OCR FS1 AS 2021 June — Question 2 6 marks

Exam BoardOCR
ModuleFS1 AS (Further Statistics 1 AS)
Year2021
SessionJune
Marks6
TopicBivariate data
TypeInterpret correlation coefficient value
DifficultyModerate -0.8 This is a straightforward Further Statistics 1 question testing standard correlation coefficient calculation using given summary statistics and basic interpretation. The calculation is routine formula application, and the follow-up parts involve standard assumptions and hypothesis testing procedures that are well-rehearsed at this level. While it's Further Maths content, it requires no novel insight or complex multi-step reasoning.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02n Sum of Poisson variables: is Poisson5.08a Pearson correlation: calculate pmcc
2 In the manufacture of fibre optical cable (FOC), flaws occur randomly. Whether any point on a cable is flawed is independent of whether any other point is flawed. The number of flaws in 100 m of FOC of standard diameter is denoted by \(X\).
  1. State a further assumption needed for \(X\) to be well modelled by a Poisson distribution. Assume now that \(X\) can be well modelled by the distribution \(\operatorname { Po } ( 0.7 )\).
  2. Find the probability that in 300 m of FOC of standard diameter there are exactly 3 flaws. The number of flaws in 100 m of FOC of a larger diameter has the distribution \(\mathrm { Po } ( 1.6 )\).
  3. Find the probability that in 200 m of FOC of standard diameter and 100 m of FOC of the larger diameter the total number of flaws is at least 4 . Judith believes that mathematical ability and chess-playing ability are related. She asks 20 randomly chosen chess players, with known British Chess Federation (BCF) ratings \(X\), to take a mathematics aptitude test, with scores \(Y\). The results are summarised as follows. $$n = 20 , \Sigma x = 3600 , \Sigma x ^ { 2 } = 660500 , \Sigma y = 1440 , \Sigma y ^ { 2 } = 105280 , \Sigma x y = 260990$$
    1. Calculate the value of Pearson's product-moment correlation coefficient \(r\).
    2. State an assumption needed to be able to carry out a significance test on the value of \(r\).
    3. Assume now that the assumption in part (b) is valid. Test at the \(5 \%\) significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
    4. There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by $$\text { ELO rating } = 8 \times \text { BCF rating } + 650$$ Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion.
    5. Calculate the value of Pearson's product-moment correlation coefficient \(r\).
    6. State an assumption needed to be able to carry out a significance test on the value of \(r\).
    7. Assume now that the assumption in part (b) is valid. Test at the \(5 \%\) significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
    8. There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by $$\mathrm { ELO } \text { rating } = 8 \times \mathrm { BCF } \text { rating } + 650 .$$ Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion. An environmentalist measures the mean concentration, \(c\) milligrams per litre, of a particular chemical in a group of rivers, and the mean mass, \(m\) pounds, of fish of a certain species found in those rivers. The results are given in the table.
      QuestionAnswerMarksAOGuidance
      1(a)\(\begin{aligned}0.25 + 0.36 + x + x ^ { 2 } = 1
      x ^ { 2 } + x - 0.39 = 0
      x = 0.3 \text { (or } - 1.3 \text { ) }
      x \text { cannot be negative }
      \mathrm { E } ( W ) = 2.23
      \mathrm { E } \left( W ^ { 2 } \right) = \Sigma w ^ { 2 } \mathrm { p } ( w ) \quad [ = 5.83 ]
      \text { Subtract } [ \mathrm { E } ( W ) ] ^ { 2 } \text { to get } \mathbf { 0 . 8 5 7 1 } \end{aligned}\)\(\begin{gathered} \text { M1 }
      \text { A1 }
      \text { A1 }
      \text { B1ft }
      \text { B1 }
      \text { M1 }
      \text { A1 }
      { [ 7 ] } \end{gathered}\)
      3.1a
      1.1b
      1.1b
      2.3
      1.1b
      1.1
      2.1
      Equation using \(\Sigma p = 1\)
      Correct simplified quadratic Correctly obtain \(x = 0.3\)
      Explicitly reject other solution
      2.23 or exact equivalent only Use \(\Sigma w ^ { 2 } \mathrm { p } ( w )\)
      Correctly obtain given answer, www
      Can be implied
      Method needed ft on their quadratic Allow for \(\mathrm { E } ( W ) ^ { 2 } = 4.9729\)
