Question

Following table shows the number of traffic fatalities (in a state) resulting from drunken driving for the years 1975 to 1983:

Years	1975	1976	1977	1978	1979	1780	1981	1982	1983
No. of deaths	0	6	3	8	2	9	4	5	10

 Fit a trend line to the above data by the method of least squares.

Answer 1

In the given problem, n = 9 (odd)

From the given table, the middle t value is 1979, h = 1.

u = `(t - "middle value")/h`

u = `(t - 1979)/1`

∴ u = t − 1979

We can tabulate the given data as follows:

Year (t)	No. of deaths (y)	u = t − 1979	u2	uy	Trend Value
1975	0	−4	16	0	2.5554
1976	6	−3	9	−18	3.2221
1977	3	−2	4	−6	3.8888
1978	8	−1	1	−8	4.5555
1979	2	0	0	0	5.2222
1980	9	1	1	9	5.8887
1981	4	2	4	8	6.556
1982	5	3	9	15	7.2223
1983	10	4	16	40	7.8890
Total	47	0	60	40

From the table, n = 9, ∑y = 47, ∑u = 0, ∑u² = 60, ∑uy = 40

The two normal equations are

∑y = na + b∑u and ∑uy = a∑u + b∑u²

Substituting the values in the normal equations, we get

47 = 9a + b × 0   ...(i) and

40 = a(0) + b(60)  ...(ii)

From (i), a = `47/9 = 5.222`

From (ii), a = `40/60 = 0.6667`

Equation of the trend line is y = a + bu

i.e., y = 5.2222 + 0.6667u,

where u = t − 1979

Substituting u = −4, −3, −2, −1, 0, 1, 2, 3, 4, we will get the trend values and put them in the table as in the last column.