Date | November 2020 | Marks available | 3 | Reference code | 20N.1.AHL.TZ0.F_13 |
Level | Additional Higher Level | Paper | Paper 1 | Time zone | Time zone 0 |
Command term | Calculate | Question number | F_13 | Adapted from | N/A |
Question
Observations on 1212 pairs of values of the random variables X, YX, Y yielded the following results.
∑x=76.3, ∑x2=563.7, ∑y=72.2, ∑y2=460.1, ∑xy=495.4∑x=76.3, ∑x2=563.7, ∑y=72.2, ∑y2=460.1, ∑xy=495.4
Calculate the value of rr, the product moment correlation coefficient of the sample.
Assuming that the distribution of X, YX, Y is bivariate normal with product moment correlation coefficient ρρ, calculate the pp-value of your result when testing the hypotheses H0 : ρ=0 ; H1 : ρ>0H0: ρ=0 ; H1: ρ>0.
State whether your pp-value suggests that XX and YY are independent.
Given a further value x=5.2x=5.2 from the distribution of XX, YY, predict the corresponding value of yy. Give your answer to one decimal place.
Markscheme
use of
r=∑xy-nˉx ˉy√(∑x2-nˉx2)(∑y2-nˉy2)r=∑xy−n¯x¯y√(∑x2−n¯x2)(∑y2−n¯y2) M1
=495.4-12×76.312×72.212√(563.7-12×76.32122)(460.1-12×72.22122)=495.4−12×76.312×72.212√(563.7−12×76.32122)(460.1−12×72.22122) A1
=0.809=0.809 A1
Note: Accept any answer that rounds to 0.810.81.
[3 marks]
t=0.80856…√101-0.80856…2t=0.80856…√101−0.80856…2 (M1)
=4.345…=4.345… A1
pp-value =7.27×10-4=7.27×10−4 A1
Note: Accept any answer that rounds to 7.27.2 or 7.3×10-47.3×10−4.
Note: Follow through their pp-value
[3 marks]
this value indicates that X,YX,Y are not independent A1
[1 mark]
use of
y-ˉy=∑xy-nˉx ˉy∑x2-nˉx2(x-ˉx)y−¯y=∑xy−n¯x¯y∑x2−n¯x2(x−¯x) M1
y-72.212=(495.4-12×76.312×72.212563.7-12×76.32122)(x-76.312)y−72.212=(495.4−12×76.312×72.212563.7−12×76.32122)(x−76.312) A1
putting x=5.2x=5.2 gives y=5.5y=5.5 A1
[3 marks]