| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unbiased estimator from raw data |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for unbiased estimators and standard error. Part (a) requires basic understanding of stratified sampling, parts (b) and (c) are direct formula substitutions with given summary statistics, and part (d) requires simple interpretation. No problem-solving insight or complex manipulation needed—purely routine S3 content. |
| Spec | 2.01a Population and sample: terminology2.01c Sampling techniques: simple random, opportunity, etc2.01d Select/critique sampling: in context5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Label staff (from \(1-16\)) and children (from \(1-40\)) | B1 | For labelling/numbering/listing staff and children |
| Use random numbers to select | B1 | For use of random numbers or "randomly select" in each group (may be implied) |
| 4 staff and 10 children | B1 | For selecting correct number of staff and children |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \hat{\mu} = 31.2142...\) awrt \(\mathbf{31.2}\) | B1 | For awrt 31.2 |
| \(s^2 = \frac{26983 - 14 \times \text{"31.2..."}^2}{13}\) | M1 | For correct expression ft their \(\bar{x}\), allow transcription error in \(\sum x^2\) e.g. 29683 |
| A1ft | Fully correct expression ft their \(\bar{x}\) only | |
| \(= 1026.33...\) awrt \(\mathbf{1030}\) | A1 | For awrt 1030 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sqrt{1026.33...}}{\sqrt{14}} = 8.562...\) awrt \(\mathbf{8.56}\) | M1, A1 | M1 for attempting \(\frac{\text{"their }s"}{\sqrt{14}}\) (must have 14). A1 for awrt 8.56 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The variation within each stratum is quite small (o.e.) | B1 | For suitable comment about variation (se) suggesting that variation (se) within strata is less than that overall |
| The difference in the means will be quite large, so variations from the overall mean will be large giving a larger overall s.e. | B1 | For suitable reason about means, pointing out that individuals' weights will vary a lot from the overall mean and so overall s.e. will be higher |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Label staff (from $1-16$) and children (from $1-40$) | B1 | For labelling/numbering/listing staff and children |
| Use random numbers to select | B1 | For use of random numbers or "randomly select" in each group (may be implied) |
| 4 staff and 10 children | B1 | For selecting correct number of staff and children |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \hat{\mu} = 31.2142...$ awrt $\mathbf{31.2}$ | B1 | For awrt 31.2 |
| $s^2 = \frac{26983 - 14 \times \text{"31.2..."}^2}{13}$ | M1 | For correct expression ft their $\bar{x}$, allow transcription error in $\sum x^2$ e.g. 29683 |
| | A1ft | Fully correct expression ft their $\bar{x}$ only |
| $= 1026.33...$ awrt $\mathbf{1030}$ | A1 | For awrt 1030 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sqrt{1026.33...}}{\sqrt{14}} = 8.562...$ awrt $\mathbf{8.56}$ | M1, A1 | M1 for attempting $\frac{\text{"their }s"}{\sqrt{14}}$ (must have 14). A1 for awrt 8.56 |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The variation within each stratum is quite small (o.e.) | B1 | For suitable comment about variation (se) suggesting that variation (se) within strata is less than that overall |
| **The difference in the means will be quite large**, so variations from the overall mean will be large giving a larger overall s.e. | B1 | For suitable reason about means, pointing out that individuals' weights will vary a lot from the overall mean and so overall s.e. will be higher |
---3. A nursery has 16 staff and 40 children on its records. In preparation for an outing the manager needs an estimate of the mean weight of the people on its records and decides to take a stratified sample of size 14 .
\begin{enumerate}[label=(\alph*)]
\item Describe how this stratified sample should be taken.
The weights, $x \mathrm {~kg}$, of each of the 14 people selected are summarised as
$$\sum x = 437 \text { and } \sum x ^ { 2 } = 26983$$
\item Find unbiased estimates of the mean and the variance of the weights of all the people on the nursery's records.
\item Estimate the standard error of the mean.
The estimates of the standard error of the mean for the staff and for the children are 5.11 and 1.10 respectively.
\item Comment on these values with reference to your answer to part (c) and give a reason for any differences.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2015 Q3 [11]}}