IB DP Maths Topic 7.7 Covariance and (population) product moment correlation coefficient ρ HL Paper 3

 

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Question

If \(X\) and \(Y\) are two random variables such that \({\text{E}}(X) = {\mu _X}\) and \({\text{E}}(Y) = {\mu _Y}\) then \({\text{Cov}}(X,{\text{ }}Y) = {\text{E}}\left( {(X – {\mu _X})(Y – {\mu _Y})} \right)\).

Prove that if \(X\) and \(Y\) are independent then \({\text{Cov}}(X,{\text{ }}Y) = 0\).

[3]
a.

In a particular company, it is claimed that the distance travelled by employees to work is independent of their salary. To test this, 20 randomly selected employees are asked about the distance they travel to work and the size of their salaries. It is found that the product moment correlation coefficient, \(r\), for the sample is \( – 0.35\).

You may assume that both salary and distance travelled to work follow normal distributions.

Perform a one-tailed test at the \(5\% \) significance level to test whether or not the distance travelled to work and the salaries of the employees are independent.

[8]
b.
Answer/Explanation

Markscheme

METHOD 1

\({\text{Cov}}(X,{\text{ }}Y) = {\text{E}}\left( {(X – {\mu _X})(Y – {\mu _Y})} \right)\)

\( = {\text{E}}(XY – X{\mu _Y} – Y{\mu _X} + {\mu _X}{\mu _Y})\)     (M1)

\( = {\text{E}}(XY) – {\mu _Y}{\text{E}}(X) – {\mu _X}{\text{E}}(Y) + {\mu _X}{\mu _Y}\)

\( = {\text{E}}(XY) – {\mu _X}{\mu _Y}\)     A1

as \(X\) and \(Y\) are independent \({\text{E}}(XY) = {\mu _X}{\mu _Y}\)     R1

\({\text{Cov}}(X,{\text{ }}Y) = 0\)     AG

METHOD 2

\({\text{Cov}}(X,{\text{ }}Y) = {\text{E}}\left( {(X – {\mu _x})(Y – {\mu _y})} \right)\)

\( = {\text{E}}(X – {\mu _x}){\text{E}}(Y – {\mu _y})\)     (M1)

since \(X,Y\) are independent     R1

\( = ({\mu _x} – {\mu _x})({\mu _y} – {\mu _y})\)     A1

\( = 0\)     AG

[3 marks]

a.

\({H_0}:\rho  = 0\;\;\;{H_1}:\rho  < 0\)     A1

Note:     The hypotheses must be expressed in terms of \(\rho \).

test statistic \({t_{test}} =  – 0.35\sqrt {\frac{{20 – 2}}{{1 – {{( – 0.35)}^2}}}} \)     (M1)(A1)

\( =  – 1.585 \ldots \)     (A1)

\({\text{degrees of freedom}} = 18\)     (A1)

EITHER

\(p{\text{ – value}} = 0.0652\)     A1

this is greater than \(0.05\)     M1

OR

\({t_{5\% }}(18) =  – 1.73\)     A1

this is less than \( – {\text{1.59}}\)     M1

THEN

hence accept \({H_0}\) or reject \({H_1}\) or equivalent or contextual equivalent     R1

Note:     Allow follow through for the final R1 mark.

[8 marks]

Total [11 marks]

b.

Examiners report

Solutions to (a) were often disappointing with few candidates gaining full marks, a common error being failure to state that

\(E(XY) = E(X)E(Y)\) or \({\text{E}}\left( {(X – {\mu _x})(Y – {\mu _y})} \right) = {\text{E}}(X – {\mu _x}){\text{E}}(Y – {\mu _y})\) in the case of independence.

a.

In (b), the hypotheses were sometimes given incorrectly. Some candidates gave \({H_1}\) as \(\rho  \ne 0\), not seeing that a one-tailed test was required. A more serious error was giving the hypotheses as \({H_0}:r = 0,{\text{ }}{H_1}:r < 0\) which shows a complete misunderstanding of the situation. Subsequent parts of the question were well answered in general.

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))).

State suitable hypotheses to investigate whether or not a negative linear association exists between X and Y.

[1]
a.

Determine the p-value.

[3]
b.i.

State your conclusion at the 1 % significance level.

[1]
b.ii.

Show that Cov(U, V) = E(UV) − E(U)E(V).

[3]
c.i.

Hence show that if U, V are independent random variables then the population product moment correlation coefficient, ρ, is zero.

[3]
c.ii.
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.

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.

[1]
a.

Determine the p-value.

[3]
b.i.

State your conclusion at the 1 % significance level.

[1]
b.ii.

Show that Cov(U, V) = E(UV) − E(U)E(V).

[3]
c.i.

Hence show that if U, V are independent random variables then the population product moment correlation coefficient, ρ, is zero.

[3]
c.ii.
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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