Method of Semi Averages

The secular trends 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 an 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 an even number of observations, the slope of the 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 the sum of $y$-values for the first half of the period, $S_2$ is the sum of $y$-values of the second half of the period, and $n$ is the number of time units covered by the time series.

The following merits and demerits of the Method of Semi Averages are as described:

Merits of Method 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.

Demerits of Method 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: Method of Semi Averages

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

Year1973197419751976197719781979
Property Damage201238392507484648742
  1. Obtain the semi-averages trend line
  2. Find out the trend values.

Solution

Let $x=t-1973$

YearProperty DamagedSemi TotalSemi AverageCoded YearTrend Values
1973201  0$y’=190+87(0)=190$
19742388312771$y’=190+87(1)=277$
1975392  2$y’=190+87(2)=364$
1976507  3$y’=190+87(3)=451$
1977484  4$y’=190+87(4)=538$
197854918756255$y’=190+87(5)=625$
1979742  6$y’=190+87(6)=712$
method of semi-averages (trend values)

\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: Method of Semi Averages

The following table gives the number of books in thousands sold at a bookstore for the years 1973 to 1981

Year197319741975197619771978197919801981
No. of Books Sold423835253224201917
  1. Find the equation of the semi-average trend line
  2. Compute the trend values
  3. Estimate the number of books sold for the year 1982.

Solution

Let $x=t-1973$

YearNo. of books (y)Semi TotalSemi AverageCoded yearTrend Values
197342  0$y’=39.5 – 3(0)=39.5$
197438140351$y’=39.5 – 3(1)=36.5$
1975352$y’=39.5 – 3(2)=33.5$
197625  3$y’=39.5 – 3(3)=30.5$
197732  4$y’=39.5 – 3(4)=27.5$
197824  5$y’=39.5 – 3(5)=24.5$
19792080206$y’=39.5 – 3(6)= 21.5$
198019  7$y’=39.5 – 3(7)=18.5$
198117  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 $y’=39.5-3(9)=12.5$.

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