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Question
You are required to calculate the coefficient of correlation for the above data set and
comment what conclusion Kushal should draw from the sample

Size(x)           Price(y)          X2                 Y2                 xy
108               12                11664              144                1296
39                13                1521               169                507
68.40             19                4678.56            361                1299.6
51                12                2601               144                612
10.40             6.50              108.16             42.25              67.6
4.40              4                 19.36              16                 17.6
Total=281.2       Total=66.5        Total=20592.08     Total=876.25       Total=3799.8



R=n∑xy –∑x*∑y /(n∑x2 – (∑x)2)1/2 *(n ∑y2 - (∑y)2)1/2

=6*3799.8 -281.2*66.5/(6*20592.08 –(281.2)2)*(6*876.25 –(66.5)2)

=22798.8 -18699.8 / 210.9 * 28.9

4099/6095.01

=0.6725

R2=0.452

Since it is less than 0.5 but more than 0 it is easy to draw a conclusion that sample is
moderately correlated.

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Stats

  • 1. Question You are required to calculate the coefficient of correlation for the above data set and comment what conclusion Kushal should draw from the sample Size(x) Price(y) X2 Y2 xy 108 12 11664 144 1296 39 13 1521 169 507 68.40 19 4678.56 361 1299.6 51 12 2601 144 612 10.40 6.50 108.16 42.25 67.6 4.40 4 19.36 16 17.6 Total=281.2 Total=66.5 Total=20592.08 Total=876.25 Total=3799.8 R=n∑xy –∑x*∑y /(n∑x2 – (∑x)2)1/2 *(n ∑y2 - (∑y)2)1/2 =6*3799.8 -281.2*66.5/(6*20592.08 –(281.2)2)*(6*876.25 –(66.5)2) =22798.8 -18699.8 / 210.9 * 28.9 4099/6095.01 =0.6725 R2=0.452 Since it is less than 0.5 but more than 0 it is easy to draw a conclusion that sample is moderately correlated.