Question 1197140
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I'll use x and y in place of n and V
x = n
y = V


x = number of years after 1990
y = value of an investment (in dollars)


Given Data:<table border = "1" cellpadding = "5"><tr><td>x</td><td>y</td></tr><tr><td>1</td><td>21120</td></tr><tr><td>3</td><td>20148.8</td></tr><tr><td>7</td><td>19440</td></tr><tr><td>12</td><td>18581.2</td></tr><tr><td>14</td><td>17480</td></tr><tr><td>19</td><td>15276</td></tr></table>There are n = 6 rows of data to represent the sample size.


We'll form the following columns:
x^2
xy

The x^2 column is where we square each x value
eg: 19 squares to 19^2 = 19*19 = 361

The xy column has us multiply each x and y value together (separately per row).
Eg: 3*20148.8 = 60446.4 in the second row of this column.

<table border = "1" cellpadding = "5"><tr><td>x</td><td>y</td><td>x^2</td><td>xy</td></tr><tr><td>1</td><td>21120</td><td>1</td><td>21120</td></tr><tr><td>3</td><td>20148.8</td><td>9</td><td>60446.4</td></tr><tr><td>7</td><td>19440</td><td>49</td><td>136080</td></tr><tr><td>12</td><td>18581.2</td><td>144</td><td>222974.4</td></tr><tr><td>14</td><td>17480</td><td>196</td><td>244720</td></tr><tr><td>19</td><td>15276</td><td>361</td><td>290244</td></tr></table>Use of a spreadsheet program is strongly recommended.


Then compute the following sums
P = sum of the x values = 56
Q = sum of the y values = 112046
R = sum of the x^2 values = 760
S = sum of the xy values = 975584.8


The regression line is of the form y = mx+b
m = slope = {{{(n*S - P*Q)/(n*R - P^2)}}}


b = y intercept = {{{(Q*R - P*S)/(n*R - P^2)}}}


Let's find the slope
{{{m = (n*S - P*Q)/(n*R - P^2)}}}


{{{m = (6*975584.8 - 56*112046)/(6*760 - 56^2)}}}


{{{m = -295.693258426966}}}


{{{m = -295.69}}} approximately


Now compute the y intercept
{{{b = (Q*R - P*S)/(n*R - P^2)}}}


{{{b = (112046*760 - 56*975584.8)/(6*760 - 56^2)}}}


{{{b = 21434.1370786517}}}


{{{b = 21434.14}}} approximately


The regression line is y = -295.69x + 21434.14 approximately
As a quick shortcut, you can use technology (eg: calculator or spreadsheet) to calculate the equation for the regression line.


x = 0 corresponds to the year 1990
Plug it in to find that y = 21434.14 which is exactly the y intercept.


Based on the regression model, the estimated value of the investment was about $21434.14 in the year 1990.


The slope tells us how much the investment goes up or goes down. 
In this case, a negative slope means the investment is decreasing in value. As x goes up y goes down.
The slope is roughly m = -295.69 to represent a decrease of $295.69 per year.



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Answers: 
<font color=red>$21434.14</font> was the approximate value of the investment in the year 1990 (estimate according to the regression line)


The investment decreases at a rate of about <font color=red>$295.69</font> per year (estimate according to the regression line)
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