Question

4. Scatter plots and calculating correlation 

Suppose you are given the following five pairs of scores: 

X	Y
6	1
9	2
6	3
8	4
1	10

Create a scatter plot 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. 

Based on your scatter plot, you would expect the correlation to be _______ .

The mean X score is Mx = _______ , and the mean core is My = _______ .

Now, using the values for the means that you just calculated, fill 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. Deviations Squared Deviations 

Scores		Deviations		Squared Deviations		Products
X	Y	X-MX	Y-MY	(X-MX)2	(Y-MY)2	(X-MX)(Y-MY)
6	1
9	2
6	3
8	4
1	10

The sum of squares for X is SSx = _______  The sum of squares for Y is SSy = _______ . The sum of products is SP = _______ .

Because the sign of the sum of products is _______ ,the sign of the Pearson correlation coefficient _______ .

The Pearson correlation coefficient is r = _______ .

Look at your scatter plot again. If you excluded the point (1, 10), you would expect the recalculated Pearson correlation coefficient to be _______ . because _______ 

• correlation is not the same as causation 

• the relationship between X and Y is nonlinear 

• (1, 10) is an outlier 

• X and Y are correlated

Answer 1

Result Details & Calculation Кеу X Values = 30 Mean = 6 EX - MX)2 = SSX = 38 X: X Values Y: Y Values My: Mean of X Values Mji Mean of Y Values X - MX & Y - Myi Deviation scores (X - MX) & (Y - My: Deviation Squared (X - MX)(Y - My): Product of Deviation Scores Y Values = 20 Mean = 4 ETY - My)2 = 55y = 50 X and Y Combined N= 5 E(X - MX/Y - My) = -36 R Calculation r= E(X - My/Y - Mx))/ VI(55x)(ssy)) r= -36 / ((38)(50)) = -0.8259 Meta Numerics (cross-check) r= -0.8259

Y Values X Values

X-MX Y - My (Y - My)2 0.000 3.000 0.000 2.000 -5.000 -3.000 -2.000 - 1.000 0.000 6.000 (X - MXJ2 0.000 9.000 0.000 4.000 25.000 9.000 4.000 1.000 0.000 36.000 (X-Mx)(Y - My) 0.000 -6.000 0.000 0.000 -30.000 Mx: 6.000 My: 4.000 Sum: 38.000 Sum: 50.000 Sum: -36.000
*Expect correlation would be 0.80

*mean of X = 6

mean of y= 4

*table in image

*

SSx=38

SSy=50

*SSp= -36 is negative so sign of correlation is is minus (negative )

*r=-0.8259

Answer 2

12 10 . 8 > 6 4 2 0 0 2 4 6 8 10 We would expect the correlation to be negative. Sample size, n = 5 Ex = 30 Ey = 25 X = Ex/n = 30/5 = 6 y = [y/n = 25/5 = 5 х у (x-x) (y-ỹ) (x-x)? (y-ỹ) (x-8)(y-ỹ) 6 2 0 -3 0 9 0 9 3 3 -2 9 4 -6 6 4 0 -1 0 1 0 8 6 2 4 1 2 1 10 -5 5 25 -25 SSxx = {(x-x)2 = 38 SSyy = {(y-7)2 = 40 SP = {(x-x)(y-7) = -29 Because the sign of the sum of products is negative, the sig of the correlation coefficient is negative. The correlation coefficient is r = SP/V(SSx*SSy) = -29/V(38*40) = -0.7438 If you excluded the point (1, 10), you would expect the recalculated correlation coefficient to be positive because it is an outlier or influential point. 1 25 Scanned with CamScanner

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