An auto manufacturing company wanted to investigate how the price of one of its car models depreciates with age. The research de
1 answer:
Answer:
a)
Price=150.414-16.626 Age
b)
$34.032
c)
We can't estimate the price of 18 year old car using the above regression equation because the price of 18 year old car falls outside the scope of the model
Step-by-step explanation:
Age(X) Price(Y)
8 18
3 94
6 50
9 21
2 145
5 42
6 36
3 99
a)
The regression equation can be written as
y=a+bx
a=ybar-bxbar
xbar=sumx/n
xbar=(8+3+6+9+2+5+6+3)/8=5.25
ybar=sumy/n
ybar=(18+94+50+21+145+42+36+99)/8=63.125
x y x-xbar y-ybar (x-xbar)^2 (x-xbar)(y-ybar)
8 18 2.75 -45.125 7.5625 -124.09375
3 94 -2.25 30.875 5.0625 -69.46875
6 50 0.75 -13.125 0.5625 -9.84375
9 21 3.75 -42.125 14.0625 -157.96875
2 145 -3.25 81.875 10.5625 -266.09375
5 42 -0.25 -21.125 0.0625 5.28125
6 36 0.75 -27.125 0.5625 -20.34375
3 99 -2.25 35.875 5.0625 -80.71875
Total 43.5 -723.25
b=-723.25/43.5
b=-16.626
a=ybar-bxbar
a=63.125-(16.626)*5.25
a=150.414
Thus, the required regression equation is
y=150.414-16.626 x
Price=150.414-16.626 Age
b)
For predicting price of 7 years old car we put x=7 in the estimated regression equation
y=150.414-16.626 x
y=150.414-16.626(7)
y=150.414-116.382
y=34.032
The predicted price of a 7 year old car is $34.032.
c)
We can't estimate the price of 18 year old car using the above regression equation because the price of 18 year old car falls outside the scope of the model. The estimate of price is only valid for 2 to 9 year old car.
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000001
000002
...
099998
099999
100000
There are obviously 100,000 strings in this set, so we have a one-to-one correspondence to the integers between 1 and 100,000. Think of any string starting with 0s as the number with the leading 0s chopped off.
There are two choices for the first digit, either 0 or 1, but a number can only contain a 6 if the first digit is 0; otherwise, the number would exceed 100,000. For every digits place afterward, if a given digits place contains a 6, then the remaining four places have 9 possible choices each, choosing from 0-9 excluding 6. If we fix the 6 in, say, the second digits place, then the number of integers between 1 and 100,000 containing exactly one 6 is
where the first 1 refers to the only choice of 0 in the first digits place, the second 1 refers to the unique 6 in the next place, and the remaining four places are filled with one of 9 possible choices.
Now, notice that we can permute the digits of such a number in 5 possible ways. That is, there are 5 choices for the placement of the 6 in the number, so we multiply this count by 5.
000001
000002
...
099998
099999
100000
There are obviously 100,000 strings in this set, so we have a one-to-one correspondence to the integers between 1 and 100,000. Think of any string starting with 0s as the number with the leading 0s chopped off.
There are two choices for the first digit, either 0 or 1, but a number can only contain a 6 if the first digit is 0; otherwise, the number would exceed 100,000. For every digits place afterward, if a given digits place contains a 6, then the remaining four places have 9 possible choices each, choosing from 0-9 excluding 6. If we fix the 6 in, say, the second digits place, then the number of integers between 1 and 100,000 containing exactly one 6 is
where the first 1 refers to the only choice of 0 in the first digits place, the second 1 refers to the unique 6 in the next place, and the remaining four places are filled with one of 9 possible choices.
Now, notice that we can permute the digits of such a number in 5 possible ways. That is, there are 5 choices for the placement of the 6 in the number, so we multiply this count by 5.