Question

For the following data, the slope $(m)$ and intercept $(c)$ of the least squares fitted straight line $(Y=mX+c)$ are given as:

• A. Slope $=0.750$, Intercept $=5.167$
• B. Slope $=0.850$, Intercept $=6.180$
• C. Slope $=0.650$, Intercept $=5.558$
• D. Slope $=0.750$, Intercept $=5.555$

Hint:

For a straight-line fit, center the data: compute sums of \((X-\bar X)(Y-\bar Y)\) and \((X-\bar X)^2\). It avoids large-number roundoff and simplifies arithmetic.

Answer 1

The Correct Option is A

Compute sample means.
$\bar{X} = \dfrac{12+14+16}{3} = 14, \quad \bar{Y} = \dfrac{14+16+17}{3} = \dfrac{47}{3} = 15.\overline{6}$.
Least-squares formulas. \[ m = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{\sum (X_i - \bar{X})^2}, \quad c = \bar{Y} - m\bar{X}. \] Evaluate the sums.
\[ \begin{aligned} \sum (X_i - \bar{X})(Y_i - \bar{Y}) &= (-2)(-1.\overline{6}) + 0(0.\overline{3}) + 2(1.\overline{3}) \\ &= 3.\overline{3} + 0 + 2.\overline{6} = 6, \\ \sum (X_i - \bar{X})^2 &= 4 + 0 + 4 = 8. \end{aligned} \] Hence, $m = \dfrac{6}{8} = 0.75$, and \[ c = \bar{Y} - m\bar{X} = 15.\overline{6} - 0.75 \times 14 = 15.\overline{6} - 10.5 = 5.166\overline{6} \approx 5.167. \] \[ \boxed{m = 0.750, \quad c = 5.167} \]