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Introduction to Statistics

V - Significance of Correlation Coefficient

Real Life Example

Here we compare the frequency of storms off the east coast of the United States with ocean temperatures off the coast of Peru.

Year Anomaly Niño 3 Frequency anomaly
of all storms
1) 79-80 +0.44 +1.2
2) 80-81 -.006 +3.2
3) 81-82 +0.35 -1.8
4) 83-84 +3.04 +5.2
5) 84-85 -0.19 -2.8
6) 85-86 -0.98 -3.8
7) 86-87 -0.45 -1.8
8) 87-88 +1.21 +4.2
9) 88-89 +0.68 +3.2
10) 89-90 -1.11 -6.8

Since the points are pretty close to a straight line, and the line goes up to the right, it's a good guess that "r" is positive and close to 1.

Let's calculate "r":

Niño 3 Anomaly

S = Square root of (1/(N-1)) [sum of (each data point - mean)2]

S = Square root of 1/(10-1) * [(0.44 - 0.29)2 + (-0.006 - 0.29)2 + (0.35 - 0.29)2 + (3.04 - 0.29)2 + (-0.19 - 0.29)2 + (-0.98 -0.29)2 + (-0.45 -0.29)2 + (1.21 - 0.29)2 + (0.68 - 0.29)2 + (-1.11 - 0.29)2]

S = Square root of 13.06/9

S = 1.2046

All Storms Anomaly

S = Square root of ((1/9) * (141.6))

S = 3.9665

r = [ (0.44-0.29) (1.2-0) (-.006-0.29) (3.2-0) +...] / [ (10-1) (1.2046) (3.9665) ]

r = [ +0.18-1.12-0.11+1.34+4.38+1.33+3.86+1.25+9.52 ] / [ (9) (1.20) (3.97) ]

r = 0.825

This is a very high r! But with only 10 data points, is it statistically significant? Use the Fisher test to find out:

1/2 ln ( (1 + .83)/(1 - .83)) ± 1.96 * Square root of (1/7) = 1/2 ln ((1 + ρ)/(1 - ρ))

1.19 ± 0.74 = 1/2 ln ((1 + ρ)/(1 - ρ))

1.93 = 1/2 ln ((1 + ρ)/(1 - ρ))

47.47 = (1 + ρ)/(1 - ρ)

ρ high = 0.96

.45 = 1/2 ln ((1 + ρ)/(1 - ρ))

2.46 = (1 + ρ)/(1 - ρ)

ρ low = 0.42

Since 0 is not even close to the interval for ρ, then we can conclude that there is a statistically significant relationship between El Niño and storm frequency off the East coast of the United States. Since r is positive, the relationship is: Strong El Niño means more storms. Our figure shows this relationship, but the statistical test confirms that it is a significant relationship. That's why statistics is important and useful!

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