Question

correlation coefficient for the data. round all values to the nearest hundredth. the equation of the line of best fit of the data is y = . the correlation coefficient is r = .

Explanation:

Step1: List the data points

The data points from the graph are: $(1,1),(2,3),(3,2),(4,5),(5,3),(6,4)$. Let $x_i$ be the $x -$ values and $y_i$ be the $y -$ values.

Step2: Calculate the means of $x$ and $y$

$n = 6$.
$\bar{x}=\frac{1 + 2+3+4+5+6}{6}=\frac{21}{6}=3.5$
$\bar{y}=\frac{1 + 3+2+5+3+4}{6}=\frac{18}{6}=3$

Step3: Calculate the numerator of the correlation - coefficient formula

$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(1 - 3.5)(1 - 3)+(2 - 3.5)(3 - 3)+(3 - 3.5)(2 - 3)+(4 - 3.5)(5 - 3)+(5 - 3.5)(3 - 3)+(6 - 3.5)(4 - 3)$
$=(-2.5)\times(-2)+(-1.5)\times0+(-0.5)\times(-1)+0.5\times2 + 1.5\times0+2.5\times1$
$=5 + 0+0.5 + 1+0 + 2.5=9$

Step4: Calculate the denominator of the correlation - coefficient formula

$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(1 - 3.5)^2+(2 - 3.5)^2+(3 - 3.5)^2+(4 - 3.5)^2+(5 - 3.5)^2+(6 - 3.5)^2$
$=(-2.5)^2+(-1.5)^2+(-0.5)^2+0.5^2+1.5^2+2.5^2$
$=6.25+2.25 + 0.25+0.25+2.25+6.25 = 17.5$
$\sum_{i = 1}^{n}(y_i-\bar{y})^2=(1 - 3)^2+(3 - 3)^2+(2 - 3)^2+(5 - 3)^2+(3 - 3)^2+(4 - 3)^2$
$=(-2)^2+0^2+(-1)^2+2^2+0^2+1^2$
$=4 + 0+1+4+0+1 = 10$
The denominator is $\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}=\sqrt{17.5\times10}=\sqrt{175}\approx13.23$

Step5: Calculate the correlation coefficient $r$

$r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}}=\frac{9}{13.23}\approx0.68$

Step6: Calculate the slope $m$ and intercept $b$ of the line of best - fit

$m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}=\frac{9}{17.5}\approx0.51$
$b=\bar{y}-m\bar{x}=3-0.51\times3.5=3 - 1.785 = 1.215\approx1.22$
The equation of the line of best - fit is $y=0.51x + 1.22$

Answer:

The equation of the line of best fit of the data is $y = 0.51x+1.22$.
The correlation coefficient is $r = 0.68$.