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Question: The present cost of the car is \[\$ 15,000\] . Its value depreciates at the rate shown above. Based ...

The present cost of the car is $15,000\$ 15,000 . Its value depreciates at the rate shown above. Based on a least-squares linear regression, what is the value, to the nearest hundred dollars of the car when t=4t = 4 ?

t=0t=1t=2t=4t=5
15,00013,00010,900?3000
Explanation

Solution

Hint : In this question, we are going to use the concept of depreciation. Here, we are going to use the least square linear regression. This least square linear regression is done by using the principal equation used to represent the equation of a line. Then we put in the values in the formula and evaluate the answer.
Formula Used:
Let the equation of the derived relation be y=mx+by = mx + b
Then, the slope (m)\left( m \right) is calculated by, m=N(xy)xyN(x2)(x)2m = \dfrac{{N\sum {\left( {xy} \right) - \sum {x\sum y } } }}{{N\sum {\left( {{x^2}} \right) - {{\left( {\sum x } \right)}^2}} }} (where \sum {} means “sums up”)
And y-intercept (b)\left( b \right) is calculated by, b=ymxNb = \dfrac{{\sum {y - m\sum x } }}{N}

Complete step-by-step answer :
The way to solve this type of question is given in a step-by-step method below:
Let xx and yy denote the two arguments of the question given. Let the equation of the derived relation of the question be y=mx+by = mx + b .
Step 1
For each (x,y)\left( {x,y} \right) point, find the value of x2{x^2} and xyxy and write them down.
Step 2
Sum all x,y,x2,xyx,y,{x^2},xy which gives us x,y,x2,xy\sum {x,\sum {y,\sum {{x^2},\sum {xy} } } }
x=8\sum x = 8
y=41900\sum {y = 41900}
x2=30\sum {{x^2} = 30}
xy=49800\sum {xy = 49800}
Step 3
Then, calculate the slope (m)\left( m \right) using the following formula:
m=N(xy)xyN(x2)(x)2\Rightarrow m = \dfrac{{N\sum {\left( {xy} \right) - \sum {x\sum y } } }}{{N\sum {\left( {{x^2}} \right) - {{\left( {\sum x } \right)}^2}} }}
In the question, N=4N = 4 (the number of given points)
So, we have m=4×4980041900×84×3082=2428.57143m = \dfrac{{4 \times 49800 - 41900 \times 8}}{{4 \times 30 - {8^2}}} = - 2428.57143
Step 4
Then, calculate the y-intercept (b)\left( b \right) using the following formula:
b=ymxN\Rightarrow b = \dfrac{{\sum {y - m\sum x } }}{N}
So, we have b=41900(2428.57143)×84=15332.1429b = \dfrac{{41900 - \left( { - 2428.57143} \right) \times 8}}{4} = 15332.1429
Step 5
Now, we assemble the equation of the line:
y=mx+by = mx + b
y=(2428.57143)x+15332.1429\Rightarrow y = \left( { - 2428.57143} \right)x + 15332.1429
For x=4x = 4 , we have:
y=(2428.57143)×4+15332.1429y = \left( { - 2428.57143} \right) \times 4 + 15332.1429
y=5617.857185600\therefore y = 5617.85718 \approx 5600
Hence, the value of a car after 4 years is going to be approximately 5,6005,600 .
So, the correct answer is “5,6005,600 ”.

Note : So, we saw that in solving questions like these, it is best to write the formula first, then the things which are given to us, then what is needed to be found. This approach makes the things very organized and the chances of making an error go down quite a lot. It is very important to follow all the steps and use the formula religiously as any small error will lead to an error, and hence making the answer all the way wrong. So, attention is needed to pay for this thing.