what is the value of the correlation coefficient r of the data set?\nvariable x 3 6 7 11 5 1 11\nvariable y…

Explanation:

Step1: Calculate the means

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

Step2: Calculate numerator and denominator components

Calculate $(x_i-\bar{x})(y_i - \bar{y})$, $(x_i-\bar{x})^2$ and $(y_i-\bar{y})^2$ for each $i$ from 1 to 7. Sum of $(x_i-\bar{x})(y_i - \bar{y})$: $(3 - 6.286)(10 - 26.143)+(6 - 6.286)(21 - 26.143)+(7 - 6.286)(18 - 26.143)+(11 - 6.286)(38 - 26.143)+(5 - 6.286)(21 - 26.143)+(1 - 6.286)(15 - 26.143)+(11 - 6.286)(60 - 26.143)$ $=(- 3.286)(-16.143)+(-0.286)(-5.143)+(0.714)(-8.143)+(4.714)(11.857)+(-1.286)(-5.143)+(-5.286)(-11.143)+(4.714)(33.857)$ $=53.04+1.47+(-5.82)+55.84 + 6.62+58.97+159.64$ $=330.76$

Sum of $(x_i-\bar{x})^2$: $(3 - 6.286)^2+(6 - 6.286)^2+(7 - 6.286)^2+(11 - 6.286)^2+(5 - 6.286)^2+(1 - 6.286)^2+(11 - 6.286)^2$ $=(-3.286)^2+(-0.286)^2+(0.714)^2+(4.714)^2+(-1.286)^2+(-5.286)^2+(4.714)^2$ $=10.79+0.08+0.51+22.22+1.65+27.94+22.22$ $=85.41$

Sum of $(y_i-\bar{y})^2$: $(10 - 26.143)^2+(21 - 26.143)^2+(18 - 26.143)^2+(38 - 26.143)^2+(21 - 26.143)^2+(15 - 26.143)^2+(60 - 26.143)^2$ $=(-16.143)^2+(-5.143)^2+(-8.143)^2+(11.857)^2+(-5.143)^2+(-11.143)^2+(33.857)^2$ $=260.59+26.45+66.32+140.59+26.45+124.18+1146.3$ $=1750.88$

Step3: Calculate correlation coefficient

The formula for the correlation coefficient $r$ is $r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}}$ $r=\frac{330.76}{\sqrt{85.41\times1750.88}}=\frac{330.76}{\sqrt{149535.58}}=\frac{330.76}{386.7}$ $r\approx0.84$

Answer:

B. 0.84