Method of Semi-Averages

The secular trend can also be measured by the method of semi-averages. The steps are:

• Divide the time series data into two equal portions. If observations are odd then either omit the middle value or include the middle value in each half.
• Take the average of each part and place these average values against the midpoints of the two parts.
• Plot the semi-averages in the graph of the original values.
• Draw the required trend line through these two potted points and extend it to cover the whole period.
• It is simple to compute the slope and $y$-intercept of the line drawn from two points. The trend values can be found from the semi-average trend line or by estimated straight line as explained:

Let $y’_1$ and $y’_2$ be the semi-averages placed against the times $x_1$ and $x_2$. Let the estimated straight line $y’=a+bx$ is to pass through the points ($x_1$, $y’_1$) and ($x_2$, $y’_2$). The constant “$a$” and “$b$” can easily be determined. the equation of the line passing through the points ($x_1$, $y’_1$) and ($x_2$, $y’_2$) can be written as:

\begin{align*}
y’ – y’_1 &= \frac{y’_2-y’_1}{x_2-x_1}(x-x_1)\\
&= b(x-x_1)\\
\Rightarrow y’ &= (y’_1 – bx_1) + bx\\
&= a+bx, \quad \text{ where $a=y’_1-bx_1$}
\end{align*}

For even number of observations the slope of trend line can be found as:

\begin{align*}
b&=\frac{1}{n/2}\left(\frac{S_2}{n/2} – \frac{S_1}{n/2} \right)\\
&= \frac{1}{n/2} \left(\frac{S_2-S_1}{n/2}\right)\\
&= \frac{4(S_2-S_1)}{n^2},
\end{align*}

where $S_1$ is sum of $y$-values for the first half of the period, $S_2$ is sum of $y$-values of the second half of the period, and $n$ is the number of time units covered by the time-series.

Merits of Semi-Averages

• The method of semi-averages is simple, easy, and quick.
• It smooths out seasonal variations
• It gives a better approximation to the trend because it is based on a mathematical model.

Demeris of Semi-Averages

• It is a rough and objective method.
• The arithmetic mean used in Semi Average is greatly affected by very large or by very small values.
• The method of semi-averages is applicable when the trend is linear. This method is not appropriate if the trend is not linear.

Numerical Example 1:

The following table shows the property damaged by road accidents in Punjab for the year 1973 to 1979.

Year	1973	1974	1975	1976	1977	1978	1979
Property Damage	201	238	392	507	484	648	742

1. Obtain the semi-averages trend line
2. Find out the trend values.

Solution

Let $x=t-1973$

Year	Property Damaged	Semi Total	Semi Average	Coded Year	Trend Values
1973	201			0	$y’=190+87(0)=190$
1974	238	831	277	1	$y’=190+87(1)=277$
1975	392			2	$y’=190+87(2)=364$
1976	507			3	$y’=190+87(3)=451$
1977	484			4	$y’=190+87(4)=538$
1978	549	1875	625	5	$y’=190+87(5)=625$
1979	742			6	$y’=190+87(6)=712$

\begin{align*}
y’_1 &= 277, x_1 = 1, y’_2 = 625, x_2=5\\
b&=\frac{y’_2-y’_1}{x_2-x_1}=\frac{625-277}{5-1}=87\\
a&=y’_1 – bx_1 = 277-87(1)=190
\end{align*}

The semi-average trend line $y’=190+87x$ (with the origin at 1973)

Numerical Example 2:

The following table gives the number of books in thousands sold at a book store for the year 1973 to 1981

Year	1973	1974	1975	1976	1977	1978	1979	1980	1981
No. of Books Sold	42	38	35	25	32	24	20	19	17

1. Find the equation of the semi-averages trend line
2. Compute the trend values
3. Estimate the number of books sold for the year 1982.

Solution

Let $x=t-1973$

Year	No. of books (y)	Semi Total	Semi Average	Coded year	Trend Values
1973	42			0	$y’=39.5 – 3(0)=39.5$
1974	38	140	35	1	$y’=39.5 – 3(1)=36.5$
1975	35	2	$y’=39.5 – 3(2)=33.5$
1976	25			3	$y’=39.5 – 3(3)=30.5$
1977	32			4	$y’=39.5 – 3(4)=27.5$
1978	24			5	$y’=39.5 – 3(5)=24.5$
1979	20	80	20	6	$y’=39.5 – 3(6)= 21.5$
1980	19			7	$y’=39.5 – 3(7)=18.5$
1981	17			8	$y’=39.5 – 3(8)=15.5$

\begin{align*}
y’_1 &= 35, x_1=1.5, y’_2=20, x_2=6.5\\
b &= \frac{y’_2 – y’_1}{x_2-x_1} = \frac{20-35}{6.5-1.5} =-3\\
a &= y’_1 – bx_1 = 35 – (-3)(1.5) = 39.5\\
y’&= 39.5 – 3x (\text{with origin at 1973})
\end{align*}

For the year 1982, the estimated number of books sold is: $y’=39.5-3(9)=12.5$.