Woodland**__
 * Population || Brown || Green || Orange || Red || Yellow || Total (row) ||
 * 1 || 6 || 1 || 6 || 0 || 7 || 20 ||
 * 6 || 6 || 3 || 6 || 0 || 5 || 20 ||
 * Total (column) || 12 || 4 || 12 || 0 || 12 || Grand total= 40 ||


 * Expected Values || 6 || 2 || 6 || 0 || 6 ||


 * Population || Brown || Orange || Green+ Yellow || Total (row) ||
 * 1 || 6 || 6 || 8 || 20 ||
 * 6 || 6 || 6 || 8 || 20 ||
 * Total (column) || 12 || 12 || 16 || Grand total= 40 ||


 * Expected Values || 6 || 6 || 8 ||


 * Population || Brown || Orange || Yellow + Green || Total (row) ||
 * //X2// 1 || 0 || 0 || 0 || 0 ||
 * //X2// 6 || 0 || 0 || 0 || 0 ||
 * Total (column) || 0 || 0 || 0 || 0 ||

Degrees of freedom= 2
 * Chi square is slightly less than 0.02 (p=0.99). Values equal to or greater than 0.02 (p=0.99) are expected to occur 99% of the time based on random chance alone.
 * 99% of the time we expect values between 0.02 and infinity to occur just as a matter of random sampling error.
 * There are no deviations between observed and expected values and so having the same exact frequencies suggests that selection is the same as the phenotypes are in the same frequencies.
 * To reject hypothesis, chi squared should be greater than 5.991 and our value is far less so we accept the null hypothesis that selection has great a significance in woodland.

__**Grassland**__
 * Population || Brown || Green || Orange || Red || Yellow || Total (row) ||
 * 2 || 5 || 2 || 1 || 9 || 3 || 20 ||
 * 3 || 5 || 2 || 4 || 4 || 5 || 20 ||
 * 4 || 7 || 1 || 3 || 3 || 6 || 20 ||
 * 5 || 5 || 3 || 4 || 3 || 5 || 20 ||
 * Total (column) || 22 || 8 || 12 || 19 || 19 || Grand total= 80 ||


 * Expected Values || 5.5 || 2 || 3 || 4.75 || 4.75 ||


 * Population || Brown || Orange || Red || Green + Yellow || Total (row) ||
 * 2 || 5 || 1 || 9 || 5 || 20 ||
 * 3 || 5 || 4 || 4 || 7 || 20 ||
 * 4 || 7 || 3 || 3 || 7 || 20 ||
 * 5 || 5 || 4 || 3 || 8 || 20 ||
 * Total (column) || 22 || 12 || 19 || 27 || Grand total= 80 ||


 * Expected Values || 5.5 || 3 || 4.75 || 6.75 ||


 * Population || Brown || Orange || Red || Green + Yellow || Total (row) ||
 * //X2// 2 || 0.04545 || 1.333 || 3.8 || 0.45 || 5.63 ||
 * //X2// 3 || 0.04545 || 0.3333 || 0.118 || 0.01 || 0.51 ||
 * //X2// 4 || 0.40909 || 0 || 0.64 || 0.01 || 1.06 ||
 * //X2// 5 || 0.04545 || 0.3333 || 0.64 || 0.231 || 1.25 ||
 * Total (column) || 0.5455 || 2 (0.999) || 5.2 || 0.7 || 8.45 ||

Degree of freedom= 9
 * Chi square value slightly greater than 8.343 (0.50).
 * Values equal to or greater than 8.343 are expected to occur 50% of the time based on random chance alone if the null hypothesis was true.
 * 50% of the time we expect values from 8.343 to infinity just as a matter of sampling error.
 * Possible that deviations between observed and expected can be explained by random sampling error.
 * To reject hypothesis the chi square value would have to be greater than 16.919 however our value is significantly smaller therefore we accept the hypothesis that selection has a significance in grassland.

__**Combining all habitats**__


 * Population || Brown || Orange || Red || Green + Yellow || Total (row) ||
 * //X2// 1 || 0 || 0 || 0 || 0 || 0 ||
 * //X2// 2 || 0.04545 || 1.333 || 3.8 || 0.45 || 5.63 ||
 * //X2// 3 || 0.04545 || 0.3333 || 0.118 || 0.01 || 0.51 ||
 * //X2// 4 || 0.40909 || 0 || 0.64 || 0.01 || 1.06 ||
 * //X2// 5 || 0.04545 || 0.3333 || 0.64 || 0.231 || 1.25 ||
 * //X2// 6 || 0 || 0 || 0 || 0 || 0 ||
 * Total (column) || 0.5455 || 2 (1.999) || 5.2 || 0.7 || 8.45 ||

Degree of freedom= 15


 * Chi squared (8.45) slightly greater than 7.261(0.95)
 * Values equal to or greater than 7.261are expected to occur 95% of the time based on random chance alone if the null hypothesis was true.
 * 95% of the time we expect values from 7.261to infinity just as a matter of sampling error.
 * Possible that deviations between observed and expected can be explained by random sampling error.
 * To reject hypothesis the chi square value would have to be greater than 24.996 however our value is significantly smaller therefore it is more likely for the null hypothesis (that selection has a greater effect in all habitats) to occur. This is because the phenotypes are in similar frequencies.