      Need 2.23 or 4.9729 and 5.83 or full numerical \(\Sigma w ^ { 2 } \mathrm { p } ( w )\)
      1(b)\(9 \times 0.8571 = 7.7139\)
      B1
      [1]
      		1.1bAllow 7.71 or 7.714
      2(a)Flaws must occur at constant average rate (uniform rate)
      B1
      [1]
      		1.2
      Context (e.g. "flaws") needed
      Extra answers, e.g. "singly": B0
      		Not "constant rate" or "average constant rate".
      2(b)\(\operatorname { Po(2.1)~or~ } e ^ { - \lambda } \frac { \lambda ^ { 3 } } { 3 ! }\)
      M1
      A1
      [2]
      1.1
      1.1b
      		Po(2.1) stated or implied, or formula with \(\lambda = 2.1\) stated Awrt 0.189
      2(c)
      Po(3)
      \(1 - \mathrm { P } ( \leq 3 )\)
      M1
      M1
      A1
      [3]
      1.1
      1.1
      1.1b
      \(\operatorname { Po } ( 2 \times 0.7 + 1.6 )\) stated or implied
      Allow \(1 - \mathrm { P } ( \leq 4 ) = 0.1847\), or from wrong \(\lambda\)
      Awrt 0.353
      Or all combinations \(\leq 3\)
      \(1 -\) above, not just \(= 3\)
      QuestionAnswerMarksAOGuidance
      3(a)0.4(00)
      B2
      [2]
      1.1
      1.1b
      		SC: if B0, give SC B1 for two of \(S _ { x x } = 12500 , S _ { y y } = 1600 , S _ { x y } = 1790\) and \(S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)\)Also allow SC B1 for equivalent methods using Covariance \SDs
      3(b)Data needs to have a bivariate normal distribution
      B1
      [1]
      		1.2Needs "bivariate normal" or clear equivalent. Not just "both normally distributed"Allow "scatter diagram forms ellipse"
      3(c)
      \(\mathrm { H } _ { 0 }\) : higher maths scores are not associated with higher BCF grading; \(\mathrm { H } _ { 1 }\) : positively associated
      CV 0.3783
      \(0.400 > 0.3783\) so reject \(\mathrm { H } _ { 0 }\)
      Significant evidence that higher maths scores are associated with higher BCF grading
      B1
      B1
      M1ft
      A1ft
      [4]
      2.5
      1.1b
      2.2b
      3.5a
      Needs context and clearly onetailed \(O R \rho\) used and defined Not "evidence that ..."
      Allow 0.378
      Reject/do not reject \(\mathrm { H } _ { 0 }\)
      Contextualised, not too definite Needn't say "positive" if \(\mathrm { H } _ { 1 } \mathrm { OK }\)
      SC 2-tail: B0; 0.4438, or 0.3783 B1; then M1A0
      \(\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho > 0\) where \(\rho\) is population pmcc (not \(r\) )
      FT on their \(r\), but not CV
      Not "scores are associated
      ...". FT on their \(r\) only
      3(d)It makes no difference as this is a linear transformation
      B1
      [1]
      		2.2aNeed both "unchanged" oe and reason, need "linear" or exact equivalent"oe" includes "their 0.4"
      4(a)Neither
      B1
      [1]
      		2.5OENot "neither is independent of the other"
      4(b)\(c = 2.848 - 0.1567 m\)
      B1
      B1
      B1
      [3]
      1.1
      1.1
      1.1
      Correct \(a\), awrt 2.85
      Correct \(b\), awrt 0.157
      Letters correct from correct method
      (If both wrongly rounded, e.g. \(c = 2.84 - 0.156 m\), give B2)
      \(\mathrm { SC } : m\) on \(c\) :
      \(m = 15.65 - 4.832 c\) : B2
      \(y = 15.65 - 4.832 x\) : B1
      \(c = 15.65 - 4.832 m : \mathrm { B } 1\)
      If B0B0, give B1 for correct letters from valid working
      QuestionAnswerMarksAOGuidance
      4(c)\(a\) unchanged, \(b\) multiplied by 2.2 (allow " \(a\) unchanged, \(b\) increases", etc)B1 [1]2.2aoe, e.g. \(c = 2.848 - 0.345 m\); \(m = 7.114 - 2.196 c\)SC: \(m\) on \(c\) in (b): Both divided by 2.2 B1
      4(d)
      Draw approximate line of best fit
      Draw at least one vertical from line to point
      Say that "Best fit" line minimises the sum of squares of these distances
      M1
      M1
      A1
      [3]
      1.1
      2.4
      2.4
      		Needs M2 and "minimises" and "sums of squares" oe
      SC: Horizontal(s):
      full marks (indept of (b))
2 In the manufacture of fibre optical cable (FOC), flaws occur randomly. Whether any point on a cable is flawed is independent of whether any other point is flawed. The number of flaws in 100 m of FOC of standard diameter is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item State a further assumption needed for $X$ to be well modelled by a Poisson distribution.

