Question
The random variables \(U,{\text{ }}V\) follow a bivariate normal distribution with product moment correlation coefficient \(\rho \).
A random sample of 12 observations on U, V is obtained to determine whether there is a correlation between U and V. The sample product moment correlation coefficient is denoted by r. A test to determine whether or not U, V are independent is carried out at the 1% level of significance.
a.State suitable hypotheses to investigate whether or not \(U\), \(V\) are independent.[2]
b.Find the least value of \(|r|\) for which the test concludes that \(\rho \ne 0\).[6]
▶️Answer/Explanation
Markscheme
\({{\text{H}}_0}:\rho = 0;{\text{ }}{{\text{H}}_1}:\rho \ne 0\) A1A1
[2 marks]
\(\nu = 10\) (A1)
\({t_{0.005}} = 3.16927 \ldots \) (M1)(A1)
we reject \({{\text{H}}_0}:\rho = 0\) if \(\left| t \right| > 3.16927 \ldots \) (R1)
attempting to solve \(\left| r \right|\sqrt {\frac{{10}}{{1 – {r^2}}}} > 3.16927 \ldots \) for \(\left| r \right|\) M1
Note: Allow = instead of >.
(least value of \(\left| r \right|\) is) 0.708 (3 sf) A1
Note: Award A1M1A0R1M1A0 to candidates who use a one-tailed test. Award A0M1A0R1M1A0 to candidates who use an incorrect number of degrees of freedom or both a one-tailed test and incorrect degrees of freedom.
Note: Possible errors are
10 DF 1-tail, \(t = 2.763 \ldots \), least value \( = \) 0.658
11 DF 2-tail, \(t = 3.105 \ldots \), least value \( = \) 0.684
11 DF 1-tail, \(t = 2.718 \ldots \), least value \( = \) 0.634.
[6 marks]
Examiners report
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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))).
a.State suitable hypotheses to investigate whether or not a negative linear association exists between X and Y.[1]
b.i.Determine the p-value.[3]
b.ii.State your conclusion at the 1 % significance level.[1]
c.i.Show that Cov(U, V) = E(UV) − E(U)E(V).[3]
c.ii.Hence show that if U, V are independent random variables then the population product moment correlation coefficient, ρ, is zero.[3]
▶️Answer/Explanation
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]
Examiners report
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