R 711 chancerossman 2015 iscam iii inve n does your

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Chance/Rossman, 2015 ISCAM III Investigation 5.8 369 (n) Does your line provide a better fit than your neighbor’s? Suggest a criterion for deciding which line “best” summarizes the relationship. (o) Check the Show Residuals box to visually represent these residuals for your line on the scatterplot. The applet also reports the sum of the absolute residuals (or SAE, the sum of absolute errors). Record this SAE value for your line. What is the best SAE in the class? (Does “best” correspond to the smallest or the largest value of SAE?) (p) A more common criterion for determining the “best” line is to instead look at the sum of the squared residuals (or sum of squared errors, SSE). Check the Show Squared Residuals box to visually represent them and to determine the SSE value for your line. What is the best SSE in the class? (q) Continue to adjust your line until you think you have minimized the sum of the squared residuals. Report your new equation and new SSE value. (r) Now check the Show Regression Line box to determine and display the equation for the line that actually does produce the smallest possible sum of squared residuals. Report its equation and SSE value. Did everyone obtain the same line/equation this time? How does it compare to your line? (You can also display the residuals and the squared residuals for this line.) Equation: SSE: Comparisons: Definition : The line that minimizes the sum of squared residuals is called the least squares line , or simply the regression line , or even the least squares regression line. (s) Suggest a technique for determining (based on the observed data x i ’s and y i ’s) the values of the slope and the intercept that minimize the SSE.
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Chance/Rossman, 2015 ISCAM III Investigation 5.8 370 Least Squares Regression Line: Derivation of Coefficients The least squares line y ˆ = b 0 + b 1 x is determined by finding the values of the coefficients b 0 and b 1 that minimize the sum of the squared residuals, SSE = 6 ( y i - i y ˆ ) 2 = ¦ º ´ ± ± ³ n i i o i x b b y 1 1 . (t) Take the derivative with respect to b 0 of the expression on the right. Then take the derivative of the original expression with respect to b 1 . [ Hints : Use the chain rule, and remember to treat the data values x i ’s and y i ’s as constants.] (u) Set these (partial) derivatives equal to zero and solve simultaneously for the values of b 0 and b 1 . [ Hints : Solve the first equation for b 0 . Then solve the second equation for b 1 (substituting in the expression for b 0 using the summation notation).]
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Chance/Rossman, 2015 ISCAM III Investigation 5.8 371 You should have found expressions equivalent to the following. n x b y b n i n i i i ¦ ¦ ± 1 1 1 0 and 2 1 1 2 1 1 1 1 ¸ ¹ · ¨ © § ± ± ¦ ¦ ¦ ¦ ¦ n i i n i i n i n i i i n i i i x x n y x y x n b With a little bit more algebra, you can show that the formulas for the least squares estimates of the intercept coefficient b 0 and slope coefficient b 1 simplify to: b 0 = y - b 1 b 1 = r s y / s x (v) For the sample data on students’ foot lengths ( x ) and heights ( y ), we can calculate these summary
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