CAIE FP2 2008 November — Question 8 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2008
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear regression
TypeFind means from regression lines
DifficultyModerate -0.3 This is a straightforward application of standard regression line properties: both lines pass through (x̄, ȳ) giving a simple simultaneous equation, and the correlation coefficient formula r = ±√(b₁b₂) is direct recall. The hypothesis test is routine with n=25. Slightly easier than average due to minimal problem-solving required, just applying memorized formulas.
Spec5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09a Dependent/independent variables5.09c Calculate regression line

8 The equations of the regression lines for a random sample of 25 pairs of data \(( x , y )\) from a bivariate population are $$\begin{array} { c c } y \text { on } x : & y = 1.28 - 0.425 x , \\ x \text { on } y : & x = 1.05 - 0.516 y . \end{array}$$
  1. Find the sample means, \(\bar { x }\) and \(\bar { y }\).
  2. Find the product moment correlation coefficient for the sample.
  3. Test at the \(5 \%\) significance level whether the population correlation coefficient differs from zero.

Question 8:
AnswerMarks Guidance
Answer/WorkingMark Guidance
(i) \(\bar{y} + 0.425\bar{x} = 1.28\) and \(\bar{x} + 0.516\bar{y} = 1.05\)M1 Formulate two equations for means
\(\bar{x} = 0.499\), \(\bar{y} = 1.068\) or \(1.07\)M1 A1 Solve for means; Part total: 3
(ii) \(r^2 = 0.425 \times 0.516\); \(r = -0.468\)M1; *A1 Find correlation coefficient for sample; Part total: 2
(iii) \(H_0: \rho = 0\), \(H_1: \rho \neq 0\)B1 State hypotheses
\(\rho \neq 0\) if \(r >\) tabular value
\(\rho_{25, 2.5\%} = 0.396\)*B1 Use of correct tabular value
Coefficient does differ from zeroA1 Correct conclusion (A.E.F., dep *A1, *B1); Part total: 4
Total: 9
## Question 8:

| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $\bar{y} + 0.425\bar{x} = 1.28$ and $\bar{x} + 0.516\bar{y} = 1.05$ | M1 | Formulate two equations for means |
| $\bar{x} = 0.499$, $\bar{y} = 1.068$ or $1.07$ | M1 A1 | Solve for means; **Part total: 3** |
| **(ii)** $r^2 = 0.425 \times 0.516$; $r = -0.468$ | M1; *A1 | Find correlation coefficient for sample; **Part total: 2** |
| **(iii)** $H_0: \rho = 0$, $H_1: \rho \neq 0$ | B1 | State hypotheses |
| $\rho \neq 0$ if $|r| >$ tabular value | M1 | Valid method for reaching conclusion |
| $\rho_{25, 2.5\%} = 0.396$ | *B1 | Use of correct tabular value |
| Coefficient does differ from zero | A1 | Correct conclusion (A.E.F., dep *A1, *B1); **Part total: 4** |
| | | **Total: 9** |

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8 The equations of the regression lines for a random sample of 25 pairs of data $( x , y )$ from a bivariate population are

$$\begin{array} { c c } 
y \text { on } x : & y = 1.28 - 0.425 x , \\
x \text { on } y : & x = 1.05 - 0.516 y .
\end{array}$$

(i) Find the sample means, $\bar { x }$ and $\bar { y }$.\\
(ii) Find the product moment correlation coefficient for the sample.\\
(iii) Test at the $5 \%$ significance level whether the population correlation coefficient differs from zero.

\hfill \mbox{\textit{CAIE FP2 2008 Q8 [9]}}