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Question
The sales of a commodity in tones varied from January 2010 to December 2010 as follows:
| In Year 2010 | Sales (in tones) |
| Jan | 280 |
| Feb | 240 |
| Mar | 270 |
| Apr | 300 |
| May | 280 |
| Jun | 290 |
| Jul | 210 |
| Aug | 200 |
| Sep | 230 |
| Oct | 200 |
| Nov | 230 |
| Dec | 210 |
Fit a trend line by the method of semi-average
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Solution
Since the number of months is even (12), we can equally divide the given data into two equal parts and obtain the averages of the first six months and last six months
| In Year 2010 |
Sales (in tones) |
Average |
| Jan | 280 | `(280 + 240 + 270 + 300 + 280 + 290)/6` = 276.667 |
| Feb | 240 | |
| Mar | 270 | |
| Apr | 300 | |
| May | 280 | |
| Jun | 290 | |
| Jul | 210 | `(210 + 200 + 230 + 200 + 230 + 210)/6` = 213.33 |
| Aug | 200 | |
| Sep | 230 | |
| Oct | 200 | |
| Nov | 230 | |
| Dec | 210 |
Thus we obtain semi-average I = 276.667 and semi-average II = 213.333
To fit a trend line we plot each value at the mid-point (month) of each half.
i.e we plot 276.667 in the middle of March and April
we plot 213.333 in the middle of September and October.
We join the two points by a straight line.
This is the required line.
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