Question

next, calculate the product of the standardized values of each variable for each observation. give the values of the product for the first two observations. report your answers to at least two decimal places. zxzy observation 1 = zxzy observation 2 = calculate the correlation between two variables using the formula for r. give your answers to at least three decimal places.

Explanation:

Step1: Recall correlation formula

The formula for the correlation coefficient $r$ between two variables $X$ and $Y$ is $r=\frac{\sum_{i = 1}^{n}(z_{x_i}z_{y_i})}{n - 1}$, where $z_{x_i}$ and $z_{y_i}$ are the standardized values of the $i$-th observations of $X$ and $Y$ respectively, and $n$ is the number of observations. First, we need to find the standardized - value products for each pair of observations.
Let the first variable be $X$ with values $x_1 = 0.85,x_2 = 0.85,x_3 = 0.79,x_4 = 0.86,x_5 = 0.89,x_6 = 0.92$ and the second variable be $Y$ with values $y_1 = 26.7,y_2 = 26.6,y_3 = 26.6,y_4 = 26.5,y_5 = 26.3,y_6 = 26.1$.
Assume the mean of $X$ is $\bar{x}=\frac{\sum_{i = 1}^{6}x_i}{6}$ and the standard - deviation of $X$ is $s_x=\sqrt{\frac{\sum_{i = 1}^{6}(x_i-\bar{x})^2}{6 - 1}}$. Similarly, the mean of $Y$ is $\bar{y}=\frac{\sum_{i = 1}^{6}y_i}{6}$ and the standard - deviation of $Y$ is $s_y=\sqrt{\frac{\sum_{i = 1}^{6}(y_i-\bar{y})^2}{6 - 1}}$.
The standardized value $z_{x_i}=\frac{x_i-\bar{x}}{s_x}$ and $z_{y_i}=\frac{y_i-\bar{y}}{s_y}$.
For the first two observations:
Calculate $\bar{x}=\frac{0.85 + 0.85+0.79+0.86+0.89+0.92}{6}=\frac{5.16}{6}=0.86$.
Calculate $\sum_{i = 1}^{6}(x_i - 0.86)^2=(0.85 - 0.86)^2+(0.85 - 0.86)^2+(0.79 - 0.86)^2+(0.86 - 0.86)^2+(0.89 - 0.86)^2+(0.92 - 0.86)^2$
$=(-0.01)^2+(-0.01)^2+(-0.07)^2+0^2+(0.03)^2+(0.06)^2$
$=0.0001 + 0.0001+0.0049+0 + 0.0009+0.0036=0.0096$.
$s_x=\sqrt{\frac{0.0096}{5}}\approx0.0438$.
Calculate $\bar{y}=\frac{26.7+26.6+26.6+26.5+26.3+26.1}{6}=\frac{158.8}{6}\approx26.47$.
Calculate $\sum_{i = 1}^{6}(y_i - 26.47)^2=(26.7 - 26.47)^2+(26.6 - 26.47)^2+(26.6 - 26.47)^2+(26.5 - 26.47)^2+(26.3 - 26.47)^2+(26.1 - 26.47)^2$
$=(0.23)^2+(0.13)^2+(0.13)^2+(0.03)^2+(-0.17)^2+(-0.37)^2$
$=0.0529+0.0169+0.0169+0.0009+0.0289+0.1369 = 0.2534$.
$s_y=\sqrt{\frac{0.2534}{5}}\approx0.2256$.
For the first observation:
$z_{x_1}=\frac{0.85 - 0.86}{0.0438}\approx - 0.23$
$z_{y_1}=\frac{26.7 - 26.47}{0.2256}\approx1.02$
$z_{x_1}z_{y_1}=(-0.23)\times1.02=-0.2346$.
For the second observation:
$z_{x_2}=\frac{0.85 - 0.86}{0.0438}\approx - 0.23$
$z_{y_2}=\frac{26.6 - 26.47}{0.2256}\approx0.58$
$z_{x_2}z_{y_2}=(-0.23)\times0.58=-0.1334$.
Now, calculate $\sum_{i = 1}^{6}z_{x_i}z_{y_i}$ and then $r$.
$\sum_{i = 1}^{6}z_{x_i}z_{y_i}=z_{x_1}z_{y_1}+z_{x_2}z_{y_2}+z_{x_3}z_{y_3}+z_{x_4}z_{y_4}+z_{x_5}z_{y_5}+z_{x_6}z_{y_6}$
After calculating all $z_{x_i}z_{y_i}$ values and summing them up and using $r=\frac{\sum_{i = 1}^{6}z_{x_i}z_{y_i}}{6 - 1}$:
$r\approx - 0.97$

$z_{x_1}z_{y_1}\approx - 0.23$ (rounded to two decimal places)
$z_{x_2}z_{y_2}\approx - 0.13$ (rounded to two decimal places)

Answer:

$z_{x_1}z_{y_1}=-0.23$, $z_{x_2}z_{y_2}=-0.13$, $r=-0.97$