An appliance repair shop owner has fitted the quadratic trend equation ŷ = 90 + 0.9x + 3x² to a time series of annual repair orders, with y= the number of repair orders and x= 1 for 2000. Forecast the number of repair orders for 2008; for 2010
This is really a problem of evaluation of functions.
As indicated, for year
2000, x=1, so for
2008, x=2008-1999=9, and for
2010, x=2010-1999=11
Let
y = f(x) = 90+0.9x+3x²
for 2000, f(1) = 90 + 0.9*(1) + 3*1² = 93.9
for 2008, f(9) = 90 + 0.9*(9) + 3*9² = 341.1
for 2010, f(11)= 90 + 0.9*(11) + 3*11² = 462.9
Note that in general, the number of repairs is an integer (whole number), so the estimate should be rounded to the nearest integer.
To forecast the number of repair orders for 2008 and 2010 based on the quadratic trend equation provided, you need to substitute the corresponding values of x into the equation and compute the corresponding values of ŷ.
For 2008:
Since x = 1 in 2000, for 2008 x = (2008 - 2000) = 8 (the number of years elapsed).
To forecast the number of repair orders for 2008 (ŷ), substitute x = 8 into the equation:
ŷ = 90 + 0.9x + 3x²
ŷ = 90 + 0.9(8) + 3(8)²
ŷ = 90 + 7.2 + 3(64)
ŷ = 90 + 7.2 + 192
ŷ = 289.2
Therefore, the forecasted number of repair orders for 2008 is 289.2.
For 2010:
Similar to the previous calculation, find the value of x for 2010.
Since x = 1 in 2000, for 2010 x = (2010 - 2000) = 10.
To forecast the number of repair orders for 2010 (ŷ), substitute x = 10 into the equation:
ŷ = 90 + 0.9x + 3x²
ŷ = 90 + 0.9(10) + 3(10)²
ŷ = 90 + 9 + 3(100)
ŷ = 90 + 9 + 300
ŷ = 399
Therefore, the forecasted number of repair orders for 2010 is 399.