Mash+R

Wood Wood Grass Grass Grass Wood

1 Black Green Orange Red Yellow

4 4 1 3 8 2 Green Orange Red Yellow

8 1 3 8 3 Green Orange Yellow

5 4 11 4 Black Green Yellow

2 8 10 5 Black Green Orange Red Yellow

1 9 1 1 8 6 Black Green Orange Red Yellow

4 5 3 2 6

The first two chi square tests were conducted in order to see wheather there is any difference between samples of jelly babies in the same habitat ( first in woodland, than in grassland). The third test is done in order to investigate whether the samples of jelly babies differ significantly between the woodland and grassland habitats.

=**Chi square test 1**=


 * __Null hypothesis :__** The proportion of black, green, orange, red and yellow jelly babies does not differ between the three populations of woodland.
 * Observed frequencies in woodland ||
 * || Black || Green || Orange || Red || Yellow ||
 * Pop 1 || 4 || 4 || 1 || 3 || 8 ||
 * Pop2 || 0 || 8 || 1 || 3 || 8 ||
 * Pop3 || 4 || 5 || 3 || 2 || 6 ||
 * **Total** || 8 || 17 || 5 || 8 || 22 ||

Since the expected frequency of black, orange and red jelly babies is less than three, black and green jelly babies should be combined together in one group and orange, red and yellow jelly babies into another group due to their physiology and predation similarities.
 * Expected frequencies in woodland ||
 * || Black || Green || Orange || Red || Yellow ||
 * Pop1 || 8/3 || 17/3 || 5/3 || 8/3 || 22/3 ||
 * Pop2 || 8/3 || 17/3 || 5/3 || 8/3 || 22/3 ||
 * Pop3 || 8/3 || 17/3 || 5/3 || 8/3 || 22/3 ||
 * Observed frequencies after combining ||
 * || Black & Green || Orange, Red & Yellow ||
 * Pop1 || 8 || 12 ||
 * Pop2 || 8 || 12 ||
 * Pop3 || 9 || 11 ||
 * **Total** || 25 || 35 ||

Our calculated value (χ² ) = 67/525 = 0.12761904 Degrees of freedom (df) = (3-1) x (2-1) = 2 Critical value (P=0.05, 2 df) = 5.99
 * Expected frequencies after combining ||
 * || Black & Green || Orange, Red & Yellow ||
 * Pop1 || 25/3 || 35/3 ||
 * Pop2 || 25/3 || 35/3 ||
 * Pop3 || 25/3 || 35/3 ||


 * __Conclusion:__** As our calculated χ² value is smaller than the critical value we will accept our Null hypothesis which states that there is no difference in the proportion of differently coloured jelly babies between the three populations of woodland. We can conclude that we would see deviations from the expected values as large as this in more than 5% occasions if we repeated our experiment.

=Chi square test 2=


 * __Null hypothesis:__** The proportion of black, green, orange, red and yellow jelly babies does not differ between the three populations of grassland.
 * Observed frequencies in grassland ||
 * || Black || Green || Orange || Red || Yellow ||
 * Pop 1 || 0 || 5 || 4 || 0 || 11 ||
 * Pop2 || 2 || 8 || 0 || 0 || 10 ||
 * Pop3 || 1 || 9 || 1 || 1 || 8 ||
 * **Total** || 3 || 22 || 5 || 1 || 29 ||

As the expected frequencies of black, orange and red jelly babies are less than three these will be combined with green and yellow jelly babies according to their physiology and predation.
 * Expected frequencies in grassland ||
 * || Black || Green || Orange || Red || Yellow ||
 * Pop1 || 1 || 22/3 || 5/3 || 1/3 || 29/3 ||
 * Pop2 || 1 || 22/3 || 5/3 || 1/3 || 29/3 ||
 * Pop3 || 1 || 22/3 || 5/3 || 1/3 || 29/3 ||
 * Observed frequencies after combining ||
 * || Black & Green || Orange, Red & Yellow ||
 * Pop1 || 5 || 15 ||
 * Pop2 || 10 || 10 ||
 * Pop3 || 10 || 10 ||
 * **Total** || 25 || 35 ||


 * Expected frequencies after combining ||
 * || Black & Green || Orange, Red & Yellow ||
 * Pop1 || 25/3 || 35/3 ||
 * Pop2 || 25/3 || 35/3 ||
 * Pop3 || 25/3 || 35/3 ||

Our calculated value ( χ²) = 3.428571 Degrees of freedom (df) = (3-1) x (2-1) = 2 Critical value (P=0.05, 2 df) = 5.99

__ **Conclusion**: __As our calculated χ² value is smaller than the critical value we will again accept our Null hypothesis.

=**Chi square test 3**=


 * __Null Hypothesis:__** There is no difference in the proportion of black, green, orange, red and yellow jelly babies between the woodland and grassland.
 * Observed Frequency ||
 * || Black || Green |||| Orange || Red || Yellow ||
 * Wood || 4 || 4 |||| 1 || 3 || 8 ||
 * Wood || 0 || 8 |||| 1 || 3 || 8 ||
 * Grass || 0 || 5 |||| 4 || 0 || 11 ||
 * Grass || 2 || 8 |||| 0 || 0 || 10 ||
 * Grass || 1 || 9 |||| 1 || 1 || 8 ||
 * Wood || 4 || 5 |||| 3 || 2 || 6 ||
 * **Total** || **11** |||| **39** || **10** || **9** || **51** ||

As the expected frequencies of black, orange and red jelly babies are all less than 3.0 these will be combined with green and yellow once based on their physiology and predation.
 * Expected Frequency ||
 * || Black || Green || Orange || Red || Yellow ||
 * Wood || 11/6 || 6.5 || 10/6 || 9/6 || 8.5 ||
 * Wood || 11/6 || 6.5 || 10/6 || 9/6 || 8.5 ||
 * Grass || 11/6 || 6.5 || 10/6 || 9/6 || 8.5 ||
 * Grass || 11/6 || 6.5 || 10/6 || 9/6 || 8.5 ||
 * Grass || 11/6 || 6.5 || 10/6 || 9/6 || 8.5 ||
 * Wood || 11/6 || 6.5 || 10/6 || 9/6 || 8.5 ||
 * Observed Frequencies after combining ||
 * || Black & Green || Orange, Red &Yellow ||
 * Wood || 8 || 12 ||
 * Wood || 8 || 12 ||
 * Grass || 5 || 15 ||
 * Grass || 10 || 10 ||
 * Grass || 10 || 10 ||
 * Wood || 9 || 11 ||
 * **Total** || ** 50 ** || ** 70 ** ||


 * Expected Frequencies after combining ||
 * || Black & Green || Orange, Red &Yellow ||
 * Wood || 50/6 || 70/6 ||
 * Wood || 50/6 || 70/6 ||
 * Grass || 50/6 || 70/6 ||
 * Grass || 50/6 || 70/6 ||
 * Grass || 50/6 || 70/6 ||
 * Wood || 50/6 || 70/6 ||



Our calculated value ( χ²) = 3.56571428

Degrees of freedom (df) = (6-1) x (2-1) = 5

Critical value from the statistical table ( P=0.05 and 1 df ) is 11.07


 * __Conclusion__**: Since our calculated χ² value is less than the critical value we can accept our Null Hypothesis.