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Consider n pairs of numbers. Suppose bar(x)=4,s_(x)=3, bar(y)=2, and s_(y)=5. Of the following which could be the least squares line? (A) y=2+x (B) y=-6+2x (C) y=-10+3x (D) y=5//3-x (E) y=6-x

Consider n n pairs of numbers. Suppose xˉ=4,sx=3,yˉ=2 \bar{x}=4, s_{x}=3, \bar{y}=2 , and sy=5 s_{y}=5 .\newlineOf the following which could be the least squares line?\newline(A) y=2+x y=2+x \newline(B) y=−6+2x y=-6+2 x \newline(C) y=−10+3x y=-10+3 x \newline(D) y=5/3−x y=5 / 3-x \newline(E) y=6−x y=6-x

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Full solution

Q. Consider n n pairs of numbers. Suppose xˉ=4,sx=3,yˉ=2 \bar{x}=4, s_{x}=3, \bar{y}=2 , and sy=5 s_{y}=5 .\newlineOf the following which could be the least squares line?\newline(A) y=2+x y=2+x \newline(B) y=−6+2x y=-6+2 x \newline(C) y=−10+3x y=-10+3 x \newline(D) y=5/3−x y=5 / 3-x \newline(E) y=6−x y=6-x
  1. Calculate slope using formula: Calculate the slope mm of the least squares line using the formula m=nΣ(xy)−ΣxΣynΣ(x2)−(Σx)2m = \frac{n\Sigma(xy) - \Sigma x\Sigma y}{n\Sigma(x^2) - (\Sigma x)^2}. We don't have Σ(xy)\Sigma(xy) or Σ(x2)\Sigma(x^2), but we can use the correlation coefficient formula r=sxysx∗syr = \frac{s_{xy}}{s_x * s_y} where sxys_{xy} can be derived as sxy=r∗sx∗sys_{xy} = r * s_x * s_y. Assume r=1r = 1 for maximum positive correlation, then sxy=1∗3∗5=15s_{xy} = 1 * 3 * 5 = 15.
  2. Calculate Σx\Sigma x and Σy\Sigma y: Calculate Σx\Sigma x and Σy\Sigma y using nn, xˉ\bar{x}, and yˉ\bar{y}. Σx=n×xˉ=n×4\Sigma x = n \times \bar{x} = n \times 4, Σy=n×yˉ=n×2\Sigma y = n \times \bar{y} = n \times 2. We still need nn to find exact values, but we can use these expressions in our slope formula.
  3. Substitute values into formula: Substitute sxys_{xy}, Σx\Sigma x, and Σy\Sigma y into the slope formula. Assuming n=10n = 10 for calculation: m = rac{(10 imes 15 - 4 imes 10 imes 2 imes 10)}{(10 imes 9 - (4 imes 10)^2)} = rac{(150 - 800)}{(90 - 1600)} = rac{-650}{-1510} = 0.43 approximately.

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