| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Moderate -0.8 This is a straightforward confidence interval calculation with standard normal distribution (large sample, n=100) followed by basic interpretation. The question requires only routine application of the CI formula and simple comparison to the hypothesized mean—no complex reasoning or novel insight needed. Easier than average A-level statistics questions. |
| Spec | 2.01a Population and sample: terminology5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Pop too big or takes too long oe or testing destroys articles oe | B1 [1] | or too expensive oe; or pop inaccessible oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = 1.96\) | B1 | seen |
| \(65.7 \pm z \times \frac{\sqrt{15}}{10}\) | M1 | Expression of correct form (must be \(z\), must be 65.7) |
| \(= 64.9\) to \(66.5\) (3 sf) | A1 [3] | Must be an interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CI does not include 64.7; Probably has affected (or increased) mean bounce ht. | B1\(\checkmark\) [1] | allow 64.7 not within CI; both needed. ft their CI ft 65.7/64.7 mix |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Pop too big or takes too long oe or testing destroys articles oe | B1 [1] | or too expensive oe; or pop inaccessible oe |
### Part (ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 1.96$ | B1 | seen |
| $65.7 \pm z \times \frac{\sqrt{15}}{10}$ | M1 | Expression of correct form (must be $z$, must be 65.7) |
| $= 64.9$ to $66.5$ (3 sf) | A1 [3] | Must be an interval |
### Part (ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CI does not include 64.7; Probably has affected (or increased) mean bounce ht. | B1$\checkmark$ [1] | allow 64.7 not within CI; both needed. ft their CI ft 65.7/64.7 mix |3 (i) Give a reason for using a sample rather than the whole population in carrying out a statistical investigation.\\
(ii) Tennis balls of a certain brand are known to have a mean height of bounce of 64.7 cm , when dropped from a height of 100 cm . A change is made in the manufacturing process and it is required to test whether this change has affected the mean height of bounce. 100 new tennis balls are tested and it is found that their mean height of bounce when dropped from a height of 100 cm is 65.7 cm and the unbiased estimate of the population variance is $15 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate a $95 \%$ confidence interval for the population mean.
\item Use your answer to part (ii) (a) to explain what conclusion can be drawn about whether the change has affected the mean height of bounce.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2016 Q3 [5]}}