IB DP Maths Topic 7.7 Topics based on the assumption of bivariate normality: knowledge of the facts that the regression HL Paper 3

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 UV 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]

a.

\(\nu  = 10\)     (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]

b.

Examiners report

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a.

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b.

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]

a.

\(t =  – 0.708\sqrt {\frac{{11 – 2}}{{1 – {{\left( { – 0.708} \right)}^2}}}} \,\, = \,\,\left( { – 3.0075 \ldots } \right)\)       (M1)

degrees of freedom = 9        (A1)

P(T < −3.0075…) = 0.00739       A1

Note: Accept any answer that rounds to 0.0074.

[3 marks]

b.i.

reject H0 or equivalent statement       R1

Note: Apply follow through on the candidate’s p-value.

[1 mark]

b.ii.

Cov(U, V) + E((U − E(U))(V − E(V)))

= E(UV − E(U)V − E(V)+ 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]

c.i.

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]

c.ii.

Examiners report

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.
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