6. Correlation Regression Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Correlation & Regression T1 Correlation and Linear Regression 1 ( ) ( ) 2 2 2 2 2 2 36021830 56308 1403500 10508 (360)(140)3985 2315 8 2315 0.952 5630 1050 xx n yy n xyxy n r  − = − =  − = − =  − = − =− − = =−  In general, as x increases, y decreases in an almost linear pattern. However, it would be better to sketch a scatter diagram based on the 8 pairs of data to verify. 2(i) (ii)

River Valley High School, Mathematics Department 2023 Correlation & Regression T2 (iv) 3(i) ( ) ( )y y b x x y bx y bx− = −  = + − Since y = –0.8x + 13.6, by comparing coefficients, b = –0.8, and 13.6 0.8(4.5) 13.6 10 10 8 80 y bx y y − =  =− + =  =  = ( ) ( ) 2 2 2 2 2 0.8 0.8 0.8 36 80 36 0.8 20488 326.4 xyxy nb xx n xxyxy x nn xy xy − =−  =− −   − =−  −    − =− −   = (ii) Correct: 80y= , 326.4xy= From data: 82.7y= , 348xy= Difference in is 2.7y Difference in is 21.6xy x(2.7) = 21.6 => x = 8 9.6 is wrong. When x = 8, y = 6.9.

River Valley High School, Mathematics Department 2023 Correlation & Regression T3 4(i) − 0.912 (ii) Although B 0.912r =− is close to −1, which indicate a strong negative linear correlation between X and Y, the scatter diagram shows that the data can be better represented by a non-linear curve (iii) y = 0.0721(x − 69)2 + 46.2 2( 69) ,xyr − = 0.9981; proposed model is better since |r| is closer to 1 here (iv) 54.9, As x = 80 is out of the range, this estimate is not reliable 5(a)(i) 0.89241 0.892r == (3 s.f.) (ii) 0.95956 0.960r== (3 s.f.) Since 0.960r= is closer to 1 than 0.892r= , the model in (i) is less suitable than the model in (ii). (iii) Regression line of F on x: 0.35903 0.029245Fx=+ 0.359 0.0292Fx=+ Regression line of x on F: 204.51 31.484xF=+ 205 31.5xF=+ (iv) Using 0.35903 0.029245Fx=+ , 2100 0.35903 0.029245 t=+ 58.37047 58.4t == s (3 s.f.) 6(i) Based on the scatter diagram, the linear model is not suitable even though r (= 0.906) is quite close to 1. (ii) Choose Model (b): y = axb Reason: The graph of y = axb fits the scatter diagram better. : From the scatter diagram, we see that as x increases, y increases at a decreasing rate. (iii) r = 0.932 y = axb ln y = ln(axb) ln y = ln a + b ln x ln y = 4.1912 + 0.3056 ln x ln a = 4.1912  a = 66.1 b = 0.306 (iv) when y = 110, x = 5.29. Extrapolation hence not reliable.

River Valley High School, Mathematics Department 2023 Correlation & Regression T4 7(i) Product moment correlation coefficient r = 0.979. There exists a strong positive linear correlation between x and y. (ii) Equation of least square regression line is y = 18.5 + 0.564 x (iii) Given that 30,y= x = 20.4 from the equation. She left at 7am. This estimate is reliable since r  1 so we can use the equation of y on x to estimate x, giving y and y = 30 is within the given data range. ( 2 reasons) (iv) z = time available – time taken = 50 – x – y = (50 – x) – (a+ bx) = (50 – x) – (18.5 + 0.564x) = 31.5 – 1.564x (v) For z = 0, x = 31.5 1.564 = 20.14  20min The latest time when Ms Chan leaves her house is 7 a.m 8(i) r = -0.952 For the depths sampled the moisture content decreases approximately linearly with an increase in the depth of the sample. Since r is close to 1 , this implies that the regression line of m on x and x on m are almost identical or close to each other (ii) m = - 2.2048x + 83.583. When m = 50 , x = 2048.2 50583.83 − = 15.232 = 15.2 Reliable since the value of m is within the range of the given data. Not extrapolating. 9(i) cPV k= ln ln lnP c V k+= ln 1ln ln kVP cc=−  y a bx=+ is a straight line, where ln ka c= and 1b c=− are constants (ii) Using G.C., enter values of P and V in list L1 and L2 . Let L3 = lnP and let L4 = lnV

