For each cell, calculate the Expected value (E) by multiplying the row total by the column total and dividing your result by the overall total. For example, multiply the row total for scree particles between 0–50 cm 20–24.9 m from the free face (46) by the column total of scree particles 20–24.9 m from the free face (100), and divide this figure by the total number of scree particles (200). This gives an expected value of 23. Scree size ROW TOTAL (long axis cm) Observed Expected Observed Expected 20–24.9 m lower scree 0–4.9 m from free upper from free face scree face 6 23 40 23 46 0–50 12 21.5 31 21.5 43 51–100 29 23 17 23 46 101–150 53 32.5 12 32.5 65 151–200 100 100 200 Column Total X2 (Observed Expected = – Expected) 2 = (6 – 23)2 (12– 21.5)2 (29 – 23)2 (53 – 32.5)2 + 23 21.5 23 32.5 X2 + + + (40 – 23)2 (31 – 21.5)2 (17 – 23)2 (12 – 32.5)2 23 21.5 23 32.5 + + + = 12.56 + 4.20 + 1.56 + 12.93 + 12.56 + 4.20 + 1.56 + 12.93 X2 = X2 62.5 Degrees of Freedom = number of rows –1 × columns –1 = 3 × 1 = 3 The tabulated figure at 3 degrees of freedom at the 99% confidence level is 11.34. As the calculated value of is above this, it can be stated with 99% confidence that there is a 62.5 statistically significant difference between the size of scree deposits on the upper part and lower part of the scree.
For each cell, calculate the Expected value (E) by multiplying the row total by the column total and dividing your result by the overall total. For example, multiply the row total for scree particles between 0–50 cm 20–24.9 m from the free face (46) by the column total of scree particles 20–24.9 m from the free face (100), and divide this figure by the total number of scree particles (200). This gives an expected value of 23. Scree size (long axis cm) ROW TOTAL Observed 20–24.9 m from free face Expected lower scree Observed 0–4.9 m from free face Expected upper scree 0–50 6 23 40 23 46 51–100 12 21.5 31 21.5 43 101–150 29 23 17 23 46 151–200 53 32.5 12 32.5 65 Column Total 100 100 200 X2 = (Observed – Expected) 2 Expected X2 = (6 – 23)2 23 + (12– 21.5)2 21.5 + (29 – 23)2 23 + (53 – 32.5)2 32.5 + (40 – 23)2 23 + (31 – 21.5)2 21.5 + (17 – 23)2 23 + (12 – 32.5)2 32.5 X2 = 12.56 + 4.20 + 1.56 + 12.93 + 12.56 + 4.20 + 1.56 + 12.93 X2 = 62.5 Degrees of Freedom = number of rows –1 × columns –1 = 3 × 1 = 3 The tabulated figure at 3 degrees of freedom at the 99% confidence level is 11.34. As the calculated value of 62.5 is above this, it can be stated with 99% confidence that there is a statistically significant difference between the size of scree deposits on the upper part and lower part of the scree.
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