CAIE FP2 2015 June — Question 7 11 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear regression
TypeRelate two regression lines
DifficultyStandard +0.8 This question requires understanding the relationship between two regression lines via the correlation coefficient formula (b_yx × b_xy = r²), algebraic manipulation to find the second regression line, and then a hypothesis test for correlation. While the concepts are standard Further Maths statistics content, the multi-step nature combining regression theory, the use of means as a constraint, and hypothesis testing makes it moderately challenging but still within typical FM scope.
Spec5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line
7 For a random sample of 10 observations of pairs of values \(( x , y )\), the equation of the regression line of \(y\) on \(x\) is \(y = 3.25 x - 4.27\). The sum of the ten \(x\) values is 15.6 and the product moment correlation coefficient for the sample is 0.56 . Find the equation of the regression line of \(x\) on \(y\). Test, at the \(5 \%\) significance level, whether there is evidence of non-zero correlation between the variables.
Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(3.25\ b_2 = 0.56^2\), \(b_2 = 0.0965\)M1 A1 Find gradient \(b_2\) of line \(x\) on \(y\) from \(b_1 b_2 = r^2\)
\(\bar{y} = 3.25 \times (15.6/10) - 4.27 = 0.8\)M1 A1 Find \(\bar{y}\) from line \(y\) on \(x\)
\((x - \bar{x}) = b_2(y - \bar{y})\) Find regression line of \(x\) on \(y\) (M0 for \(y\) on \(x\))
\((x - 1.56) = 0.0965(y - 0.8)\)M1 A1
\(x = 0.0965y + 1.48\)A1
\(H_0\): \(\rho = 0\), \(H_1\): \(\rho \neq 0\)B1 State both hypotheses
\(r_{10,\ 5\%} = 0.632\)*B1 State or use correct tabular two-tail \(r\)-value
Accept \(H_0\) if \(0.56 < r\)-valueM1 Valid method for conclusion (AEF)
There is no non-zero correlationA1 Correct conclusion (AEF, dep *B1)
Part marks: 7, 4Total: 11 
## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3.25\ b_2 = 0.56^2$, $b_2 = 0.0965$ | M1 A1 | Find gradient $b_2$ of line $x$ on $y$ from $b_1 b_2 = r^2$ |
| $\bar{y} = 3.25 \times (15.6/10) - 4.27 = 0.8$ | M1 A1 | Find $\bar{y}$ from line $y$ on $x$ |
| $(x - \bar{x}) = b_2(y - \bar{y})$ | | Find regression line of $x$ on $y$ (M0 for $y$ on $x$) |
| $(x - 1.56) = 0.0965(y - 0.8)$ | M1 A1 | |
| $x = 0.0965y + 1.48$ | A1 | |
| $H_0$: $\rho = 0$, $H_1$: $\rho \neq 0$ | B1 | State both hypotheses |
| $r_{10,\ 5\%} = 0.632$ | *B1 | State or use correct tabular two-tail $r$-value |
| Accept $H_0$ if $0.56 < r$-value | M1 | Valid method for conclusion (AEF) |
| There is no non-zero correlation | A1 | Correct conclusion (AEF, dep *B1) |

**Part marks: 7, 4 | Total: 11**

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7 For a random sample of 10 observations of pairs of values $( x , y )$, the equation of the regression line of $y$ on $x$ is $y = 3.25 x - 4.27$. The sum of the ten $x$ values is 15.6 and the product moment correlation coefficient for the sample is 0.56 . Find the equation of the regression line of $x$ on $y$.

Test, at the $5 \%$ significance level, whether there is evidence of non-zero correlation between the variables.

\hfill \mbox{\textit{CAIE FP2 2015 Q7 [11]}}
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