CAIE FP2 2013 November — Question 10 11 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBivariate data
TypeHypothesis test for correlation
DifficultyStandard +0.3 This is a standard three-part correlation question requiring routine application of formulas: calculating PMCC from summary statistics, finding a regression line estimate, and performing a hypothesis test for correlation. While it involves multiple steps and careful arithmetic, all techniques are textbook procedures with no novel problem-solving required, making it slightly easier than average for Further Maths.
Spec5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09c Calculate regression line

10 The lengths, \(x \mathrm {~m}\), and masses, \(y \mathrm {~kg}\), of 12 randomly chosen babies born at a particular hospital last year are summarised as follows. $$\Sigma x = 7.50 \quad \Sigma x ^ { 2 } = 4.73 \quad \Sigma y = 38.6 \quad \Sigma y ^ { 2 } = 124.84 \quad \Sigma x y = 24.25$$ Find the value of the product moment correlation coefficient for this sample. Obtain an estimate for the mass of a baby, born last year at the hospital, whose length is 0.64 m . Test, at the \(2 \%\) significance level, whether there is non-zero correlation between the two variables.

Question 10:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(r = \frac{(24.25 - 7.5 \times 38.6/12)}{\sqrt{\{(4.73 - 7.5^2/12)(124.84 - 38.6^2/12)\}}}\)M1 A1 Find correlation coefficient \(r\) (A0 if only 3 s.f. used)
\(= \frac{0.125}{\sqrt{(0.0425 \times 0.6767)}}\)
\(= \frac{0.125}{(0.2062 \times 0.8226)}\)
\(= 0.737\)*A1
\(b = 0.125/0.0425 = 2.94[1]\)B1 Calculate gradient \(b\) in \(y - \bar{y} = b(x - \bar{x})\)
\(y = 38.6/12 + 2.94(x - 7.5/12)\)M1 Find regression line of \(y\) on \(x\) (A.E.F., allow use of \(x\) on \(y\))
\(= 3.21[7] + 2.94(x - 0.625)\)
or \(1.38 + 2.94x\)A1
\(y = 3.26\) [kg]B1 Find \(y\) when \(x = 0.64\)
\(H_0: \rho = 0,\ H_1: \rho \neq 0\)B1 State both hypotheses
\(r_{12,\ 2\%} = 0.658\)*B1 State or use correct tabular two-tail \(r\) value
Reject \(H_0\) if \(r >\) tabular value
There is non-zero correlationA1 Correct conclusion (AEF, dep *A1, *B1)
## Question 10:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = \frac{(24.25 - 7.5 \times 38.6/12)}{\sqrt{\{(4.73 - 7.5^2/12)(124.84 - 38.6^2/12)\}}}$ | M1 A1 | Find correlation coefficient $r$ (A0 if only 3 s.f. used) |
| $= \frac{0.125}{\sqrt{(0.0425 \times 0.6767)}}$ | | |
| $= \frac{0.125}{(0.2062 \times 0.8226)}$ | | |
| $= 0.737$ | *A1 | |
| $b = 0.125/0.0425 = 2.94[1]$ | B1 | Calculate gradient $b$ in $y - \bar{y} = b(x - \bar{x})$ |
| $y = 38.6/12 + 2.94(x - 7.5/12)$ | M1 | Find regression line of $y$ on $x$ (A.E.F., allow use of $x$ on $y$) |
| $= 3.21[7] + 2.94(x - 0.625)$ | | |
| or $1.38 + 2.94x$ | A1 | |
| $y = 3.26$ [kg] | B1 | Find $y$ when $x = 0.64$ |
| $H_0: \rho = 0,\ H_1: \rho \neq 0$ | B1 | State both hypotheses |
| $r_{12,\ 2\%} = 0.658$ | *B1 | State or use correct tabular two-tail $r$ value |
| Reject $H_0$ if $|r| >$ tabular value | M1 | Valid method for reaching conclusion |
| There is non-zero correlation | A1 | Correct conclusion (AEF, dep *A1, *B1) |

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10 The lengths, $x \mathrm {~m}$, and masses, $y \mathrm {~kg}$, of 12 randomly chosen babies born at a particular hospital last year are summarised as follows.

$$\Sigma x = 7.50 \quad \Sigma x ^ { 2 } = 4.73 \quad \Sigma y = 38.6 \quad \Sigma y ^ { 2 } = 124.84 \quad \Sigma x y = 24.25$$

Find the value of the product moment correlation coefficient for this sample.

Obtain an estimate for the mass of a baby, born last year at the hospital, whose length is 0.64 m .

Test, at the $2 \%$ significance level, whether there is non-zero correlation between the two variables.

\hfill \mbox{\textit{CAIE FP2 2013 Q10 [11]}}