question 4\n1 pts\nwhat is the value of the correlation coefficient r of the data set?\nvariable x 3 6 7 11…

Explanation:

Step1: Calculate means of x and y

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

Step2: Calculate numerator and denominator components

$n = 7$. For numerator: $\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(3 - 6.286)(10 - 26.143)+(6 - 6.286)(21 - 26.143)+(7 - 6.286)(18 - 26.143)+(11 - 6.286)(38 - 26.143)+(5 - 6.286)(21 - 26.143)+(1 - 6.286)(15 - 26.143)+(11 - 6.286)(60 - 26.143)$ $=(- 3.286)(-16.143)+(-0.286)(-5.143)+(0.714)(-8.143)+(4.714)(11.857)+(-1.286)(-5.143)+(-5.286)(-11.143)+(4.714)(33.857)$ $=53.04+1.47+(-5.81)+55.83+6.62+58.94+159.67$ $=330.76$

For denominator: $\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}$ $\sum_{i = 1}^{n}(x_i-\bar{x})^2=(3 - 6.286)^2+(6 - 6.286)^2+(7 - 6.286)^2+(11 - 6.286)^2+(5 - 6.286)^2+(1 - 6.286)^2+(11 - 6.286)^2$ $=(-3.286)^2+(-0.286)^2+(0.714)^2+(4.714)^2+(-1.286)^2+(-5.286)^2+(4.714)^2$ $=10.8+0.08+0.51+22.22+1.65+27.94+22.22$ $=85.42$

$\sum_{i = 1}^{n}(y_i - \bar{y})^2=(10 - 26.143)^2+(21 - 26.143)^2+(18 - 26.143)^2+(38 - 26.143)^2+(21 - 26.143)^2+(15 - 26.143)^2+(60 - 26.143)^2$ $=(-16.143)^2+(-5.143)^2+(-8.143)^2+(11.857)^2+(-5.143)^2+(-11.143)^2+(33.857)^2$ $=260.59+26.45+66.32+140.59+26.45+124.12+1146.2$ $=1740.72$

$\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}=\sqrt{85.42\times1740.72}=\sqrt{148751.98}=385.68$

Step3: Calculate correlation coefficient r

$r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}}=\frac{330.76}{385.68}\approx0.84$

Answer:

D. 0.84