Question

2.40 Stretching a scatterplot. Changing the units of

measurement can greatly alter the appearance of a scat-
terplot. Consider the following data: STRETCH

x −4 −4 −3 3 4 4
y 0.5 −0.6 −0.5 0.5 0.5 −0.6
(a) Draw x and y axes each extending from −6 to 6. Plot
the data on these axes.
(b) Calculate the values of new variables x* = x/10 and
y* = 10y, starting from the values of x and y. Plot y*
against x*
on the same axes using a different plotting
symbol. The two plots are very different in appearance.

(c) Find the correlation between x and y. Then find
the correlation between x*
and y*
. How are the two
correlations related? Explain why this isn’t surprising.

Answer 1

solution:

a)

Scatter plot of x and y:

b).

Scatter plot of x*and y*:

c).

For x and y:

X	Y	XY	X²	Y²
-4	0.5	-2	16	0.25
-4	-0.6	2.4	16	0.36
-3	-0.5	1.5	9	0.25
3	0.5	1.5	9	0.25
4	0.5	2	16	0.25
4	-0.6	-2.4	16	0.36
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
0	-0.2	3	82	1.72

Sample size, n = 6

SSxx = Ʃx² - (Ʃx)²/n = 82 - (0)²/6 = 82

SSyy = Ʃy² - (Ʃy)²/n = 1.72 - (-0.2)²/6 = 1.71333333

SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 3 - (0)(-0.2)/6 = 3

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 3/√(82*1.71333) = 0.2531

-----------------

For x* and y*:

X	Y	XY	X²	Y²
-0.4	5	-2	0.16	25
-0.4	-6	2.4	0.16	36
-0.3	-5	1.5	0.09	25
0.3	5	1.5	0.09	25
0.4	5	2	0.16	25
0.4	-6	-2.4	0.16	36
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
0	-2	3	0.82	172

Sample size, n = 6

SSxx = Ʃx² - (Ʃx)²/n = 0.82 - (0)²/6 = 0.82

SSyy = Ʃy² - (Ʃy)²/n = 172 - (-2)²/6 = 171.333333

SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 3 - (0)(-2)/6 = 3

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 3/√(0.82*171.33333) = 0.2531

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