Solutions to Statistics 8 APQ (RVHS)

Solutions to Statistics 8 Tutorial: Correlation & Linear Regression Additional Practice Questions 1(i) The product moment correlation coefficient between x and y, r is -0.943(3 s.f.). (ii) Since r is close to 1, it suggests a linear model is appropriate. However the scatter diagram shows the relationship is non-linear. (iii) The scatter diagram shows that as x increases, the rate of decrease in y becomes smaller and y appears to approach a value. For a linear model, the rate of decrease is constant. OR The product moment correlation coefficient between 1/x and y, 2r is 0.993 where 2r is almost 1 as compared to r =0.943. Therefore the model where 0 0by a a ,bx= +   is better. (iv) 2.16 (3s.f.), 2.33 (3s.f.)ab== (v) When 45x . , y== 2.68 (3 s.f..) Since 4.5x= is within the data range of the x values and value of 2r is close to 1, the estimate is reliable. 2(i) r = 0.972 Even though r is close to 1, it does not mean that there is a cause and effect relationship between x and y.

(ii) The points on the scatter diagram lie close to a straight line with positive gradient. This agrees with the value of r obtained in part (i). (iii) For the model y = abx, i.e. ln y = ln a + x ln b, r = 0.98759 For the model y = axb, i.e. ln y = ln a + b ln x, r = 0.97789 Since | r | = 0.98759 is closest to 1 for the model y = abx, this is the most suitable model. (iv) ln y = 0.128367x + 0.9994 ln a = 0.9994  a = 2.72 ln b = 0.128367  b = 1.14 3. Incorrect result is x = 10, y = 42. From the scatter diagram, x = 10, y = 42 is an outlier. 14.16830 1.61302 and 9.97265 0.59168y x x y=− + = + Using GC to solve the above simultaneous equations, 34.8522x= 42.04899y = Let k be the correct value when x =10, ( )42.04899 7 (1.3 14.8 30.1 60.8 81.3 98.6)k = − + + + + + 10 5 x 1 0 5 y

= 7.4 (to 1 decimal place) As x is the independent variable, y on x should be used. The estimate is reliable since r = 0.977 is close to 1, indicates a strong positive linear correlation between x and y and x = 40 is within data range. c is the sum of least square deviation between the observed value y and the predicted value on the regression line y on x. Using GC, c = 401.541488 = 402 (to 3 sig figs) (ans) 4. (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. 5(i) There will be no difference as the product moment correlation coefficient is independent of the units in which the data is measured. (ii) The regression line of t on x should be used because the running time t is dependent on the leg length, x. (iii) t 13.90 10.80

(iv) Yes. Aaron has reason to disagree because the scatter diagram suggests that t and x has a curvilinear relationship rather than a linear one. (v)(a) Product moment correlation coefficient between t and is 0.992 (3 s.f.) The new model is a better model because is closer to 1 than . (v)(b) Regression line is (3 s.f.) When t = 10, (to 2 dec places) since x > 0 Thus minimum length of leg required is 1.16m. This estimate may not be reliable as t = 10 is outside the sample data range for t. OR Extrapolated values are unreliable. 2 1 x 0.992 0.963 0.963−= 2 17.8603 2.8616t x=+ 2 17.86 2.86t x=+ 2 110 7.8603 2.8616 x=+ 2 2.8616 2.1397x = 1.16x=

6(i) (ii) Using GC, r = 0.989. Using regression line of y on t, 0.041899 0.94916yt=+ When 7t= , 6.6860 6.69y== cm2 (3s.f.) Since 7t= is within the data range and r = 0.989 is close to 1, the answer is reliable. (iii) When 80t = , 75.974 76.0y== cm2 (3s.f) Note that 76.0cm2 > 64cm2 (the total area of the slice of bread.) The regression line may not be suitable as it is impossible for the bread to keep growing mould. Also, from the scatter diagram, it shows that as t increases (after 8 days, the mould starts to grow at a decreasing rate. Hence, a linear model may not be appropriate. t y 10 0 8.8 0.3