| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question requiring basic recall and simple calculations: stating assumptions, reading off values, calculating SE = s/√n, and verifying the margin of error. All parts are routine textbook exercises with no problem-solving or novel insight required, making it slightly easier than average despite being Further Maths content. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| Mean | 2.6025 |
| s | 0.2793 |
| SE | |
| N | 8 |
| df | 7 |
| Interval | \(2.6025 \pm 0.2335\) |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | Underlying distribution of caesium levels needs to |
| be Normal | E1 | |
| [1] | Context not required | |
| 4 | (b) | 2.369 < μ < 2.836 |
| [1] | . | |
| 4 | (c) | 0.2793 |
| Answer | Marks |
|---|---|
| = 0.09875 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | Allow 0.0987 or 0.0988 | |
| 4 | (d) | t value = 2.365 |
| t value × SE = 2.365 × 0.09875 = 0.2335 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | AG | |
| 4 | (e) | By using a higher confidence level |
| [1] | Do NOT allow ‘a stricter interval’ |
Question 4:
4 | (a) | Underlying distribution of caesium levels needs to
be Normal | E1
[1] | Context not required
4 | (b) | 2.369 < μ < 2.836 | B1
[1] | .
4 | (c) | 0.2793
SE
8
= 0.09875 | M1
A1
[2] | Allow 0.0987 or 0.0988
4 | (d) | t value = 2.365
t value × SE = 2.365 × 0.09875 = 0.2335 | B1
B1
[2] | AG
4 | (e) | By using a higher confidence level | E1
[1] | Do NOT allow ‘a stricter interval’
Do not allow higher significance level.
Do not allow higher confidence level if
then contradicted by writing eg 90%4 Shellfish in the sea near nuclear power stations are regularly monitored for levels of radioactivity. On a particular occasion, the levels of caesium-137 (a radioactive isotope) in a random sample of 8 cockles, measured in becquerels per kilogram, were as follows.\\
$\begin{array} { l l l l l l l l } 2.36 & 2.97 & 2.69 & 3.00 & 2.51 & 2.45 & 2.21 & 2.63 \end{array}$
Software is used to produce a 95\% confidence interval for the level of caesium-137 in the cockles. The output from the software is shown in Fig. 4. The value for 'SE' has been deliberately omitted.
T Estimate of a Mean\\
Confidence Level 0.95
Sample\\
Mean 2.6025\\
s 0.2793\\
□\\
0.2793
N □ 8
Result
T Estimate of a Mean
\begin{table}[h]
\begin{center}
\begin{tabular}{ l l }
Mean & 2.6025 \\
s & 0.2793 \\
SE & \\
N & 8 \\
df & 7 \\
Interval & $2.6025 \pm 0.2335$ \\
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item State an assumption necessary for the use of the $t$ distribution in the construction of this confidence interval.
\item State the confidence interval which the software gives in the form $a < \mu < b$.
\item In the software output shown in Fig. 4, SE stands for standard error. Find the standard error in this case.
\item Show how the value of 0.2335 in the confidence interval was calculated.
\item State how, using this sample, a wider confidence interval could be produced.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2019 Q4 [7]}}