Explanation Calculation: Least squares estimate: In a linear equation y = b 0 + b 1 x i the constant b 1 be the slope and b 0 be the y -intercept form and x is the independent variable and y is the independent variable. The error function for the equation is, f ( b 0 , b 1 ) = ∑ i = 1 n ( y i − ( b 0 + b 1 x i ) ) 2 Normal equations of Y i = β 0 + β 1 x i + ε i : The first normal equations of the error function will be obtained by finding the first derivative of f ( b 0 , b 1 ) with respect to b 0 and equating it to 0. Hence, the first normal equation is n b 0 + ( ∑ x i ) b 1 = ∑ y i The second normal equations of the error function will be obtained by finding the first derivative of f ( b 0 , b 1 ) with respect to b 1 and equating it to 0. Hence, the second normal equation is ( ∑ x i ) b 0 + ( ∑ x i 2 ) b 1 = ∑ x i y i In the linear equation the point estimates of the b 0 and b 1 are β ^ 0 and β ^ 1 . These point estimates are the values that minimize the equation f ( b 0 , b 1 ) . That is, f ( β ^ 0 , β ^ 1 ) ≤ f ( b 0 , b 1 ) . Then the values of β ^ 0 and β ^ 1 are the least squares estimates and the equation y ^ = β ^ 0 + β ^ 1 x i Slope: In a linear equation y ^ = β ^ 0 + β ^ 1 x the constant b 1 be the slope and b 0 be the y -intercept form and x is the independent variable and y is the independent variable. The general formula to obtain slope is, b 1 = β ^ 1 = S x y S x x = ∑ ( x i − x ¯ ) ( y i − y ¯ ) ∑ ( x i − x ¯ ) 2 Y-intercept: In a linear equation y ^ = β ^ 0 + β ^ 1 x the constant b 1 be the slope and b 0 be the y -intercept form and x is the independent variable and y is the independent variable. The general formula to obtain y- intercept is, b 0 = β ^ 0 = y ¯ − β ^ 1 x ¯ = ∑ y i − β ^ 1 ∑ x i n Here, the objective is to prove that b 1 = β ^ 1 = S x y S x x and b 0 = β ^ 0 = y ¯ − β ^ 1 x ¯ satisfies the normal equations of Y i = β 0 + β 1 x i + ε i . First normal equation: To prove that the value n b 0 + ( ∑ x i ) b 1 is ∑ y i with the substitution of b 1 = β ^ 1 = S x y S x x and b 0 = β ^ 0 = y ¯ − β ^ 1 x ¯ . The computation procedure goes as follows: n b 0 + ( ∑ x i ) b 1 = n β ^ 0 + ( ∑ x i ) β ^ 1 = n × ( y ¯ − β ^ 1 x ¯ ) + ( ∑ x i ) × S x y S x x = n y ¯ − n S x y S x x x ¯ + n × x ¯ × S x y S x x = n y ¯ = ∑ y i Thus, the first normal equation is satisfied by b 1 = β ^ 1 = S x y S x x and b 0 = β ^ 0 = y ¯ − β ^ 1 x ¯

Probability and Statistics for Engineering and the Sciences

9th Edition

ISBN: 9781305251809

Author: Jay L. Devore

Publisher: Cengage Learning

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Chapter 12.2, Problem 25E

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Show that b1=β^1=SxySxx and b0=β^0=y¯−β^1x¯ satisfies the normal equations of Yi=β0+β1xi+εi.

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Chapter 12.2, Problem 24E

Chapter 12.2, Problem 26E

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Probability and Statistics for Engineering and the Sciences

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Show that b 1 = β ^ 1 = S x y S x x and b 0 = β ^ 0 = y ¯ − β ^ 1 x ¯ satisfies the normal equations of Y i = β 0 + β 1 x i + ε i .

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Show that b1=β^1=SxySxx and b0=β^0=y¯−β^1x¯ satisfies the normal equations of Yi=β0+β1xi+εi.

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