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Primary Succession Tropical Rainforest GPP ATP / NADPH Forest fire

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Presentation on theme: "Primary Succession Tropical Rainforest GPP ATP / NADPH Forest fire"— Presentation transcript:

1 Primary Succession Tropical Rainforest GPP ATP / NADPH Forest fire
Photosynthesis Secondary Succession Light dependent NPP Climax Community

2 Pearsons Correlation Coefficient
C/W 14/06/16 Pearsons Correlation Coefficient To recap primary and secondary succession To calculate Pearsons Correlation coefficient using an example set of data and our own data.

3 KARL PEARSON ( ) British mathematician, ‘father’ of statistics and a pioneer of eugenics! 3

4 Pearson’s Correlation Coefficient
What can this test tell you? If there is a statistically significant correlation between two measured variables, X and Y, and…. If that correlation is negative or positive Note: The data must show normal distribution

5 What is the correlation coefficient r ?

6 Mass (arbitrary units) Length (arbitrary units)
Is there a significant correlation between an animal’s nose-to-tail length and its body mass? Animal Mass (arbitrary units) Length (arbitrary units) 1 2 4 5 3 8 12 14 6 9 19 7 22 If yes, then is the correlation positive (does a long tail mean a larger mass)?

7 Pearson’s Correlation Coefficient
Where: n = the number of values of X and Y ∑XY – [(∑X)(∑Y)]/n  {∑X2-[(∑X)2/n]} {∑Y2-[(∑Y)2/n]} r = r will always be a number between –1 and +1 This number can be compared with those in a table of critical values using: n – 2 degrees of freedom.

8 A negative value for r implies a negative correlation
What is the null hypothesis (H0)? H0 = there is no statistically significant correlation between length and body mass HA = there is a statistically significant correlation between length and body mass If the value for r exceeds the critical value, then you can reject the null hypothesis. A negative value for r implies a negative correlation A positive value for r implies a positive correlation

9 Construct the following results table
Animal (n = 7) Mass (X) Length (Y) X2 Y2 XY 1 2 4 5 3 8 12 14 6 9 19 7 22 Total Mean 9

10 Construct the following results table
Animal (n = 7) Mass (X) Length (Y) X2 Y2 XY 1 2 4 5 3 8 12 14 6 9 19 7 22 Total Mean 10

11 Construct the following table:
Animal Mass (X) Length (Y) X2 Y2 XY 1 2 4 5 16 3 8 9 12 14 64 6 19 81 7 22 Total X = 37 Mean 5.29

12 Calculate values for X:
Animal Mass (X) Length (Y) X2 Y2 XY 1 2 4 5 16 3 8 9 12 14 64 6 19 81 7 22 Total X = 37 X2 = 251 Mean 5.29 12

13 Calculate values for Y:
Animal Mass (X) Length (Y) X2 Y2 XY 1 2 4 5 16 25 3 8 9 64 12 244 14 196 6 19 81 361 7 22 484 Total X = 37 Y = 82 X2 = 251 Y2 = 1278 Mean 5.29 11.71 13

14 Calculate values for XY:
Animal Mass (X) Length (Y) X2 Y2 XY 1 2 4 5 16 25 20 3 8 9 64 24 12 244 48 14 196 112 6 19 81 361 152 7 22 484 176 Total X = 37 Y = 82 X2 = 251 Y2 = 1278 XY = 553 Mean 5.29 11.71 14

15 Use values obtained to populate the equation: X = 37 Y = 82
∑XY – [(∑X)(∑Y)]/n  {∑X2-[(∑X)2/n]} {∑Y2-[(∑Y)2/n]} r = 553 – (37 x 82)/7  {251-[372/7]} {1278-[822/7]} r = 119.57 = 132.60 r = 15

16 Compare your calculated value of r with the relevant critical value in your stats table
The + sign shows that any correlation is positive. Our value of r = Degrees of freedom = n = 5 D.F. Critical Value (P = 0.05) 3 0.88 4 0.81 5 0.75 6 0.71 7 0.67 Our value for r exceeds the critical value, so we can reject the null hypothesis. We can conclude that there is a significant positive correlation between the length of an animal and its body mass i.e. a long tail is associated with a large body mass!

17 Construct the following results table
Person (n = 8) Units of alcohol (X) Reaction time (s) (Y) X2 Y2 XY 1 2 3 4 5 4.5 Total Mean 17

18 Construct the following results table
Person (n = 8) Units of alcohol (X) Reaction time (s) (Y) X2 Y2 XY 1 2 3 4 5 4.5 Total Mean 18

19 Step 1: Rank each set of data (lowest to highest)
Student Maths exam score Maths rank Biology exam score Biology rank Alex 57 83 Bernard 45 37 Charlotte 72 41 Demi 78 85 Eustace 53 56 Ferdinand 63 Gemma 86 77 Hector 98 87 Ivor 59 70 Jasmine 71 3 7 1 1 7 2 8 8.5 8 2 3 5 8.5 9 9 6 10 10 4 5 6 4 Where two or more scores are tied... ... each is assigned an average rank


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