For a certain data \( (x_i,y_i) \) \( i=1,2,\ldots,7 \), \( \sum_{i=1}^{7}y_i=246 \) and \( \sum_{i=1}^{7}u_iy_i=29 \), \( \sum_{i=1}^{7}u_i^2=28 \) where \( u_i=x_i-2004 \), then straight line trend by the method of least square is .
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( y=35.14+1.036(x-2004) \)
The trend line is fitted in the shifted variable \( u=x-2004 \), as \( y=a+bu \). The normal equations are
\( \displaystyle \sum y=na+b\sum u \)
\( \displaystyle \sum uy=a\sum u+b\sum u^2 \)
The 7 time points are centred on 2004 (2001 to 2007), so u takes the values −3, −2, −1, 0, 1, 2, 3. Then \( \sum u=0 \), and indeed \( \sum u^2=9+4+1+0+1+4+9=28 \), as given. With \( \sum u=0 \) the normal equations separate:
\( \displaystyle a=\frac{\sum y}{n}=\frac{246}{7}\approx 35.14 \)
\( \displaystyle b=\frac{\sum uy}{\sum u^2}=\frac{29}{28}\approx 1.036 \)
\( \displaystyle y=35.14+1.036u \)
\( \displaystyle y=35.14+1.036(x-2004) \)
Why the other options are incorrect
Option (1) \( y=35.14+1.036x \) — Incorrect. The trend was fitted in u, so u must be replaced by \( x-2004 \), not by x itself.
Option (3) \( y=4+35.14(x-2004) \) — Incorrect. 35.14 is the constant term a, and the slope is \( \frac{29}{28}\approx 1.036 \).
Option (4) \( y=6+1.2x \) — Incorrect. Neither the constant nor the slope matches \( \frac{246}{7}\approx 35.14 \) and \( \frac{29}{28}\approx 1.036 \).