Question

The table below represents the closing prices of stock ABC for the last 5 days. What is the r-value of the linear regression that fits these data:
Day Value
1 472.08
2 454.26
3 444.95
4 439.49
5 436.55

Answer 1

To answer this one, you may use scientific calculator that is capable of generating the value of r in the regression. We let the day be the values of x and the value be the values of y. By performing the task in the scientific calculator, it was found out that the value of r is -0.947.

Answer 2

Answer:

The r-value of the linear regression that fits these data is -0.947110707.

Step-by-step explanation:

The linear regression equation is in the form of

$y=bx+a$

Where,

$b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}$

$a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}$

The formula of r is

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n(\sum x^2)-(\sum x)^2][n(\sum y^2)-(\sum y)^2]}}$

The values are

$\sum x=15,\sum y=2247.35,\sum x^2=55,\sum y^2=1010938.423,\sum xy=6656.18, n=5$

Using above formula, we get

$r=-0.947110707$

Therefore the r-value of the linear regression that fits these data is -0.947110707.