For a straight line trend \( y_{t}=a+bx \) for \( n \) data by the method of least squares if, \( n=8,\sum y=47.8,\sum xy=4.3 \) and \( \sum x^{2}=42 \), then \( y_{t} \) at \( x=-3.5 \) is_____.
[where \( y_{t} \) represents trend value of \( y \)]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5.62
For 8 years the time variable x is taken as deviations from the middle: −3.5, −2.5, …, 2.5, 3.5. Then \( \sum x=0 \), and indeed \( \sum x^2=2(0.25+2.25+6.25+12.25)=42 \), as given.
With \( \sum x=0 \) the normal equations \( \sum y=na+b\sum x \) and \( \sum xy=a\sum x+b\sum x^2 \) reduce to:
\( \displaystyle a=\frac{\sum y}{n}=\frac{47.8}{8}=5.975 \)
\( \displaystyle b=\frac{\sum xy}{\sum x^2}=\frac{4.3}{42}=0.10238 \)
The trend value at x = −3.5:
\( \displaystyle y_t=5.975+0.10238\times(-3.5) \)
\( \displaystyle =5.975-0.3583 \)
\( \displaystyle =5.6167\approx 5.62 \)