Assume now that $X$ can be well modelled by the distribution $\operatorname { Po } ( 0.7 )$.
\item Find the probability that in 300 m of FOC of standard diameter there are exactly 3 flaws.

The number of flaws in 100 m of FOC of a larger diameter has the distribution $\mathrm { Po } ( 1.6 )$.
\item Find the probability that in 200 m of FOC of standard diameter and 100 m of FOC of the larger diameter the total number of flaws is at least 4 .

Judith believes that mathematical ability and chess-playing ability are related. She asks 20 randomly chosen chess players, with known British Chess Federation (BCF) ratings $X$, to take a mathematics aptitude test, with scores $Y$. The results are summarised as follows.

$$n = 20 , \Sigma x = 3600 , \Sigma x ^ { 2 } = 660500 , \Sigma y = 1440 , \Sigma y ^ { 2 } = 105280 , \Sigma x y = 260990$$

(a) Calculate the value of Pearson's product-moment correlation coefficient $r$.\\
(b) State an assumption needed to be able to carry out a significance test on the value of $r$.\\
(c) Assume now that the assumption in part (b) is valid. Test at the $5 \%$ significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
\item There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by

$$\text { ELO rating } = 8 \times \text { BCF rating } + 650$$

Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion.\\
(a) Calculate the value of Pearson's product-moment correlation coefficient $r$.\\
(b) State an assumption needed to be able to carry out a significance test on the value of $r$.\\
(c) Assume now that the assumption in part (b) is valid. Test at the $5 \%$ significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.\\
(d) There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by

$$\mathrm { ELO } \text { rating } = 8 \times \mathrm { BCF } \text { rating } + 650 .$$

Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion.

