Spearman’s Rank Correlation Coefficient - AQA
Purpose: To test the strength of the correlation (relationship) between two variables.
Equation: rs = 1 - 6∑d2
n3-n
Where:
 rs = Spearman’s Rank Value
 ∑ = the sum of
 d = difference in rank of the values of each matched pair
 n = number of ranked pairs
Null Hypothesis: There is no statistically significant relationship between
______________ and _______________.
Alternative Hypothesis: There is a statistically significant relationship between
______________ and _______________.
2) Rank your “x” data, giving the largest measurement the lowest
rank. If some measurements have the same value, give them the
average of the ranks. Repeat this procedure to rank the “y” data.
1) Place your values for “x”
(independent) and “y” (dependent
variable) into the table.
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Variable x
x
Rank x (Rx)
3) Calculate the difference
between “Rank x” & “Rank y”.
Variable y
y
Rank y (Ry)
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difference
d (Rx-Ry)
4) Square the
difference.
difference2
d2
Tip: If the sum of Rx – Ry = 0
then the ranking is correct
∑ Rx – Ry = ________
5) Calculate ∑d2 = __________
6) Using the formula below, calculate Spearman’s Rank Coefficient (rs).
rs = 1 - 6∑d2
n3-n
=
Critical Values Table
Number of
pairs (n)
5
6
7
8
9
10
11
12
13
14
15
16
17
1.000
0.886
0.786
0.738
0.683
0.649
0.618
0.587
0.560
0.539
0.521
0.503
0.488
Number of
pairs (n)
18
19
20
21
22
23
24
25
26
27
28
29
30
0.472
0.460
0.447
0.436
0.425
0.416
0.407
0.398
0.390
0.382
0.376
0.369
0.362
Rs =
n=
Critical Value =
If the absolute value of the Spearman’s Rank Coefficient (rs) is less than the critical
value, you accept the null hypothesis.
If the absolute value of the Spearman’s Rank Coefficient (rs) is greater than the critical
value, you reject the null hypothesis.
Write a sentence explaining your result. Include whether you accept or reject your
null hypothesis, the rs value, the critical value, significance level and degrees of
freedom.
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