In fitting a straight-line trend $Y = a + bX$ by the method of least squares, the time deviations are chosen so that $\sum X = 0$. The value of $b$ is then:
(a) $\dfrac{\sum Y}{n}$
(b) $\dfrac{\sum XY}{\sum X^2}$
(c) $\dfrac{\sum X}{n}$
(d) $\dfrac{\sum XY}{n}$

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Accepted Answer

The normal equations are $\sum Y = na + b\sum X$ and $\sum XY = a\sum X + b\sum X^2$. Setting $\sum X = 0$ removes the last term of each:
$$\sum Y = na \Rightarrow a = \frac{\sum Y}{n}, \qquad \sum XY = b\sum X^2 \Rightarrow b = \frac{\sum XY}{\sum X^2}.$$

- **(a) $\sum Y / n$** — wrong: this is $a$, the trend at the origin, not $b$.
- **(b) $\sum XY / \sum X^2$** — **correct**: this is $b$, the slope, when $\sum X = 0$. …

Q.In fitting a straight-line trend $Y = a + bX$ by the method of least squares, the time deviations are chosen so that $\sum X = 0$ . The value of $b$ is then:

Concept understanding — Method of Least Squares

Q.In fitting a straight-line trend Y=a+bXY = a + bXY=a+bX by the method of least squares, the time deviations are chosen so that ∑X=0\sum X = 0∑X=0. The value of bbb is then:

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Q.In fitting a straight-line trend $Y = a + bX$ by the method of least squares, the time deviations are chosen so that $\sum X = 0$ . The value of $b$ is then:

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