SOLUTION: Linear Regression The table below shows the value, V , of an investment (in dollars) n years after 1990. n 1 3 7 12 14 19 V 21120 20148.8 19440 18581.2 17480

Algebra ->  Linear-equations -> SOLUTION: Linear Regression The table below shows the value, V , of an investment (in dollars) n years after 1990. n 1 3 7 12 14 19 V 21120 20148.8 19440 18581.2 17480       Log On


   

Question 1197140: Linear Regression
The table below shows the value,
V
, of an investment (in dollars)
n
years after 1990.
n
1 3 7 12 14 19
V
21120 20148.8 19440 18581.2 17480 15276
Determine the linear regression equation that models the set of data above, and use this equation to answer the questions below. Round to the nearest hundredth as needed.
Based on this regression model, the value of this investment was $
in the year 1990.

Based on the regression model, the value of this investment is
Select an answer
at a rate of $
per year.
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

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:
xy
121120
320148.8
719440
1218581.2
1417480
1915276
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.
xyx^2xy
121120121120
320148.8960446.4
71944049136080
1218581.2144222974.4
1417480196244720
1915276361290244
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 = %28n%2AS+-+P%2AQ%29%2F%28n%2AR+-+P%5E2%29

b = y intercept = %28Q%2AR+-+P%2AS%29%2F%28n%2AR+-+P%5E2%29

Let's find the slope
m+=+%28n%2AS+-+P%2AQ%29%2F%28n%2AR+-+P%5E2%29

m+=+%286%2A975584.8+-+56%2A112046%29%2F%286%2A760+-+56%5E2%29

m+=+-295.693258426966

m+=+-295.69 approximately

Now compute the y intercept
b+=+%28Q%2AR+-+P%2AS%29%2F%28n%2AR+-+P%5E2%29

b+=+%28112046%2A760+-+56%2A975584.8%29%2F%286%2A760+-+56%5E2%29

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:
$21434.14 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 $295.69 per year (estimate according to the regression line)