Correlation: From Picture to Formula

Transcription

1 Correlation: From icture to Formula KEYWORDS: Teaching; Scatter diagrams; Correlation coe cients. eter Holmes RSS Centre for Statistical Education, Nottingham Trent University, England. Summary Correlation is introduced intuitively early in the school curriculum by considering patterns in scatter diagrams. Later on, various formulae are used for calculating correlation coe cients. This article suggests ways in which the formulae can be related to the scatter diagrams. ^ INTRODUCTION ^ I n early secondary school, pupils are introduced to bivariate data and learn to interpret scatter diagrams. By considering patterns such as those in gure 1, they are introduced to the ideas of positive and negative correlation and whether this correlation is strong or weak. They may even see the etreme cases of perfect positive and perfect negative correlation. upils do not usually have much di culty relating to these qualitative descriptions and can usually identify, or even predict, in simple practical circumstances. Di culties arise when we try to quantify this qualitative understanding and introduce such formulae as y y r ˆ q y y 6 d t ˆ C D n n 1 which are the formulae for the earson roduct Moment Correlation Coe cient, the Spearman Rank Correlation Coe cient and the Kendall Rank Correlation Coe cient respectively. Why do these formulae give 1 for perfect positive correlation, a positive number for strong positive correlation and similar relationships to those shown in the scatter diagrams? Can we use the diagrams to derive the formulae intuitively? erfect ositive Correlation Strong ositive Correlation Around Zero Correlation Fig 1. Strong Negative Correlation erfect Negative Correlation Teaching Statistics. Volume 3, Number 3, Autumn

2 The rst problem is how to get any formula at all that relates to the diagrams. ^ A SIMLE CORRELATION ^ FORMULA Consider the scatter diagram divided into four regions by drawing lines parallel to the aes through a central point ^ see gure. The central point could be the medians or the means; for the rest of this article I shall assume that they are through the mean point ; y. Fig. y C D Any points that fall in regions A or C make the picture look more like positive correlation; any in B or D make it look more like negative correlation. Suppose there are n points altogether and that n(a) is the number in region A and similarly for n(b), n(c) and n(d). If we give a value of 1/n to every point in A or C and ^1/n to every point in B or D, and add, we obtain cor ˆ n A n C n B n D n What are the properties of cor? (i) If all the points are in A and C then cor ˆ 1. A B (ii) If all the points are in B and D then cor ˆ 1. (iii) If the points are in three or four regions then cor will lie between 1 and 1. (iv) If the points are predominantly in A and C then cor will be positive. (v) If the points are predominantly in B and D then cor will be negative. This is almost what we wanted in order to epress the qualitative ideas quantitatively. We have a formula to represent the picture. ^ THE EARSON RODUCT ^ MOMENT CORRELATION COEFFICIENT The formula for cor works, but it is rather crude. For eample, both the diagrams in gure 3 would give cor ˆ 1, but we would want to show that 3b is more positively correlated than 3a. We could do this by giving more weight to those points that are further away from the lines dividing the regions. After all, points near the lines could easily change sign by moving slightly; those further away establish the positive correlation more rmly. The distances from the lines are given by and y y (see gure 4). Now, and y y are positive or negative in the di erent regions and so is the product y y, as shown in table 1. So y y is positive in A and C, negative in B and D. The result of adding all the y y will therefore be positive or negative respectively in these regions. But the sum will not lie between 1 and 1. It depends on (i) the scale of and y (ii) the number of points. a b Fig Teaching Statistics. Volume 3, Number 3, Autumn 001

