| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type I error |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question covering standard S2 content: sampling methods, Type I error definition, one-tailed z-test with known variance, and CLT understanding. All parts are routine recall and application with no novel problem-solving required. The calculation in part (iii) is mechanical (find sample mean, compute z-statistic, compare to critical value). Slightly easier than average due to the guided structure and basic conceptual questions. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.01d Select/critique sampling: in context2.05a Hypothesis testing language: null, alternative, p-value, significance5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) 2nd more representative of all appointments or Lengths may vary during the day | B1 | Any implication that times or conditions vary throughout day, e.g. doctors get tired |
| or 1st does not include later appts so not representative | B1 [2] | |
| (ii) \(0.01\) o.e. | B1 [1] | |
| Concluding that times spent are too long when they are not. | B1 [1] | Concluding that the mean time spent is more than 10 mins when it is not. Must be in context. |
| (iii) H0: Pop mean appt time (or \(\mu\)) = 10 | B1 | Both correct. Allow \(\mu\), but not just "mean" |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{147 - 10}{\sqrt{3.4^2/12}} = (\pm)2.292\) or (\(0.0109\) if area comparison done) | M1 | Allow incorrect \(10 + 2.326 \times \frac{3.4}{12}\) M1. Must have \(\sqrt{12}\) (accept totals method) |
| A1 | = 12.28 A1 | |
| \("2.292" < 2.326\) o.e. (No evidence to reject H0.) | M1 | For valid comparison \(< 12.28\) M1. Comp "2.292" with 2.326 Or 0.0109 with 0.01 Or 147/12 with 12.28 |
| No reason to believe appts are too long | A1 [5] | Dep 2.326, ft their "2.292" |
| (iv) Normal population | B1 [1] | No contradictions. Must have "population" or equiv |
**(i)** 2nd more representative of all appointments or Lengths may vary during the day | B1 | Any implication that times or conditions vary throughout day, e.g. doctors get tired
or 1st does not include later appts so not representative | B1 [2]
**(ii)** $0.01$ o.e. | B1 [1]
Concluding that times spent are too long when they are not. | B1 [1] | Concluding that the mean time spent is more than 10 mins when it is not. Must be in context.
**(iii)** **H0:** Pop mean appt time (or $\mu$) = 10 | B1 | Both correct. Allow $\mu$, but not just "mean"
**H1:** Pop mean appt time (or $\mu$) $> 10$
$\frac{147 - 10}{\sqrt{3.4^2/12}} = (\pm)2.292$ or ($0.0109$ if area comparison done) | M1 | Allow incorrect $10 + 2.326 \times \frac{3.4}{12}$ M1. Must have $\sqrt{12}$ (accept totals method)
| A1 | = 12.28 A1
$"2.292" < 2.326$ o.e. (No evidence to reject **H0**.) | M1 | For valid comparison $< 12.28$ M1. Comp "2.292" with 2.326 Or 0.0109 with 0.01 Or 147/12 with 12.28
No reason to believe appts are too long | A1 [5] | Dep 2.326, ft their "2.292"
**(iv)** Normal population | B1 [1] | No contradictions. Must have "population" or equiv7 A researcher is investigating the actual lengths of time that patients spend with the doctor at their appointments. He plans to choose a sample of 12 appointments on a particular day.\\
(i) Which of the following methods is preferable, and why?
\begin{itemize}
\item Choose the first 12 appointments of the day.
\item Choose 12 appointments evenly spaced throughout the day.
\end{itemize}
Appointments are scheduled to last 10 minutes. The actual lengths of time, in minutes, that patients spend with the doctor may be assumed to have a normal distribution with mean $\mu$ and standard deviation 3.4. The researcher suspects that the actual time spent is more than 10 minutes on average. To test this suspicion, he recorded the actual times spent for a random sample of 12 appointments and carried out a hypothesis test at the 1\% significance level.\\
(ii) State the probability of making a Type I error and explain what is meant by a Type I error in this context.\\
(iii) Given that the total length of time spent for the 12 appointments was 147 minutes, carry out the test.\\
(iv) Give a reason why the Central Limit theorem was not needed in part (iii).
\hfill \mbox{\textit{CAIE S2 2014 Q7 [10]}}