Question

Find the value of the linear correlation coefficient r. The paired data below consist of the
temperatures on randomly chosen days and the amount a certain kind of plant grew (in
millimeters).

Temperature	62	76	50	51	71	46	51	44	79
Growth	36	39	50	13	33	33	17	6	16

A) 0.196

B) 0.256

C) 0

D) -0.210

Answer 1

X	y	(x-xbar)^2	(y-ybar)^2	(x-xbar)(y-ybar)
62 76 50 51 71 46 51 44 79	36 39 50 13 33 33 17 6 16	9.679 292.790 79.012 62.235 146.679 166.123 62.235 221.679 404.457 Sum: 1444.889	81.000 144.000 529.000 196.000 36.000 36.000 100.000 441.000 121.000 Sum: 1684.000	28.000 205.333 -204.444 110.444 72.667 -77.333 78.889 312.667 -221.222 Sum: 305.000

X Values
∑ = 530
Mean = 58.889
∑(X - Mx)2 = SSx = 1444.889

Y Values
∑ = 243
Mean = 27
∑(Y - My)2 = SSy = 1684

X and Y Combined
N = 9
∑(X - Mx)(Y - My) = 305

R Calculation
r = ∑((X - My)(Y - Mx)) / √((SSx)(SSy))

r = 305 / √((1444.889)(1684)) = 0.196

So answer is

Option (a)