3 y D - A ^ SEARMAN'S RANK ^ CORRELATION COEFFICIENT y - y The formula is y 6 d i C B where d i ˆ r i r y i, where r i denotes the rank of the i th value and similarly for y i. Fig 4. A B C D y y y y Table 1. Signs of ; y y, and y y We can allow for (i) by epressing each distance y y in standard measures, and, where sd sd sd y stands for standard deviation. We can allow for (ii) by dividing by n. utting these together, we obtain 1 y y n sd sd y or, using sd ˆ 1 and similarly for y, n we obtain y y q y y which is the earson roduct Moment Correlation Coe cient (r). It is easy to show that when the points all lie on a straight line then the coe cient is either 1 or 1. If y ˆ a b then y ˆ a b and the formula simpli es to a p a ˆ p a a which equals 1 if a is positive and equals 1 if a is negative. It is also possible to prove that when the points are not on a straight line then the coe cient lies between 1 and 1. It is well known that, assuming no ranks are tied, Spearman's coe cient is the same as the product moment coe cient but calculated using ranks. So the above argument can be etended to this case ^ but this is rather indirect. It would be nice to have a more direct argument. Gri ths (1980) showed that the formula could be obtained by considering d as a measure of disagreement of ranks and standardizing this to lie between 1 and 1. But this does not tie it in with the scatter diagram. How about a direct approach? Assume that there are no tied ranks. If we plot the ranks instead of the original values the essential picture does not change; the and y values just become equally spaced. See gure 5 for an eample. In gure 6a the points are perfectly correlated with r i ˆ r y i. Figure 6b shows what happens when two of the points are not on the line. The distance of each point from the line is d i ˆ r i r y i and this can be taken as a measure of how far the new picture is from perfect positive correlation of 1. At the other etreme we have gure 6c, where there is perfect negative correlation. So, if we start with gure 6a having correlation 1, we subtract a multiple of d, say c d, aiming to reach a value of 1 when we have gure 6c. That is, we take c d i and choose c so that when there is perfect negative correlation, as in gure 6c, r s ˆ 1. How can we nd the value of c? The proof di ers depending on whether n is odd or even. Here is a proof for n odd ˆk 1. Figure 7 shows the perfect positive and perfect negative correlation lines. It can be seen that for perfect negative correlation Teaching Statistics. Volume 3, Number 3, Autumn

4 X d ˆ f k 1 k g ˆ 8f1... k 1 k g ˆ 8 Xk 8k k 1 k 1 r ˆ 6 1 and, since k ˆ n 1 =, we obtain X d ˆ n 1 n 1 n 3 So for perfect negative correlation we have 1 c n 1 n n 1 ˆ 1 3 and so 6 c ˆ n 1 n n 1 and 6 d i as required. A similar proof holds for even values of n. ^ KENDALL'S RANK CORRELATION ^ COEFFICIENT Looking at the original scatter diagram, we can compare all possible pairs of points. If there are no tied values then each pair is in relative position either (concordant pairs) or (discordant pairs). Each concordant pair makes the diagram look as though it has positive correlation; each discordant pair negative. There are n n 1 = pairs altogether. So, if there are altogether C concordant pairs and D discordant pairs, then C D will lie between 1 and 1 and will n n 1 = be 1 when all the pairs of points are concordant, 1 when all the points are discordant. In any other case it will lie between 1 and 1. C D So has the properties we need for a n n 1 correlation coe cient, and this is the Kendall coe cient. Other simple approaches to this coe cient are given by Wilkie (1980) and Noether (1981). Note. After I had independently come up with the coe cient that I have called cor I was told by Gottfried Noether that it was rst used by a statistician in the 190s working on nonparametric statistics. Unfortunately I did not make a note of the details and I have been unable to trace a reference. References Gri ths, D. (1980). A pragmatic approach to Spearman's Rank Correlation Coe cient. Teaching Statistics, (1), 10^13. Noether, G.E. (1981). Why Kendall Tau? Teaching Statistics, 3(), 41^. Wilkie, D. (1980). ictorial representation of Kendall's Rank Correlation Coe cient. Teaching Statistics, (3), 76^7. (All three references appear in Holmes,. (ed.) The Best of Teaching Statistics. Nottingham: Teaching Statistics Trust, available at Teaching Statistics. Volume 3, Number 3, Autumn 001

5 Fig 5. Fig 6. k+1 (1,k+1) erfect positive correlation rank r(y) k+1 k 4 (k,k+) 4 d i k (k,k) 1 (1,1) erfect negative correlation Fig 7. 1 k+1 rank r() k+1 Teaching Statistics. Volume 3, Number 3, Autumn