Question

what is the value of the correlation coefficient r of the data set?
variable x 3 6 7 11 5 1 11
variable y 10 21 18 38 21 15 60
0.84
0.71
1
-0.84

Explanation:

Step1: Calculate the means

Let \(x = [3,6,7,11,5,1,11]\), \(y=[10,21,18,38,21,15,60]\).
The mean of \(x\), \(\bar{x}=\frac{3 + 6+7+11+5+1+11}{7}=\frac{44}{7}\approx6.286\).
The mean of \(y\), \(\bar{y}=\frac{10 + 21+18+38+21+15+60}{7}=\frac{183}{7}\approx26.143\).

Step2: Calculate numerator and denominators

The numerator \(S_{xy}=\sum_{i = 1}^{7}(x_i-\bar{x})(y_i - \bar{y})\).
\((3 - 6.286)(10 - 26.143)+(6 - 6.286)(21 - 26.143)+(7 - 6.286)(18 - 26.143)+(11 - 6.286)(38 - 26.143)+(5 - 6.286)(21 - 26.143)+(1 - 6.286)(15 - 26.143)+(11 - 6.286)(60 - 26.143)\)
\(=(- 3.286)(-16.143)+(-0.286)(-5.143)+(0.714)(-8.143)+(4.714)(11.857)+(-1.286)(-5.143)+(-5.286)(-11.143)+(4.714)(33.857)\)
\(=53.04+1.47+(-5.81)+55.83 + 6.62+58.97+159.62\)
\(=330.74\)

The denominator \(S_{xx}=\sum_{i = 1}^{7}(x_i-\bar{x})^2\)
\((3 - 6.286)^2+(6 - 6.286)^2+(7 - 6.286)^2+(11 - 6.286)^2+(5 - 6.286)^2+(1 - 6.286)^2+(11 - 6.286)^2\)
\(=(-3.286)^2+(-0.286)^2+(0.714)^2+(4.714)^2+(-1.286)^2+(-5.286)^2+(4.714)^2\)
\(=10.79+0.08+0.51+22.22+1.65+27.94+22.22\)
\(=85.41\)

The denominator \(S_{yy}=\sum_{i = 1}^{7}(y_i-\bar{y})^2\)
\((10 - 26.143)^2+(21 - 26.143)^2+(18 - 26.143)^2+(38 - 26.143)^2+(21 - 26.143)^2+(15 - 26.143)^2+(60 - 26.143)^2\)
\(=(-16.143)^2+(-5.143)^2+(-8.143)^2+(11.857)^2+(-5.143)^2+(-11.143)^2+(33.857)^2\)
\(=260.59+26.45+66.32+140.59+26.45+124.12+1146.32\)
\(=1740.84\)

The correlation coefficient \(r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{330.74}{\sqrt{85.41\times1740.84}}=\frac{330.74}{\sqrt{148651.19}}=\frac{330.74}{385.55}\approx0.84\)

Answer:

A. 0.84