Question

Here is a bivariate data set.xy455642524351414849584645434541503546Find the correlation coefficient and report it accurate to four decimal places.r =

Answer 1

From the table, we can deduce the following:

(x, y) ==> (45, 56), (42, 52), (43, 51), (41, 48), (49, 58), (46, 45), (43, 45), (41, 50), (35, 46)

Let's find the correlation coefficient.

To find the correlation coefficient, apply the formula:

$r=\frac{\Sigma(\bar{x}-x_i)(\bar{y}-y_i)}{\sqrt{\Sigma(\bar{x}-x_i)^2\Sigma(\bar{y}-y_i)^2}}$

Where:

xi is the mean of the x-variables

yi is the mean of the y-variables.

Thus, we have:

$\begin{gathered} x_i=\frac{45+42+43+41+49+46+43+41+35}{9}=\frac{385}{9}=42.8 \\ \\ y_i=\frac{56+52+51+48+58+45+45+50+46}{9}=\frac{451}{9}=50.1 \end{gathered}$

Now, for the correlation coefficient, we have:

$\begin{gathered} \Sigma(x-x_i)^2=(45-42.8)^2+(42-42.8)^2+(43-42.8)^2+(41-42.8)^2+(49-42.8)^2+(46-42.8)^2+(43-42.8)^2+(41-42.8)^2+(35-42.8)^2 \\ \Sigma(x-x_i)^2=121.6 \\ \\ \Sigma(y-y_i)^2=(56-50.1)^2+(52-50.1)^2+(51-50.1)^2+(48-50.1)^2+(58-50.1)^2+(45-50.1)^2+(45-50.1)^2+(46-50.1)^2 \\ \Sigma(y-y_i)^2=174.9 \end{gathered}$

Solving further:

$\begin{gathered} \Sigma(x-x_i)(y-y_1)=(45-42.8)(56-50.1)+(42-42.8)(52-50.1)+(43-42.8)(51-50.1)+(41-42.8)(48-50.1)+(49-42.8)(58-50.1)+(46-42.8)(45-50.1)+(43-42.8)(45-50.1)+(41-42.8)(50-50.1)+(35-42.8)(46-50.1) \\ \\ \Sigma(x-x_i)(y-y_i)=79.2 \end{gathered}$

Now, plug in the values in the formula and solve for r.

We have:

$\begin{gathered} r=\frac{79.2}{\sqrt{(121.6)(174.9)}} \\ \\ r=\frac{79.2}{\sqrt{21267.84}} \\ \\ r=\frac{79.2}{145.835} \\ \\ r=0.5433 \end{gathered}$

Therefore, the correlation coefficient is 0.5433

• ANSWER:

0.5433