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 u