| Date | May 2018 | Marks available | 3 | Reference code | 18M.3sp.hl.TZ0.4 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Show that and Hence | Question number | 4 | Adapted from | N/A |
Question
The random variables X , Y follow a bivariate normal distribution with product moment correlation coefficient ρ.
A random sample of 11 observations on X, Y was obtained and the value of the sample product moment correlation coefficient, r, was calculated to be −0.708.
The covariance of the random variables U, V is defined by
Cov(U, V) = E((U − E(U))(V − E(V))).
State suitable hypotheses to investigate whether or not a negative linear association exists between X and Y.
Determine the p-value.
State your conclusion at the 1 % significance level.
Show that Cov(U, V) = E(UV) − E(U)E(V).
Hence show that if U, V are independent random variables then the population product moment correlation coefficient, ρ, is zero.
Markscheme
H0 : ρ = 0; H1 : ρ < 0 A1
[1 mark]
\(t = - 0.708\sqrt {\frac{{11 - 2}}{{1 - {{\left( { - 0.708} \right)}^2}}}} \,\, = \,\,\left( { - 3.0075 \ldots } \right)\) (M1)
degrees of freedom = 9 (A1)
P(T < −3.0075...) = 0.00739 A1
Note: Accept any answer that rounds to 0.0074.
[3 marks]
reject H0 or equivalent statement R1
Note: Apply follow through on the candidate’s p-value.
[1 mark]
Cov(U, V) + E((U − E(U))(V − E(V)))
= E(UV − E(U)V − E(V)U + E(U)E(V)) M1
= E(UV) − E(E(U)V) − E(E(V)U) + E(E(U)E(V)) (A1)
= E(UV) − E(U)E(V) − E(V)E(U) + E(U)E(V) A1
Cov(U, V) = E(UV) − E(U)E(V) AG
[3 marks]
E(UV) = E(U)E(V) (independent random variables) R1
⇒Cov(U, V) = E(U)E(V) − E(U)E(V) = 0 A1
hence, ρ = \(\frac{{{\text{Cov}}\left( {U,\,V} \right)}}{{\sqrt {{\text{Var}}\left( U \right)\,{\text{Var}}\left( V \right)} }} = 0\) A1AG
Note: Accept the statement that Cov(U,V) is the numerator of the formula for ρ.
Note: Only award the first A1 if the R1 is awarded.
[3 marks]
