A straight-line trend \(y_t = a + bX\) is fitted to a time series \(y\) by the method of least squares. Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(a\) | (I) | The change in the trend value for an increase of one unit in X |
| (B) | \(b\) | (II) | The deviation of an actual value from its trend value |
| (C) | \(y_t\) | (III) | The trend value at the origin (X = 0) |
| (D) | \(y - y_t\) | (IV) | The estimated (trend) value of the variable |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A) \(a\)
This matches (III). Putting X = 0 in \(y_t = a + bX\) gives \(y_t = a\), the trend value at the origin.
(B) \(b\)
This matches (I). When X increases by 1, \(y_t\) changes by \(b\); it is the slope of the trend line.
(C) \(y_t\)
This matches (IV). It is the value given by the line, the trend value or estimated value of \(y\).
(D) \(y - y_t\)
This matches (II). It is the deviation of the actual value from the trend value; the method of least squares makes the sum of the squares of these deviations least.