Based on your scatter diagram, you would expect the correlation to be V . The mean x score is Mx = C] , and the mean y score is My = C]. Now, using the values for the means that you just calculated, ll out the following table by calculating the deviations from the means for X and Y, the squares of the deviations, and the products of the deviations. Scores Deviations Squared Deviations Products X Y XMx Y-MY (XMx)z (YMy)2 (XMx)(YMy) 3 1 .2 .3 E] e 4 z -1 -2 | 1| | 4| 2 e 3 1 -1 | 1| | 1| -1 2 4 -3 0 9 0 0 1o 10 5 6 [W W 30 The sum of squares for x is 55x = . The sum of squares for y is SSy = -. The sum of products is SP = . Because the sign of the sum of products is V , the sign of the correlation coefcient will be positive v . The correlation coefficient is r = 0.83 v . Look at your scatter diagram again. If you excluded the point (10, 10), you would expect the recalculated correlation coefficient to be different V , because 7 . Suppose you are given the following five pairs of scores: X Y 3 4 2 6 3 2 4 10 10Create a scatter diagram of these scores in the following diagram. For each of the five (X, Y) pairs, click on the plotting symbol (the black X) in the upper right corner of the tool, and drag it to the appropriate location on the grid. 10" + Based on your scatter diagram, you would expect the correlation to be 7