Question

Fit a straight line trend by the method of least squares to the following data

Year	1980	1981	1982	1983	1984	1985	1986	1987
Sales	50.3	52.7	49.3	57.3	56.8	60.7	62.1	58.7

Answer 1

Year (x)	Sales (y)	x = `((x - 1983.5))/0.5`	xy	x2	Trend values (yt)
1980	50.3	– 7	– 352.1	49	50.1775
1981	52.7	– 5	– 263.5	25	51.8375
1982	49.3	– 3	– 147.9	9	53.4975
1983	57.3	– 1	– 57.3	1	55.1575
1984	56.8	1	56.8	1	56.8175
1985	60.7	3	182.1	9	58.4775
1986	62.1	5	310.5	25	60.1375
1987	58.7	7	410.9	49	61.7975
N = 8	`sum` = 447.9	`sumx` = 0	`sumxy` = 139.5	`sumx^2` = 168	`sumy"t"` = 447.9

a = `(sumy)/"n" = 447.9/8` = 55.9875

b = `(sumxy)/(sumx^2) = 139.5/168` = 0.830

Therefore, the required equation of the straight line trend is given by

y = a + bx

y = 55.9875 + 0.830x

⇒ y = 55.9875 + 0.83`((x - 1983.5)/0.5)`

The trend values can be obtained by

When x = 1980

y = 55.9875 + 0.83 `((1980 - 1983.5)/0.5)`

= 55.9875 + 0.83(– 7)

= 55.9875 – 5.81

= 50.1775

When x = 1981

y = 55.9875 + 0.83 `((1981 - 1983.5)/0.5)`

= 55.9875 + 0.83(– 5)

= 55.9875 – 4.15

= 51.8375

When x = 1982

y = 55.9875 + 0.83 `((1981 - 1983.5)/0.5)`

= 55.9875 + 0.83(– 3)

= 55.9875 – 2.49

= 53.4975

When x = 1983

y = 55.9875 + 0.83 `((1983 - 1983.5)/0.5)`

= 55.9875 + 0.83(– 1)

= 55.9875 – 0.83

= 55.1575

When x = 1984

y = 55.9875 + 0.83 `((1984 - 1983.5)/0.5)`

= 55.9875 + 0.83(1)

= 56.8175

When x = 1985

y = 55.9875 + 0.83 `((1985 - 1983.5)/0.5)`

= 55.9875 + 0.83(3)

= 55.9875 + 2.49

= 58.4775

When x = 1986

y = 55.9875 + 0.83 `((1986 - 1983.5)/0.5)`

= 55.9875 + 0.83(5)

= 55.9875 + 4.15

= 60.1375

When x = 1987

y = 55.9875 + 0.83 `((1987 - 1983.5)/0.5)`

= 55.9875 + 0.83(7)

= 55.9875 + 5.81

= 61.7975