the following is a set of data from a sample of n = 11 items. complete parts (a) through (c).\nx 18 19 9 7…

Explanation:

Step1: Recall the formula for correlation coefficient

The formula for the correlation coefficient $r$ is $r=\frac{S_{XY}}{S_XS_Y}$, where $S_{XY}$ is the sample covariance, $S_X$ is the sample standard - deviation of $X$, and $S_Y$ is the sample standard - deviation of $Y$. First, we need to find the means of $X$ and $Y$. Let $x_i$ and $y_i$ be the individual data points of $X$ and $Y$ respectively, and $n = 11$. The mean of $X$, $\bar{X}=\frac{\sum_{i = 1}^{n}x_i}{n}$. $\sum_{i=1}^{11}x_i=18 + 19+9+7+15+12+17+1+3+16+20=137$, so $\bar{X}=\frac{137}{11}\approx12.455$. The mean of $Y$, $\bar{Y}=\frac{\sum_{i = 1}^{n}y_i}{n}$. $\sum_{i=1}^{11}y_i=54 + 57+27+21+45+36+51+3+9+48+60=361$, so $\bar{Y}=\frac{361}{11}\approx32.818$. The sample standard - deviation of $X$ is $S_X=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{X})^2}{n - 1}}$. $\sum_{i = 1}^{n}(x_i-\bar{X})^2=(18 - 12.455)^2+(19 - 12.455)^2+(9 - 12.455)^2+(7 - 12.455)^2+(15 - 12.455)^2+(12 - 12.455)^2+(17 - 12.455)^2+(1 - 12.455)^2+(3 - 12.455)^2+(16 - 12.455)^2+(20 - 12.455)^2$ $=30.747+42.847 + 11.930+29.757+6.477+0.207+20.657+131.277+89.497+12.577+56.930 = 432.800$. $S_X=\sqrt{\frac{432.800}{10}}\approx6.580$. The sample standard - deviation of $Y$ is $S_Y=\sqrt{\frac{\sum_{i = 1}^{n}(y_i-\bar{Y})^2}{n - 1}}$. $\sum_{i = 1}^{n}(y_i-\bar{Y})^2=(54 - 32.818)^2+(57 - 32.818)^2+(27 - 32.818)^2+(21 - 32.818)^2+(45 - 32.818)^2+(36 - 32.818)^2+(51 - 32.818)^2+(3 - 32.818)^2+(9 - 32.818)^2+(48 - 32.818)^2+(60 - 32.818)^2$ $=448.629+584.729+33.859+139.669+148.429+10.129+330.529+889.229+567.329+230.429+738.829 = 3981.761$. $S_Y=\sqrt{\frac{3981.761}{10}}\approx19.954$. We know that $S_{XY}=129.818$ (given).

Step2: Calculate the correlation coefficient

$r=\frac{S_{XY}}{S_XS_Y}=\frac{129.818}{6.580\times19.954}=\frac{129.818}{131.307}\approx0.989$.

Answer:

$0.989$