Class Notes (837,548)
Tony Quon (67)
Lecture 18

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Tony Quon
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If values of X are randomly related to values of Y (i.e., no relationship) then one would expect a covariance around zero as individual components of the sum in the covariance cancel each other out Correlation: The problem is that the size of the covariance will depend on units measured (i.e., you will get a different covariance if you use cm as opposed to inches) We therefore “normalize” the covariance by dividing by the standard deviations of X and Y to get the correlation: 1 ⎛ x −µ ⎞⎛ y −µ ⎞ ρ = ∑ ⎜ i X ⎟⎜ i Y⎟ n−1 ⎝ σ X ⎠⎝ σ Y ⎠ —> The population correlation coefﬁcient, , gives us a measure of the clustering around the line (i.e., the amount of proportionality between the explanatory and response variables independent of the units of measurement) Sample Estimate of the Correlation —> Of course, if we only have a sample of the population then we estimate the population correlation coefﬁcient, x, with the sample correlation coefﬁcient given by: ⎛ ⎞⎛ ⎞ r = 1 ∑ ⎜ xi− x ⎟⎜yi− y ⎟ n−1 ⎝ sX ⎠⎝ Y ⎠ —> It has the same properties as the population correlation coefﬁcient and is an unbiased estimator of x Properties of the Correlation Coefﬁcient 1.r has no units 2.-1 <= r <= 1 3. r = 1 implies perfect positive correlation and r = -1 implies perfect negative correlation 4.Its sign indicates positive or negative correlation 5. r only measures linear association, it doesn’t say anything about non-linear patterns, no matter how strong 6. Computational formula: 1 ∑ i i n(∑ xi)( i) r = n−1 s s ( ) X Y Example: Crickets and Temperature Chirps/sec 20 16 20 18 17 16 15 17 15 16 15 17 16 17 14 Temp (F)89 72 93 84 81 75 70 82 69 83 80 83 81 84 76 —> Mean chirps/sec = 16.6, standard error = 1.72 Mean temp/sec = 80.13, standard error = 6.72 1 ⎛X i x ⎞⎛Yi− x⎞ r = n−1∑ ⎜ s ⎟⎜ s ⎟ ⎝ x ⎠⎝ y ⎠ 1 ⎛⎡20−16.6 ⎤⎡−80.13 ⎤ ⎡14−16.6⎤⎡76−80.13
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