| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a straightforward application of a one-sample t-test and confidence interval with standard procedures: calculate sample mean and standard deviation, apply the t-test formula, compare to critical value, then construct the confidence interval. While it requires careful calculation and knowledge of t-distribution tables, it involves no conceptual challenges or novel problem-solving—just direct application of learned techniques with clear instructions. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 9(i) | x = 13.88/8 = 1.735 (allow 1.73 or 1.74 for this B1) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [ = 13/2500 or 0.0052 or 0.072112 ] | M1 | Estimate population variance |
| Answer | Marks | Guidance |
|---|---|---|
| 0 1 | B1 | State hypotheses (B0 for x …) |
| t = (x – 1.7) / (s/√8) = 1.37 | M1A1 | Find value of t |
| Answer | Marks | Guidance |
|---|---|---|
| 7, 0.95 | B1 | State or use correct tabular t-value |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | B1 | Consistent conclusion (FT on both t-values) |
| Answer | Marks | Guidance |
|---|---|---|
| 9(ii) | 13.88/8 ± t √(s2 /8) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7, 0.975 | A1 | State or use correct tabular value of t |
| 1.73[5] ± 0.06 or [1.67, 1.80] | A1 | Evaluate confidence interval (either form) |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 9:
--- 9(i) ---
9(i) | x = 13.88/8 = 1.735 (allow 1.73 or 1.74 for this B1) | B1 | Find sample mean
s2 = (24.1182 – 13.882 /8) / 7
[ = 13/2500 or 0.0052 or 0.072112 ] | M1 | Estimate population variance
(allow biased here: 0.00455 or 0.067452)
H : µ = 1.7, H : µ > 1.7 (AEF)
0 1 | B1 | State hypotheses (B0 for x …)
t = (x – 1.7) / (s/√8) = 1.37 | M1A1 | Find value of t
t = 1.89[5]
7, 0.95 | B1 | State or use correct tabular t-value
(or can compare x with 1.7 + 0.048 = 1.75 )
[Accept H :] Mean ht. not greater than 1.7 m (AEF)
0 | B1 | Consistent conclusion (FT on both t-values)
7
--- 9(ii) ---
9(ii) | 13.88/8 ± t √(s2 /8) | M1 | Find confidence interval
t = 2.36[5]
7, 0.975 | A1 | State or use correct tabular value of t
1.73[5] ± 0.06 or [1.67, 1.80] | A1 | Evaluate confidence interval (either form)
3
Question | Answer | Marks | GuidanceThere are a large number of students at a particular college. The heights, in metres, of a random sample of 8 students are as follows.
$$1.75 \quad 1.72 \quad 1.62 \quad 1.70 \quad 1.82 \quad 1.75 \quad 1.68 \quad 1.84$$
You may assume that heights of students are normally distributed.
\begin{enumerate}[label=(\roman*)]
\item Test, at the 5\% significance level, whether the population mean height of students at this college is greater than 1.70 metres. [7]
\item Find a 95\% confidence interval for the population mean height of students at this college. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2018 Q9 [10]}}