River Valley High School, Mathematics Department 2023 Correlation & Regression T5 0.994r =− (iii) By using G.C., the required estimated regression line of y on x is 2.1434-0.66219yx= 2.14 0.662y = x − (to 3 s.f.) ln 2.1434 0.66219lnVP=− ( )ln =2.1434 0.66219ln 8 0.76641V −= 0.76641 2.15203 2.2Ve= = = (to 1 d.p.) (ans) Since 0.994 1r =− − which indicates that the sample values of x and y are almost perfectly linearly correlated and ( ) ( )ln(1) ln 8 ln 14 , therefore the prediction is reliable. (iv) ln 1ln ln kVP cc=− 2.1434 0.66219yx=− 1 0.66219c = 1 1.51014 1.510.66219c= = = (to 3 s.f.) ln =2.1434k c ln 2.1434 1.51014 3.2368k =  = 3.2368 25.452 25.5ke= = = (to 3 s.f.) (v) For 2 ()yY − to be minimum, Then 'Y a bx=+ must be 2.1434-0.66219yx= By using G.C., let L5 = 2.1434−0.66219 L3 and ( ) 2 L6 L4 L5=− Minimum value of 2 ( ) 0.0282yY −= (to 3 s.f.)

River Valley High School, Mathematics Department 2023 Correlation & Regression T6 10(a) The least squares regression line of y on x is the line with the minimum sum of the squared vertical distances from each point to the line while the least squares regression line x on y is the line with the minimum sum of the squared horizontal distances from each point to the line. ( yx, ) is the point through which the estimated regression line of y on x and that of x on y both pass. (b)(i) From the regression line y on x, we have 7 786 7 68 += xy . Since 8=x , 1907 786)8(7 68 =+=y .  1906 136225229160 =+++++   390=+ …. (1) 7 68 )( )()( 2 = − −−= xx yyxxb        −=     − 6 )( 7 68 6 2 2 xxyxxy Hence,      −=     ++−++ 6 23044547 68 6 )750)(48()796660(   ( ) 680)750(8)796660( =++−++   20=− … (2) (1) + (2) : 4102 =  205= (1) − (2) : 3752 =  185= (ii) When y = 210, 78668)210(7 += x  05882353.10=x = 10.1 ( to 3 s.f.) The predicted value of x is likely to be reliable as estimation is done by interpolation and r = 0.985 is close to 1.

River Valley High School, Mathematics Department 2023 Correlation & Regression T7 (iii) 11(i) 161x= (from calculator or computation) when 161x= , 103.6 0.726xy=+ (161 103.6) / 0.726y=− 79.06336088= using y y n= 179.06336088 (65.1 73.2 85 80.9 89.9)6 k= + + + + + 80.3k = Use G.C. to find regression line of y on x: 97.593 1.097yx=− + (ii) Use y on x line to predict weight. When 165x= , 97.593 1.097(165)y=− + 83.4y= (1 d.p.) – using 3 d.p. of a and b to compute. or 83.5y= - using full accuracy of a and b to compute (iii) Using G.C., 0.893r= C is unusually overweight. y 65.1 73.2 80.3 80.9 85.0 89.9 150 157 160 162 167 170 x

River Valley High School, Mathematics Department 2023 Correlation & Regression T8 12 13(a)(i) For x on y, s.f.) (3 1.31309.013.313085.0 +−=+−= yxyx For y on x, (3s.f.) 0.9685.2999.958526.2 +−=++−= xyxy (ii) Since chemical Y is the controlled variable, use regression line of x on y. 91.10013.313085.00 =+−= yy The estimation is not valid as this is an extrapolation, linear relation may not hold outside the range of data (b)(i) By comparing the linear product moment correlation for the 3 models, Model C is the most appropriate with the highest value of 993.0=r as it best describes the data given.

River Valley High School, Mathematics Department 2023 Correlation & Regression T9 Using linear transformation xw ln= , Regression line of w on y is 0.026136 3.8294 0.0261 3.83 (3 s.f.)w y w y=− +  =− + (ii) Chang