# Exam-style Questions: Bivariate Data

**1.** Sketch scatter diagrams to show possible forms (or describe in words) for the following values of the product moment correlation coefficient, r.

**a)** r = 0.78

**b)** r = -1

**c)** r = -0.25

**(3 marks)**

**d)** calculate the regression line x on y for the following summarised data

n = 12 Σx = 1800 Σx2 = 336296 Σy = 36.0 Σy2 = 126.34 Σxy = 6348.6

**(5 marks)**

**(Marks available: 8)**

**Answer outline and marking scheme for question: 1**

**a)** a good positive correlation

**b)** a perfect negative correlation - all points lie on line

**c)** a poor negative correlation - a fair degree of scatter

**(3 marks)**

**d)** need S_{xy} = 6348.6 - __1800 x 36__ = **948.6**

12

S_{yy} = 126.34 - __36__^{2} = **18.34**

12

so b' = __948.6__ = **51.7**

18.34

a' = __1800__ - 51.7 x __36__ = **-5.17**

12 12

so regression line x on y is x = 51.7y - 5.17

**(5 marks)**

**(Marks available: 8)**

**2.** A wine expert grades 10 bottles of wine on a scale from 0 to 50. He records the results next to their ages

Wine | A | B | C | D | E | F | G | H | I | J |

Age | 15 | 21 | 24 | 28 | 30 | 34 | 36 | 40 | 42 | 44 |

Score | 18 | 11 | 14 | 25 | 27 | 17 | 33 | 22 | 37 | 41 |

**a)** calculate the product moment correlation coefficient for the age of the wine against the grade given

**b)** calculate Spearman's rank correlation coefficient for the same data and comment on the results

**(Marks available: 9)**

**Answer outline and marking scheme for question: 2**

**a)** data obtained from calculator

n = 10 Σx = 314 Σx2 = 10678 Σy = 245 Σy2 = 6907 Σxy = 8347 and from calculator r = 0.760 (3sf)

if you prefer to use the formulae then you should obtain the following

S_{xy} = 654 S_{xx} = 818.4 S_{yy} = 904.5 and r = __654 __= 0.760

**b)** ranking

Wine | A | B | C | D | E | F | G | H | I | J |

Age x | 15 | 21 | 24 | 28 | 30 | 34 | 36 | 40 | 42 | 44 |

Score y | 18 | 11 | 14 | 25 | 27 | 17 | 33 | 22 | 37 | 41 |

Rank x | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

Rank y | 7 | 10 | 9 | 4 | 5 | 8 | 3 | 6 | 2 | 1 |

d | 3 | 1 | 1 | 3 | 1 | 3 | 1 | 3 | 0 | 0 |

d2 | 9 | 1 | 1 | 9 | 1 | 9 | 1 | 9 | 0 | 0 |

n = 10 and Σd^{2} = 40 so r_{s} = 1 - __6 x 40__ = 0.758 (3sf)

10(102-1)

commenting; approximation r_{s} of r, is extremely close. Both show a good positive correlation meaning the older the wine the better quality it is according to this particular expert.

**(Marks available: 9)**

**3.** A physicist wants to find out what happens to a length of metal rod (cm) when put under various temperatures (°C).

She carries out an experiment and her data is as follows

Temp t | 105 | 110 | 115 | 120 | 125 | 130 |

Length x | 200.1 | 201.8 | 201.8 | 202 | 204.3 | 205.1 |

The physicist would like to calculate a line of regression for this data.

**a)** advise the physicist on which line to use 'x on t' or 't on x'

**b)** calculate this line of regression and use it to estimate the length of a bar at 145°C

**(Marks available: 9)**

**Answer outline and marking scheme for question: 3**

**a)** as the temperature appears to be controlled and independent we should calculate the line 'x on t'

**b)** summarised data from calc

n = 6 Σ t = 705 Σt2 = 83275 Σx = 1214.1 Σx2 = 245692.39 Σtx = 142746

S_{tx} = __89.25__ S_{tt} = 437.5 hence b = 89.25 = 0.204

437.5

so a = __1214.1__ - 0.204 x __705__ = 178.38

6 6

equation of 'x on t' is x = 178.38 + 0.204t

at 145^{0}C x = 178.38 + 0.204 x -145 = **207.96cm**

**(Marks available: 9)**