Question

In: Math

Here is a bivariate data set. x y 50.4 96.8 53.7 27.6 -17.6 158.6 49.9 -48.1...

Here is a bivariate data set.

x	y
50.4	96.8
53.7	27.6
-17.6	158.6
49.9	-48.1
30.5	-6.1
36.7	179
43.5	-3.9
40.5	65.1
52.1	-129.7
43.5	85.1
33.4	-41.4
63.4	-50
25.3	31.8
41.2	11.8
34.4	149.4
50.5	-50.5

Find the correlation coefficient and report it accurate to three decimal places.
r =

Expert Solution

Solution:

Correlation coefficient = r = [n∑xy - ∑x∑y]/sqrt[(n∑x^2 – (∑x)^2)*(n∑y^2 – (∑y)^2)]

The calculation table is given as below:

No.	x	y	x^2	y^2	xy
1	50.4	96.8	2540.16	9370.24	4878.72
2	53.7	27.6	2883.69	761.76	1482.12
3	-17.6	158.6	309.76	25153.96	-2791.36
4	49.9	-48.1	2490.01	2313.61	-2400.19
5	30.5	-6.1	930.25	37.21	-186.05
6	36.7	179	1346.89	32041	6569.3
7	43.5	-3.9	1892.25	15.21	-169.65
8	40.5	65.1	1640.25	4238.01	2636.55
9	52.1	-129.7	2714.41	16822.09	-6757.37
10	43.5	85.1	1892.25	7242.01	3701.85
11	33.4	-41.4	1115.56	1713.96	-1382.76
12	63.4	-50	4019.56	2500	-3170
13	25.3	31.8	640.09	1011.24	804.54
14	41.2	11.8	1697.44	139.24	486.16
15	34.4	149.4	1183.36	22320.36	5139.36
16	50.5	-50.5	2550.25	2550.25	-2550.25
Total	631.4	475.5	29846.18	128230.2	6290.97

From above table, we have

n = 16

∑x = 631.4

∑y = 475.5

∑x^2 = 29846.18

∑y^2 = 128230.2

∑xy = 6290.97

Correlation coefficient = r = [n∑xy - ∑x∑y]/sqrt[(n∑x^2 – (∑x)^2)*(n∑y^2 – (∑y)^2)]

Correlation coefficient = r = [16*6290.97 - 631.4*475.5]/sqrt[(16*29846.18 – 631.4^2)*(16*128230.2– 475.5^2)]

Correlation coefficient = r = -0.52595

r = -0.526

milcah answered 1 year ago

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