An environmentalist measures the mean concentration, $c$ milligrams per litre, of a particular chemical in a group of rivers, and the mean mass, $m$ pounds, of fish of a certain species found in those rivers. The results are given in the table.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{3}{|c|}{Question} & Answer & Marks & AO & \multicolumn{2}{|c|}{Guidance} \\
\hline
1 & (a) &  & \(\begin{aligned} & 0.25 + 0.36 + x + x ^ { 2 } = 1 \\ & x ^ { 2 } + x - 0.39 = 0 \\ & x = 0.3 \text { (or } - 1.3 \text { ) } \\ & x \text { cannot be negative } \\ & \mathrm { E } ( W ) = 2.23 \\ & \mathrm { E } \left( W ^ { 2 } \right) = \Sigma w ^ { 2 } \mathrm { p } ( w ) \quad [ = 5.83 ] \\ & \text { Subtract } [ \mathrm { E } ( W ) ] ^ { 2 } \text { to get } \mathbf { 0 . 8 5 7 1 } \end{aligned}\) & \(\begin{gathered} \text { M1 } \\ \text { A1 } \\ \text { A1 } \\ \text { B1ft } \\ \text { B1 } \\ \text { M1 } \\ \text { A1 } \\ { [ 7 ] } \end{gathered}\) & \begin{tabular}{l}
3.1a \\
1.1b \\
1.1b \\
2.3 \\
1.1b \\
1.1 \\
2.1 \\
\end{tabular} & \begin{tabular}{l}
Equation using $\Sigma p = 1$ \\
Correct simplified quadratic Correctly obtain $x = 0.3$ \\
Explicitly reject other solution \\
2.23 or exact equivalent only Use $\Sigma w ^ { 2 } \mathrm { p } ( w )$ \\
Correctly obtain given answer, www \\
\end{tabular} & \begin{tabular}{l}
Can be implied \\
Method needed ft on their quadratic Allow for $\mathrm { E } ( W ) ^ { 2 } = 4.9729$ \\
Need 2.23 or 4.9729 and 5.83 or full numerical $\Sigma w ^ { 2 } \mathrm { p } ( w )$ \\
\end{tabular} \\
\hline
1 & (b) &  & $9 \times 0.8571 = 7.7139$ & \begin{tabular}{l}
B1 \\[0pt]
[1] \\
\end{tabular} & 1.1b & Allow 7.71 or 7.714 &  \\
\hline
2 & (a) &  & Flaws must occur at constant average rate (uniform rate) & \begin{tabular}{l}
B1 \\[0pt]
[1] \\
\end{tabular} & 1.2 & \begin{tabular}{l}
Context (e.g. "flaws") needed \\
Extra answers, e.g. "singly": B0 \\
\end{tabular} & Not "constant rate" or "average constant rate". \\
\hline
2 & (b) &  & \(\operatorname { Po(2.1)~or~ } e ^ { - \lambda } \frac { \lambda ^ { 3 } } { 3 ! }\) & \begin{tabular}{l}
M1 \\
A1 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1b \\
\end{tabular} & Po(2.1) stated or implied, or formula with $\lambda = 2.1$ stated Awrt 0.189 &  \\
\hline
2 & (c) &  & \begin{tabular}{l}
Po(3) \\
$1 - \mathrm { P } ( \leq 3 )$ \\
\end{tabular} & \begin{tabular}{l}
M1 \\
M1 \\
A1 \\[0pt]
[3] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1 \\
1.1b \\
\end{tabular} & \begin{tabular}{l}
$\operatorname { Po } ( 2 \times 0.7 + 1.6 )$ stated or implied \\
Allow $1 - \mathrm { P } ( \leq 4 ) = 0.1847$, or from wrong $\lambda$ \\
Awrt 0.353 \\
\end{tabular} & \begin{tabular}{l}
Or all combinations $\leq 3$ \\
$1 -$ above, not just $= 3$ \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{3}{|c|}{Question} & Answer & Marks & AO & \multicolumn{2}{|c|}{Guidance} \\
\hline
3 & (a) &  & 0.4(00) & \begin{tabular}{l}
B2 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1b \\
\end{tabular} & SC: if B0, give SC B1 for two of $S _ { x x } = 12500 , S _ { y y } = 1600 , S _ { x y } = 1790$ and $S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)$ & Also allow SC B1 for equivalent methods using Covariance \& SDs \\
