| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Confidence interval interpretation |
| Difficulty | Moderate -0.3 This is a straightforward application of the central limit theorem to construct a confidence interval using sample statistics, followed by a simple interpretation. The calculations are routine (finding sample mean and variance, applying the z-value for 97%, checking if 267 lies in the interval), requiring no conceptual insight beyond standard A-level Further Maths statistics procedures. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \frac{35522}{100} = 355.22\) | B1 | Correct value of \(\bar{x}\) |
| \(s^2 = \frac{1}{99}\left(32902257 - \frac{35522^2}{100}\right) = 204890.2238\) | B1 | AWRT \(s^2 = 204890\) or \(s = 453\) |
| \(z = 2.1701\) | B1 | AWRT 2.17 or correct \(t\) value AWRT 2.20 |
| \(355.22 \pm 2.1701 \times \sqrt{\frac{204890.2238}{100}}\) | M1 | Uses formula for upper or lower limit of confidence interval using their values |
| \((257, 453)\) | A1 | AWRT 257 and 453, or 256 and 455 if \(t\) value used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Rebekah is correct as 267 is within the confidence interval | E1F | Infers Rebekah's statement or Mike's claim is supported by comparing proposed population mean with confidence interval, provided 267 lies within it. Condone use of "it" for 267 |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \frac{35522}{100} = 355.22$ | B1 | Correct value of $\bar{x}$ |
| $s^2 = \frac{1}{99}\left(32902257 - \frac{35522^2}{100}\right) = 204890.2238$ | B1 | AWRT $s^2 = 204890$ or $s = 453$ |
| $z = 2.1701$ | B1 | AWRT 2.17 or correct $t$ value AWRT 2.20 |
| $355.22 \pm 2.1701 \times \sqrt{\frac{204890.2238}{100}}$ | M1 | Uses formula for upper or lower limit of confidence interval using their values |
| $(257, 453)$ | A1 | AWRT 257 and 453, or 256 and 455 if $t$ value used |
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Rebekah is correct as 267 is within the confidence interval | E1F | Infers Rebekah's statement or Mike's claim is supported by comparing proposed population mean with confidence interval, provided 267 lies within it. Condone use of "it" for 267 |
---5 Rebekah is investigating the distances, $X$ light years, between the Earth and visible stars in the night sky.
She determines the distance between the Earth and a star for a random sample of 100 visible stars.
The summarised results are as follows:
$$\sum x = 35522 \quad \text { and } \quad \sum x ^ { 2 } = 32902257$$
5
\begin{enumerate}[label=(\alph*)]
\item Calculate a 97\% confidence interval for the population mean of $X$, giving your values to the nearest light year.\\
5
\item Mike claims that the population mean is 267 light years.
Rebekah says that the confidence interval supports Mike's claim.
State, with a reason, whether Rebekah is correct.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2023 Q5 [6]}}