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

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.

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]

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

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.

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.