| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample confidence interval |
| Difficulty | Standard +0.3 This is a standard paired t-test confidence interval calculation requiring students to compute differences, find mean and standard deviation, then apply the t-distribution formula. While it involves multiple computational steps and interpretation, it follows a routine procedure taught in Further Statistics with no novel insight required. The context is straightforward and the final interpretation is direct. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Twin pair | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| IQ of twin raised by natural parents | 82 | 92 | 115 | 132 | 88 | 95 | 112 | 83 | 123 |
| IQ of twin raised by adoptive parents | 92 | 88 | 115 | 134 | 97 | 94 | 107 | 88 | 130 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(-10\ \ 4\ \ 0\ \ -2\ \ -9\ \ 1\ \ 5\ \ -5\ \ -7\) | M1 | Consider differences \(N - A\) (or \(A - N\)) |
| \(\bar{x} = -23/9\ [= -2.556;\) allow \(+]\) | Estimate mean | |
| \(s^2 = (301 - 23^2/9)/8 = 30.28\) *or* \(5.503^2\) | Estimate population variance (to 3 sf) | |
| \((26.91\) *or* \(5.188^2)\) | M1 A1 | (allow biased) |
| \(\bar{x} \pm ts/\sqrt{n}\) for any \(t\) | M1 | Use valid formula (M1 needs \(n\) consistent with \(s\)) |
| \(t_{8,\,0.95} = 1.86_{[0]}\) | *B1 | Use correct tabular \(t\) value |
| \(-2.56 \pm 3.41\) *or* \([-5.97,\ 0.86]\) | *A1 | Evaluate confidence interval (dep *B1). Part mark: 6 |
| 0 lies in interval (A.E.F.) | M1 | Use valid argument consistent with interval |
| Yes, supports claim (A.E.F.) | A1 | State conclusion (dep *A1 apart from rounding). Part mark: 2. Total: [8] |
## Question 8:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $-10\ \ 4\ \ 0\ \ -2\ \ -9\ \ 1\ \ 5\ \ -5\ \ -7$ | M1 | Consider differences $N - A$ (or $A - N$) |
| $\bar{x} = -23/9\ [= -2.556;$ allow $+]$ | | Estimate mean |
| $s^2 = (301 - 23^2/9)/8 = 30.28$ *or* $5.503^2$ | | Estimate population variance (to 3 sf) |
| $(26.91$ *or* $5.188^2)$ | M1 A1 | (allow biased) |
| $\bar{x} \pm ts/\sqrt{n}$ for any $t$ | M1 | Use valid formula (M1 needs $n$ consistent with $s$) |
| $t_{8,\,0.95} = 1.86_{[0]}$ | *B1 | Use correct tabular $t$ value |
| $-2.56 \pm 3.41$ *or* $[-5.97,\ 0.86]$ | *A1 | Evaluate confidence interval (dep *B1). Part mark: 6 |
| 0 lies in interval (A.E.F.) | M1 | Use valid argument consistent with interval |
| Yes, supports claim (A.E.F.) | A1 | State conclusion (dep *A1 apart from rounding). Part mark: 2. Total: **[8]** |
---8 Part of a research study of identical twins who had been separated at birth involved a random sample of 9 pairs, in which one twin had been raised by the natural parents and the other by adoptive parents. The IQ scores of these twins were measured, with the following results.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | }
\hline
Twin pair & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
IQ of twin raised by natural parents & 82 & 92 & 115 & 132 & 88 & 95 & 112 & 83 & 123 \\
\hline
IQ of twin raised by adoptive parents & 92 & 88 & 115 & 134 & 97 & 94 & 107 & 88 & 130 \\
\hline
\end{tabular}
\end{center}
It may be assumed that the difference in IQ scores has a normal distribution. The mean IQ scores of separated twins raised by natural parents and by adoptive parents are denoted by $\mu _ { N }$ and $\mu _ { A }$ respectively. Obtain a $90 \%$ confidence interval for $\mu _ { N } - \mu _ { A }$.
One of the researchers claimed that there was no evidence of a difference between the two population means. State, giving a reason, whether the confidence interval supports this claim.
\hfill \mbox{\textit{CAIE FP2 2009 Q8 [8]}}