| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Comment on claim using CI |
| Difficulty | Moderate -0.3 This is a straightforward application of standard S3 confidence interval procedures: calculating sample statistics, applying the normal distribution formula with known population SD, and interpreting the result. All steps are routine textbook exercises requiring only direct formula application with no problem-solving insight, though the multi-part structure and calculation burden place it slightly below average difficulty. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \mu = \frac{85.2}{12} = 7.10\) | M1 A1 | |
| \(s^2 = \frac{1}{11}\left\{906.18 - \frac{(85.2)^2}{12}\right\}\) | M1 | Substitution in correct formula |
| \(= 27.3873\) | A1 ft | Complete correct expression |
| A1 | AWRT 27.4 | |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Confidence interval is given by \(\bar{x} \pm z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\) | M1 | |
| \(7.10 \pm 1.6449 \times \frac{5.1}{\sqrt{12}}\) | A1 ft | Correct expression with their values |
| B1 | 1.6449 | |
| ie:- (4.6783, 9.5216) | A1 A1 | AWRT (4.68, 9.52) |
| (5 marks) |
| Answer | Marks |
|---|---|
| The value 4 is not in the interval; | B1 |
| Thus the claim is not substantiated. | B1 |
| (2 marks) |
## (a)
$\bar{x} = \mu = \frac{85.2}{12} = 7.10$ | M1 A1 |
$s^2 = \frac{1}{11}\left\{906.18 - \frac{(85.2)^2}{12}\right\}$ | M1 | Substitution in correct formula
$= 27.3873$ | A1 ft | Complete correct expression
| | A1 | AWRT 27.4
| (5 marks) |
## (b)
Confidence interval is given by $\bar{x} \pm z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ | M1 |
$7.10 \pm 1.6449 \times \frac{5.1}{\sqrt{12}}$ | A1 ft | Correct expression with their values
| | B1 | 1.6449
ie:- (4.6783, 9.5216) | A1 A1 | AWRT (4.68, 9.52)
| (5 marks) |
## (c)
The value 4 is not in the interval; | B1 |
Thus the claim is not substantiated. | B1 |
| (2 marks) |Observations have been made over many years of $T$, the noon temperature in °C, on 21st March at Sunnymere. The records for a random sample of 12 years are given below.
5.2, 3.1, 10.6, 12.4, 4.6, 8.7, 2.5, 15.3, $-1.5$, 1.8, 13.2, 9.3.
\begin{enumerate}[label=(\alph*)]
\item Find unbiased estimates of the mean and variance of $T$. [5]
\end{enumerate}
Over the years, the standard deviation of $T$ has been found to be 5.1.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Assuming a normal distribution find a 90\% confidence interval for the mean of $T$. [5]
\end{enumerate}
A meteorologist claims that the mean temperature at noon in Sunnymere on 21st March is 4 °C.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use your interval to comment on the meteorologist's claim. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q8 [12]}}