| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Calculate probabilities using sample mean distribution |
| Difficulty | Standard +0.3 This is a straightforward application of the Central Limit Theorem with standard normal distribution calculations. Part (i) requires a routine z-score calculation for a sample mean. Part (ii) involves a standard hypothesis test with clear parameters given. The conceptual questions about one-tail tests and CLT necessity are bookwork recall. All steps are mechanical with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{4.8}{\sqrt{40}}\) | B1 | or \(\frac{4.8^2}{40}\). Accept \(4.8\sqrt{40}\) or \(4.8^2 \times 40\) for totals method |
| \(\frac{50.3 - 49.5}{\dfrac{4.8}{\sqrt{40}}}\) \((= 1.054)\) | M1 | For standardising with their SD. Accept \(\pm\). Accept totals method. No mixed methods |
| \(1 - \Phi(``1.054")\) | M1 | For use of tables and finding area consistent with their working |
| \(= 0.146\) (3 sf) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Looking for decrease | B1 | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): Pop mean time spent (or \(\mu\)) \(= 49.5\); \(H_1\): Pop mean time spent (or \(\mu\)) \(< 49.5\) | B1 | Not just "mean time spent" |
| \(\frac{\dfrac{1920}{40} - 49.5}{\dfrac{4.8}{\sqrt{40}}}\) \((= -1.976)\) | M1 | For standardising. Allow \(\div \frac{4.8}{40}\). Accept totals method; CV method. No mixed methods |
| \(`1.976' > 1.555\) (or \(`{-1.976}' < -1.555\)) | M1 | For valid comparison (area comparison \(0.024 < 0.06\)) |
| There is evidence that mean time has decreased. | A1 | CWO. No contradictions in conclusions. Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Population normally distributed, so No | B1 | Both needed. Total: 1 |
## Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{4.8}{\sqrt{40}}$ | B1 | or $\frac{4.8^2}{40}$. Accept $4.8\sqrt{40}$ or $4.8^2 \times 40$ for totals method |
| $\frac{50.3 - 49.5}{\dfrac{4.8}{\sqrt{40}}}$ $(= 1.054)$ | M1 | For standardising with their SD. Accept $\pm$. Accept totals method. No mixed methods |
| $1 - \Phi(``1.054")$ | M1 | For use of tables and finding area consistent with their working |
| $= 0.146$ (3 sf) | A1 | **Total: 4** |
## Question 5(ii)(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Looking for decrease | B1 | **Total: 1** |
## Question 5(ii)(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: Pop mean time spent (or $\mu$) $= 49.5$; $H_1$: Pop mean time spent (or $\mu$) $< 49.5$ | B1 | Not just "mean time spent" |
| $\frac{\dfrac{1920}{40} - 49.5}{\dfrac{4.8}{\sqrt{40}}}$ $(= -1.976)$ | M1 | For standardising. Allow $\div \frac{4.8}{40}$. Accept totals method; CV method. No mixed methods |
| $`1.976' > 1.555$ (or $`{-1.976}' < -1.555$) | M1 | For valid comparison (area comparison $0.024 < 0.06$) |
| There is evidence that mean time has decreased. | A1 | CWO. No contradictions in conclusions. **Total: 4** |
## Question 5(ii)(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Population normally distributed, so No | B1 | Both needed. **Total: 1** |
---5 The number of hours that Mrs Hughes spends on her business in a week is normally distributed with mean $\mu$ and standard deviation 4.8. In the past the value of $\mu$ has been 49.5.
\begin{enumerate}[label=(\roman*)]
\item Assuming that $\mu$ is still equal to 49.5 , find the probability that in a random sample of 40 weeks the mean time spent on her business in a week is more than 50.3 hours.
Following a change in her arrangements, Mrs Hughes wishes to test whether $\mu$ has decreased. She chooses a random sample of 40 weeks and notes that the total number of hours she spent on her business during these weeks is 1920.
\item (a) Explain why a one-tail test is appropriate.\\
(b) Carry out the test at the 6\% significance level.\\
(c) Explain whether it was necessary to use the Central Limit theorem in part (ii) (b).
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2014 Q5 [10]}}