A straight-line trend \(y = a + bX\), where X = year − 2021, is fitted to the following data by the method of least squares.
| Year | 2018 | 2019 | 2020 | 2021 | 2022 | 2023 | 2024 |
| Output (in tonnes) | 7 | 10 | 10 | 13 | 14 | 18 | 19 |
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | The value of \(a\) | (I) | 2 |
| (B) | The value of \(b\) | (II) | 9 |
| (C) | The trend value for the year 2025 | (III) | 13 |
| (D) | The 3-year moving average for the year 2019 | (IV) | 21 |
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)
With the origin at the middle year 2021, X takes the values −3, −2, −1, 0, 1, 2, 3 and \(\sum X = 0\), so the normal equations give
\( \displaystyle a = \frac{\sum y}{n}, \quad b = \frac{\sum Xy}{\sum X^2} \)
(A) The value of \(a\)
This matches (III) 13.
\(\sum y = 7 + 10 + 10 + 13 + 14 + 18 + 19 = 91\)
\( \displaystyle a = \frac{91}{7} = 13 \)
(B) The value of \(b\)
This matches (I) 2.
\(\sum Xy = -21 - 20 - 10 + 0 + 14 + 36 + 57 = 56\)
\(\sum X^2 = 9 + 4 + 1 + 0 + 1 + 4 + 9 = 28\)
\( \displaystyle b = \frac{56}{28} = 2 \)
(C) The trend value for the year 2025
This matches (IV) 21. For 2025, X = 2025 − 2021 = 4.
\(y = 13 + 2 \times 4 = 21\)
(D) The 3-year moving average for the year 2019
This matches (II) 9. It is the mean of 2018, 2019 and 2020, placed against the middle year 2019.
\( \displaystyle \frac{7 + 10 + 10}{3} = \frac{27}{3} = 9 \)