If \((t_1, y_1), (t_2,y_2)\ldots (t_n, y_n)\) denote the time series and \(y_t\) are the trend values of the variables y, then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\sum_{i=1}^{n}(y_i-y_t) = 0\)
The trend line \(y_t = a + bt\) fitted by the least squares method satisfies the first normal equation:
\( \displaystyle \sum_{i=1}^{n} y_i = na + b\sum_{i=1}^{n} t_i \)
The right side is the sum of the trend values, \(\sum (a + bt_i) = \sum y_t\). So
\( \displaystyle \sum_{i=1}^{n} y_i = \sum y_t \)
\( \displaystyle \sum_{i=1}^{n}(y_i - y_t) = 0 \)
The actual values lie above and below the trend line, and the deviations cancel out.
CBSE Applied Mathematics (Class XII):
Computation of Straight-line Trend by Using Method of Least Squares
