spearman's rank

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Spearman’s rank example Statistical Analysis Results from the following Spearman’s Rank’s will be used to determine whether a correlation is evident between light intensity and abundance of Yorkshire Fog, in the form of percentage coverage. Null Hypothesis There will be no correlation between Light Intensity and Percentage Coverage. Alternative Hypothesis There will be a correlation between Light Intensity and Percentage Coverage. Spearman’s Rank Results: Light Intensity (lux ) Rank 1 Percentage Coverage (%) Rank 2 D D 2 367 2 63% 6 4 16 341 1 70% 7 6 36 373 3 59% 5 2 4 716 4 49% 3 1 1 916 5 50% 4 1 1 1182 6 40% 2 4 16 1689 7 28% 1 6 36 [14] Using the Equation: Where: (R) = Spearman’s Rank Correlation Co-efficient D = Difference between Rank 1 and Rank 2 N = Number of Pairs of Observations in the Sample d 2 = Sum of the squared differences between Ranks Number of Pairs of Observation equals 7, this is due to there being 7 values of light intensity and 7 for percentage coverage; totalling 14 dividing this by two gives us our value. The value of D is obtained by subtracting Rank 2 from Rank 1.

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Page 1: Spearman's Rank

Spearman’s rank example

Statistical Analysis

Results from the following Spearman’s Rank’s will be used to determine whether a correlation is evident between light intensity and abundance of Yorkshire Fog, in the form of percentage coverage.

Null Hypothesis

There will be no correlation between Light Intensity and Percentage Coverage.

Alternative Hypothesis

There will be a correlation between Light Intensity and Percentage Coverage.

Spearman’s Rank Results:

Light Intensity (lux) Rank 1 Percentage Coverage (%) Rank 2 D D2

367 2 63% 6 4 16 341 1 70% 7 6 36 373 3 59% 5 2 4 716 4 49% 3 1 1 916 5 50% 4 1 1

1182 6 40% 2 4 16 1689 7 28% 1 6 36

[14]

Using the Equation:

Where: (R) = Spearman’s Rank Correlation Co-efficient

D = Difference between Rank 1 and Rank 2

N = Number of Pairs of Observations in the Sample

∑d2 = Sum of the squared differences between Ranks

Number of Pairs of Observation equals 7, this is due to there being 7 values of light intensity and 7 for percentage coverage; totalling 14 dividing this by two gives us our value.

The value of D is obtained by subtracting Rank 2 from Rank 1.

E.g. Rank 1 = 2 Rank 2 = 6 thus D = 2-6 =4

Square each value of D. e.g. D1 = 42 = 16

Sum of D2 (the total of each square value of D) = 16+36+4+1+1+16+36 = 110

(R) = 1 – (6x110) / 73 – 7 = -0.9642857143 = - 0.964 (3d.p.)

Page 2: Spearman's Rank

Critical Value

In this investigation 7 samples were obtained on each transect, as a result it is this number of samples which will determine the critical value for my product moment correlation coefficient calculated during my spearman's rank. The critical value is 0.786/-0.786. For the calculated product moment correlation coefficient to be statistically significant, and accepted with 95% confidence, it has to be a greater positive value than the positive critical value or a more negative value than the negative critical value. The value obtained was -0.964 (3d.p). This means that the statistical relationship between the two variables, percentage coverage and light intensity, are likely to have had a meaningful one rather than one down to chance, and that we can accept this with 95% confidence.