Question

1 8 16 Since we now have quadratic and power models that both Year PCB Conc. model curved data, reconsider the previous question. 1.0 2 1.9 a) Notice the power curve in the previous question is not that 4 4.2 great of a fit to the data (compare the fit of a quadratic 6 7.2 curve below to the power function above). Use regression 11.8 to find a quadratic equation to model the PCB Concentration in Lake Trout data. Round the numbers in your equation to 2 decimal places. b) Use your quadratic equation to estimate the age of a fish with a PCB concentration of 15 ppm, rounded to the nearest tenth. 14 12 PCB Concentration (ppm) 8 6 4 0 9 3 6 4 7 5 2 1 Age

Answer 1

= X2 = п > п n : B= Quadratic regression Exi Lyi Σα? (1) mean : ī= ý (2) trend line : y = A + Bx + Cx2 SxySex-Sx'ySex SxxSxx— (Sxx-)2 SxºySxx–SxySxx C= SxxSx-X?—(Sxx)2 A=g - Bº – Caº (3) correlation coefficient : (4-(A+Bri+Cx?))2 1 E(gi-g)? Sxx = {(xi TE) 2 · ū2 Sxy = {(xi – m)(yi – T) = Xiyi – n• ty Sxx = (xi – T)(x; – x») = x} – n• Ix? Sxx = [(x – X+)2 = {x4 - n • XX? Sxy = [(x? -x2)(yi – ) = xtyi n · X r =

Quadratic regression: y=A+BX+Cx2 (input by clicking each cell in the table below) No. х y 1 1 1 2 2 data 3 4 1.9 4.2 7.2 4 5 8 11.8 Execute Clear Store/Read Print 10digit function value mean of x 4.2 mean of y 5.22 correlation coefficient r 0.9995726032 A 0.598810468 B 0.3514869151 С 0.1299563838 20 15 10 5 0 0.21.2 2.2 3.2 4.25.26.27.28.29.2 A+BX+Cx2

quadratice 0 We have eq 201599 + 0.351 xt 0-130x2 y=0:54 PLB = 0.1322 +0.3512 + 0.599 in desund of to 2 places Pes - 0.1322 +0. 352 +0-60 PCB = 15 ppm fo 2 X 2 0:1322 +0.352 +0.6 =15 + 2.6932+ 4,61 115:38 +2.6934 110.77-0 -2.69 + |(2693 + 4x18. 140-77 2 a = 9.265, -11.95 belevent age of fish is 9.265 2 - yenes.