\hline
3 & (b) &  & Data needs to have a bivariate normal distribution & \begin{tabular}{l}
B1 \\[0pt]
[1] \\
\end{tabular} & 1.2 & Needs "bivariate normal" or clear equivalent. Not just "both normally distributed" & Allow "scatter diagram forms ellipse" \\
\hline
3 & (c) &  & \begin{tabular}{l}
$\mathrm { H } _ { 0 }$ : higher maths scores are not associated with higher BCF grading; $\mathrm { H } _ { 1 }$ : positively associated \\
CV 0.3783 \\
$0.400 > 0.3783$ so reject $\mathrm { H } _ { 0 }$ \\
Significant evidence that higher maths scores are associated with higher BCF grading \\
\end{tabular} & \begin{tabular}{l}
B1 \\
B1 \\
M1ft \\
A1ft \\[0pt]
[4] \\
\end{tabular} & \begin{tabular}{l}
2.5 \\
1.1b \\
2.2b \\
3.5a \\
\end{tabular} & \begin{tabular}{l}
Needs context and clearly onetailed $O R \rho$ used and defined Not "evidence that ..." \\
Allow 0.378 \\
Reject/do not reject $\mathrm { H } _ { 0 }$ \\
Contextualised, not too definite Needn't say "positive" if $\mathrm { H } _ { 1 } \mathrm { OK }$ \\
SC 2-tail: B0; 0.4438, or 0.3783 B1; then M1A0 \\
\end{tabular} & \begin{tabular}{l}
$\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho > 0$ where $\rho$ is population pmcc (not $r$ ) \\
FT on their $r$, but not CV \\
Not "scores are associated \\
...". FT on their $r$ only \\
\end{tabular} \\
\hline
3 & (d) &  & It makes no difference as this is a linear transformation & \begin{tabular}{l}
B1 \\[0pt]
[1] \\
\end{tabular} & 2.2a & Need both "unchanged" oe and reason, need "linear" or exact equivalent & "oe" includes "their 0.4" \\
\hline
4 & (a) &  & Neither & \begin{tabular}{l}
B1 \\[0pt]
[1] \\
\end{tabular} & 2.5 & OE & Not "neither is independent of the other" \\
\hline
4 & (b) &  & $c = 2.848 - 0.1567 m$ & \begin{tabular}{l}
B1 \\
B1 \\
B1 \\[0pt]
[3] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1 \\
1.1 \\
\end{tabular} & \begin{tabular}{l}
Correct $a$, awrt 2.85 \\
Correct $b$, awrt 0.157 \\
Letters correct from correct method \\
(If both wrongly rounded, e.g. $c = 2.84 - 0.156 m$, give B2) \\
\end{tabular} & \begin{tabular}{l}
$\mathrm { SC } : m$ on $c$ : \\
$m = 15.65 - 4.832 c$ : B2 \\
$y = 15.65 - 4.832 x$ : B1 \\
$c = 15.65 - 4.832 m : \mathrm { B } 1$ \\
If B0B0, give B1 for correct letters from valid working \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{Question} & Answer & Marks & AO & \multicolumn{2}{|c|}{Guidance} \\
\hline
4 & (c) & $a$ unchanged, $b$ multiplied by 2.2 (allow " $a$ unchanged, $b$ increases", etc) & B1 [1] & 2.2a & oe, e.g. $c = 2.848 - 0.345 m$; $m = 7.114 - 2.196 c$ & SC: $m$ on $c$ in (b): Both divided by 2.2 B1 \\
\hline
4 & (d) & \begin{tabular}{l}
Draw approximate line of best fit \\
Draw at least one vertical from line to point \\
Say that "Best fit" line minimises the sum of squares of these distances \\
\end{tabular} & \begin{tabular}{l}
M1 \\
M1 \\
A1 \\[0pt]
[3] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
2.4 \\
2.4 \\
\end{tabular} & Needs M2 and "minimises" and "sums of squares" oe & \begin{tabular}{l}
SC: Horizontal(s): \\
full marks (indept of (b)) \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}
\end{enumerate}

\hfill \mbox{\textit{OCR FS1 AS 2021 Q2 [6]}}
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