| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.3 Part (a) is a standard confidence interval calculation requiring direct application of the formula with given values. Part (b) involves understanding the relationship between confidence level, sample size, and interval width, requiring algebraic manipulation but following predictable patterns. This is slightly above average difficulty due to the inverse reasoning in part (b), but remains a textbook-style question testing understanding of confidence interval properties rather than requiring novel insight. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(63 \pm z \times \frac{9}{\sqrt{100}}\) | M1 | Expression of correct form, any \(z\) |
| \(z = 1.645\) | B1 | Seen |
| \(61.5\) to \(64.5\) (3 sf) | A1 [3] | Must be an interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z = \frac{1.96}{2}\) \((= 0.98)\) | M1 | Allow \(\frac{\text{any}\ z}{2}\) |
| \(\Phi(0.98)\) \((= 0.8365)\) | ||
| \(0.8365 - (1 - 0.8365)\) \((= 0.673)\) | M1 | |
| \(\alpha = 67.3\) (3 sf) | A1 [3] | Allow 67 from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4 = (2x'z'x'\sigma')/\sqrt{n}\) | M1 | Attempt to solve equation of correct form |
| \(n = 200\) | A1 [2] | SR B1 for \(n = 100\) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $63 \pm z \times \frac{9}{\sqrt{100}}$ | M1 | Expression of correct form, any $z$ |
| $z = 1.645$ | B1 | Seen |
| $61.5$ to $64.5$ (3 sf) | A1 [3] | Must be an interval |
### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \frac{1.96}{2}$ $(= 0.98)$ | M1 | Allow $\frac{\text{any}\ z}{2}$ |
| $\Phi(0.98)$ $(= 0.8365)$ | | |
| $0.8365 - (1 - 0.8365)$ $(= 0.673)$ | M1 | |
| $\alpha = 67.3$ (3 sf) | A1 [3] | Allow 67 from correct working |
### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4 = (2x'z'x'\sigma')/\sqrt{n}$ | M1 | Attempt to solve equation of correct form |
| $n = 200$ | A1 [2] | SR B1 for $n = 100$ |
---5
\begin{enumerate}[label=(\alph*)]
\item The masses, in grams, of certain tomatoes are normally distributed with standard deviation 9 grams. A random sample of 100 tomatoes has a sample mean of 63 grams. Find a $90 \%$ confidence interval for the population mean mass of these tomatoes.
\item The masses, in grams, of certain potatoes are normally distributed with known population standard deviation but unknown population mean. A random sample of potatoes is taken in order to find a confidence interval for the population mean. Using a sample of size 50 , a $95 \%$ confidence interval is found to have width 8 grams.
\begin{enumerate}[label=(\roman*)]
\item Using another sample of size 50 , an $\alpha \%$ confidence interval has width 4 grams. Find $\alpha$.
\item Find the sample size $n$, such that a $95 \%$ confidence interval has width 4 grams.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2016 Q5 [8]}}