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) – (