Question

For the data in Exercise 10 of Chapter 10, Section 2 “The Linear Correlation Coefficient” Compute the least squares regression line. Can you compute the sum of the squared errors S S E using the definition Σ ( y − y ^ ) 2? Explain. Compute the sum of the squared errors S S E using the formula S S E = S S y y − β ^ 1 S S x y. Compute the linear correlation coefficient for the sample data summarized by the following information: n = 10 ∑ x = - 3 ∑ x 2 = 263 ∑ y = 55 ∑ y 2 = 917 ∑ x y = - 355 - 10 ≤ x ≤ 10

Answer #1

Ʃx = -3, Ʃy = 55, Ʃxy = -355, Ʃx² = 263, Ʃy² = 917, n = 10

x̅ = Ʃx/n = -3/10 = -0.3

y̅ = Ʃy/n = 55/10 = 5.5

SSxx = Ʃx² - (Ʃx)²/n = 263 - (-3)²/10 = 262.1

SSyy = Ʃy² - (Ʃy)²/n = 917 - (55)²/10 = 614.5

SSxy = Ʃxy - (Ʃx)(Ʃy)/n = -355 - (-3)(55)/10 = -338.5

Slope, b = SSxy/SSxx = -338.5/262.1 = -1.29149

y-intercept, a = y̅ -b* x̅ = 5.5 - (-1.29149)*-0.3 = 5.11255

Least squares regression line:

**ŷ = 5.1126 + (-1.2915) x **

--

No, we cannot compute the sum of the squared errors SSE using
the definition Σ(y-ŷ)^{2}. As we do not know the value of
x and y to compute the values.

--

SSE = SSyy - β₁*SSxy = 614.5 - (-1.29149)*(-338.5) = 177.33

--

Correlation coefficient, r = SSxy/√(SSxx*SSyy) =
-338.5/√(262.1*614.5) = **-0.8435**

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2
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