Regression Equations

Question

The income of a small business for several years is given in the table below.

Year/Thousands of Dollars

1960/13.8

1961/58.6

1962/66.8

1963/61.8

1964/41.1

1965/28.8

1966/35

1967/67.4

a)  Perform regression to find a cubic polynomial that fits the data. Use years since 1960 as your inputs.

b)  Use the model to predict the income of the business in 1970.

c)  Use the model to predict when the business income will reach $800,000.

Answer 1

a)

Need to fit

y=f(x) = ax³+bx²+cx+d

Using linear algebra, I get

a=2.07

b=-22.6

c=64.297

d = 13.9773

b)

f(10) = 466.87 thousands of dollars

c)

ax³+bx²+cx+d = 800

You can try to solve for x, but it is easier to incrementally try different values for x, starting with 11

f(11) = 741

f(11.1) = 774

f(11.12) = 780.6

f(11.17) = 797.2

f(11.18) = 800.56

So, around 11.18 years or during 1971

You can also plot y=800 against f(x) with demos, and get 11.178 years at the intersection of the two curves.

Answer 2

a.) The cubic regression equation is 2.074X^3-22.6061x^2+64.297x+13.977

b.) $470,000

c.) At around 11 years.