RI S6 Corr and Reg Add Prac Solns

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 _______________________________________________________ Additional Practice Questions for Chapter S6: Correlation and Regression Page 1 of 18 Additional Practice Questions for Chapter S6: Correlation and Regression (Solution) 1 8863/2007/01/Q8 Seven cities in a certain country are linked by rail to the capital city. The table below shows the distance of each city from the capital and the rail fare from the city to the capital. City A B C D E F G Distance, x km 124 44 76 148 16 180 104 Rail fare, $y 156 53 99 169 23 177 138 (i) Give a sketch of the scatter diagram for the data as shown on your calculator. [2] (ii) Calculate the product moment correlation coefficient. [1] You are given that the regression line of y on x has equation 16.7 1.01y x  , where the coefficients are given correct to 3 significant figures. (iii) Calculate the equation of the regression line of x on y, giving your answer in the form x a by  . [1] (iv) Use the appropriate regression line to estimate (a) the rail fare from a city that is 28 km from the capital, [2] (b) the distance of a city from the capital if the rail fare is $198. [2] (v) Comment briefly on the reliability of the estimates in part (iv). [2] [ (ii) 0.973 (3sf)r (iii) 10.6 0.940 (3sf)x y  (iv)(a) $45 (b) 176 km ] Solution: (i) (iv) (a) Given 28x , use the regression line of y on x to get 44.98 45 (nearest integer)y  . So the estimated rail fare is $45. (iv) (b) Given 198y , use the regression line of x on y , i.e. 10.558 0.93976x y  (5s.f) to get 176x (3s.f.). So the estimated distance is 176 km. (v) The value 28x is within the given range of values of x where 16 180x  . Since interpolation was used, and the correlation coefficient 0.973 is close to 1, indicating a strong positive linear correlation between the variables, the estimate obtained in (a) is likely to be reliable. The value 198y is beyond the given range of values of y where 23 177y  . Since extrapolation was used, the estimate obtained in (b) is unlikely to be reliable. 177 23 16 180 x y (ii) 0.973 (3sf)r (iii) 10.6 0.940 (3sf)x y  10.558 0.93976 (5sf)x y 

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ _______________________________________________________ Additional Practice Questions for Chapter S6: Correlation and Regression Page 2 of 18 2 8863/2008/01/Q11 An engineering company makes cranes. The numbers, x, sold in each three-month period for two years, together with the profits, y thousand dollars, on the sale of these cranes are given in the following table. (i) Give a sketch of the scatter diagram for the data as shown in your calculator. [2] (ii) Find x and ,y and mark the point  ,x y on your scatter diagram. [2] (iii) Calculate the equation of the regression line of y on x, and draw this line on your scatter diagram. [2] (iv) Calculate the product moment correlation, and comment on its value in relation to your scatter diagram. [2] (v) For the next three-month period, the sales target is 20 cranes. Estimate the corresponding profit. [2] (vi) The company’s sales director uses the regression line in (iii) to predict the profit if 40 cranes were to be sold in a three-month period. Comment on the validity of this prediction. [2] [(ii) 17, 344x y  (iii) 17.1 53.3y x  (iv) 0.969 (v) $395,000] Solution: (i) (ii) _ 17xx n  _ 343.75 344yy n   (to nearest integer) OR: use GC to find _ x and _ y: x 15 17 13 21 16 22 14 18 y 290 350 270 430 340 410 300 360 430 270 13 22 x y (17, 344)

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ _______________________________________________________ Additional Practice Questions for Chapter S6: Correlation and Regression Page 3 of 18 (iii) To use GC to draw regression line : Linear regression of y on x is 17.1 53.3y x  (3 s.f) (iv) 0.969r (3 s.f) The product moment correlation coefficient is close to 1, suggesting a strong positive linear correlation between x and y. This is congruent with the scatter diagram which shows most of the points lying close to the regression line with a positive gradient. (v) When 20x , 17.083(20) 53.333 394.993y   i.e, the estimated corresponding profit is $395 000. (vi) Since 40x is outside the given range of values of x, extrapolation is used. The director’s prediction would be unreliable as the linear model may not be applicable outside the given range of data.

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ _______________________________________________________ Additional Practice Questions for Chapter S6: Correlation and Regression Page 4 of 18 3 Explain why it is advisable to draw a scatter diagram before interpreting a correlation coefficient calculated for a sample drawn from a bivariate distribution. [1] Sketch a scatter diagram that shows an obvious relation between two variables but which yields a linear product-moment correlation coefficient that is close to zero. [1] Solution: A scatter diagram helps to assess if a linear relationship exists between the 2 variables, or a non-linear relationship fits better. It also helps to identify any outliers in the data that do not fit in the general pattern. One possible scatter diagram is 4 9740/2015/NJC/02/Q12 (modified) A medical officer wishes to investigate a patient’s heart-beat rate h beats per minute (bpm) when he walks at different speeds s km/h. The data is shown below: s (km/h) 1 1.5 2 2.5 3 3.5 4 4.5 5 h (bpm) 60 63 66 75 86 99 150 110 130 (i) Sketch a scatter plot of the above data. [1] (ii) One of the values of h appears to be incorrect. Indicate the corresponding point on your diagram by labelling it P. [1] Omit P for the remainder of this question. (iii) Calculate the product moment correlation coefficient for this set of data. Use the equation of an appropriate regression line to predict the value of s when h = 100, justifying your choice of regression line. [4] It is suggested to use one of the following two models instead: Model (I): h = a + bs2, Model (II): h = a + bes, where a and b are real constants. (iv) Determine which of the two models is a better choice, giving a reason for your answer. [2] (v) Suppose a new data pair  ,s his added to the table above, where sand h are the patient’s sample mean walking speed (in km/h) and his sample mean heart-beat rate (in bpm) respectively, based on the data above. Without any calculations, explain whether the equation of the regression line you have obtained in part (iii) would change. [2]

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ _______________________________________________________ Additional Practice Questions for Chapter S6: Correlation and Regression Page 5 of 18 [(iii) 0.9812; s = 3.67 (iv) model (I) (v) ( , )s h] Solution: (i), (ii) (iii) r-value = 0.9812; We use the h on s line since s is the independent variable. h on s line: h = 35.851 + 17.486s When h = 100, s = 3.669 = 3.67 (3 s.f.) (iv) For model (I), r-value = 0.9898 For model (II), r-value = 0.9380 Hence model (I) is more suitable since the correlation coefficient is nearer to 1, which suggests a stronger li