Edexcel S1 — Question 1 7 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBivariate data
TypeCalculate r from summary statistics
DifficultyModerate -0.8 This is a straightforward application of the PMCC formula using given summary statistics. Students need only substitute values into a standard formula (no derivation required) and perform arithmetic calculations. The interpretation in part (b) is routine. This is easier than average as it requires pure recall and calculation with no problem-solving or conceptual challenges.
Spec5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation

  1. A shop recorded the number of pairs of gloves, \(n\), that it sold and the average daytime temperature, \(T ^ { \circ } \mathrm { C }\), for each month over a 12-month period.
The data was then summarised as follows: $$\Sigma T = 124 , \quad \Sigma n = 384 , \quad \Sigma T ^ { 2 } = 1802 , \quad \Sigma n ^ { 2 } = 18518 , \quad \Sigma T n = 2583 .$$
  1. Calculate the product moment correlation coefficient for these data.
  2. Comment on what your value shows and suggest a reason for this.

AnswerMarks Guidance
\(S_{TT} = 1802 - \frac{124^2}{12} = 520.667\)M1
\(S_{mm} = 18518 - \frac{384^2}{12} = 6230\)M1
\(S_{Tm} = 2583 - \frac{124 \times 384}{12} = -1385\)M1
\(r = \frac{-1385}{\sqrt{520.667 \times 6230}} = 0.7690\)M1 A1
(b) It shows –ve correlation meaning less glove sales in higher temperatures e.g. people mainly buy gloves when their hands are coldB1 B1 (7 marks)
$S_{TT} = 1802 - \frac{124^2}{12} = 520.667$ | M1 |

$S_{mm} = 18518 - \frac{384^2}{12} = 6230$ | M1 |

$S_{Tm} = 2583 - \frac{124 \times 384}{12} = -1385$ | M1 |

$r = \frac{-1385}{\sqrt{520.667 \times 6230}} = 0.7690$ | M1 A1 |

(b) It shows –ve correlation meaning less glove sales in higher temperatures e.g. people mainly buy gloves when their hands are cold | B1 B1 | (7 marks)

---
\begin{enumerate}
  \item A shop recorded the number of pairs of gloves, $n$, that it sold and the average daytime temperature, $T ^ { \circ } \mathrm { C }$, for each month over a 12-month period.
\end{enumerate}

The data was then summarised as follows:

$$\Sigma T = 124 , \quad \Sigma n = 384 , \quad \Sigma T ^ { 2 } = 1802 , \quad \Sigma n ^ { 2 } = 18518 , \quad \Sigma T n = 2583 .$$

(a) Calculate the product moment correlation coefficient for these data.\\
(b) Comment on what your value shows and suggest a reason for this.\\

\hfill \mbox{\textit{Edexcel S1  Q1 [7]}}