CAIE FP2 2012 June — Question 10 12 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeStandard CI with summary statistics
DifficultyStandard +0.3 This is a standard two-sample inference problem requiring calculation of sample means and variances from summary statistics, then applying routine formulas for confidence intervals and hypothesis testing. While it involves multiple steps and careful arithmetic, it requires no novel insight—just systematic application of A-level statistics procedures that students would have practiced extensively.
Spec5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

10 Engineers are investigating the speed of the internet connection received by households in two towns \(P\) and \(Q\). The speeds, in suitable units, in \(P\) and \(Q\) are denoted by \(x\) and \(y\) respectively. For a random sample of 50 houses in town \(P\) and a random sample of 40 houses in town \(Q\) the results are summarised as follows. $$\Sigma x = 240 \quad \Sigma x ^ { 2 } = 1224 \quad \Sigma y = 168 \quad \Sigma y ^ { 2 } = 754$$ Calculate a \(95 \%\) confidence interval for \(\mu _ { P } - \mu _ { Q }\), where \(\mu _ { P }\) and \(\mu _ { Q }\) are the population mean speeds for \(P\) and \(Q\). Test, at the \(1 \%\) significance level, whether \(\mu _ { P }\) is greater than \(\mu _ { Q }\).

Question 10:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(s_P^2 = (1224 - 240^2/50)/49 = 1.469\) or \(72/49\) or \(1.212^2\)M1 A1 Estimate population variance using \(P\)'s sample (allow biased: \(\sigma_{P,50}^2 = 1.44\) or \(1.2^2\))
\(s_Q^2 = (754 - 168^2/40)/39 = 1.241\) or \(242/195\) or \(1.114^2\)A1 Estimate population variance using \(Q\)'s sample (allow biased: \(\sigma_{Q,40}^2 = 1.21\) or \(1.1^2\))
\(s^2 = s_P^2/50 + s_Q^2/40 = 0.0604\) or \(0.246^2\)M1 Estimate population variance for combined sample (allow use of \(\sigma_{P,50}^2\), \(\sigma_{Q,40}^2\): \(0.05905\) or \(0.243^2\))
\((240/50 - 168/40) \pm zs\)M1 Find confidence interval
\(0.6 \pm 0.482\) or \([0.118,\ 1.082]\)A1 Use \(z = 1.96\) and evaluate correct to 3 d.p. (allow use of \(0.05905\) instead of \(0.0604\): \([0.124,\ 1.076]\))
\(H_0: \mu_P = \mu_Q\), \(H_1: \mu_P > \mu_Q\)B1 State hypotheses
\(z = (240/50 - 168/40)/s = 2.44\) (allow \(0.6/0.243 = 2.47\))M1 *A1 Calculate value of \(z\) (to 2 d.p.)
\(z_{0.99} = 2.326\) [or \(2.33\)]*B1 State or use correct tabular \(z\) value (or can compare \(0.6\) with \(0.572\) or \(0.56[5]\))
Reject \(H_0\) if \(z >\) tabular valueM1 Valid method for reaching conclusion
\(\mu_P\) is greater than \(\mu_Q\)A1 Correct conclusion (AEF, dep *A1, *B1)
S.R. Allow (implicit) assumption of equal variances, but deduct A1 if not explicit:
Pooled estimate \(s^2\): \((50\sigma_{P,50}^2 + 40\sigma_{Q,40}^2)/88 = 1.368\)
\(0.6 \pm 1.96s\sqrt{1/50 + 1/40} = 0.6 \pm 0.486\) or \([0.114,\ 1.086]\)
\(0.6/s\sqrt{1/50 + 1/40} = 2.42\) Value of \(z\) (to 2 dp)
Total: 12 marks
## Question 10:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $s_P^2 = (1224 - 240^2/50)/49 = 1.469$ or $72/49$ or $1.212^2$ | M1 A1 | Estimate population variance using $P$'s sample (allow biased: $\sigma_{P,50}^2 = 1.44$ or $1.2^2$) |
| $s_Q^2 = (754 - 168^2/40)/39 = 1.241$ or $242/195$ or $1.114^2$ | A1 | Estimate population variance using $Q$'s sample (allow biased: $\sigma_{Q,40}^2 = 1.21$ or $1.1^2$) |
| $s^2 = s_P^2/50 + s_Q^2/40 = 0.0604$ or $0.246^2$ | M1 | Estimate population variance for combined sample (allow use of $\sigma_{P,50}^2$, $\sigma_{Q,40}^2$: $0.05905$ or $0.243^2$) |
| $(240/50 - 168/40) \pm zs$ | M1 | Find confidence interval |
| $0.6 \pm 0.482$ or $[0.118,\ 1.082]$ | A1 | Use $z = 1.96$ and evaluate correct to 3 d.p. (allow use of $0.05905$ instead of $0.0604$: $[0.124,\ 1.076]$) |
| $H_0: \mu_P = \mu_Q$, $H_1: \mu_P > \mu_Q$ | B1 | State hypotheses |
| $z = (240/50 - 168/40)/s = 2.44$ (allow $0.6/0.243 = 2.47$) | M1 *A1 | Calculate value of $z$ (to 2 d.p.) |
| $z_{0.99} = 2.326$ [or $2.33$] | *B1 | State or use correct tabular $z$ value (or can compare $0.6$ with $0.572$ or $0.56[5]$) |
| Reject $H_0$ if $z >$ tabular value | M1 | Valid method for reaching conclusion |
| $\mu_P$ is greater than $\mu_Q$ | A1 | Correct conclusion (AEF, dep *A1, *B1) |
| **S.R.** Allow (implicit) assumption of equal variances, but deduct A1 if not explicit: | | |
| Pooled estimate $s^2$: $(50\sigma_{P,50}^2 + 40\sigma_{Q,40}^2)/88 = 1.368$ | | |
| $0.6 \pm 1.96s\sqrt{1/50 + 1/40} = 0.6 \pm 0.486$ or $[0.114,\ 1.086]$ | | |
| $0.6/s\sqrt{1/50 + 1/40} = 2.42$ | | Value of $z$ (to 2 dp) |

**Total: 12 marks**

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10 Engineers are investigating the speed of the internet connection received by households in two towns $P$ and $Q$. The speeds, in suitable units, in $P$ and $Q$ are denoted by $x$ and $y$ respectively. For a random sample of 50 houses in town $P$ and a random sample of 40 houses in town $Q$ the results are summarised as follows.

$$\Sigma x = 240 \quad \Sigma x ^ { 2 } = 1224 \quad \Sigma y = 168 \quad \Sigma y ^ { 2 } = 754$$

Calculate a $95 \%$ confidence interval for $\mu _ { P } - \mu _ { Q }$, where $\mu _ { P }$ and $\mu _ { Q }$ are the population mean speeds for $P$ and $Q$.

Test, at the $1 \%$ significance level, whether $\mu _ { P }$ is greater than $\mu _ { Q }$.

\hfill \mbox{\textit{CAIE FP2 2012 Q10 [12]}}