| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Carry out hypothesis test |
| Difficulty | Moderate -0.8 This is a straightforward question testing standard knowledge of confidence intervals for normal distributions. Part (a) requires simple recall of the normality assumption, part (b) asks for basic interpretation of a probability plot (points lie approximately on a straight line), and part (c) is a routine calculation of a t-confidence interval with small sample size. While it's Further Maths content, the question involves no problem-solving or novel insight—just application of standard procedures. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | The population from which the sample is drawn must |
| be Normally distributed | E1 | |
| [1] | 1.2 | Do not allow ‘The sample is Normally distributed’ |
| 7 | (b) | Because the Normal probability plot appears to be |
| linear. | E1 | |
| [1] | 3.5a | Do not allow ‘linear correlation’ |
| 7 | (c) | Sample mean = 285.2 |
| Answer | Marks |
|---|---|
| 269.1 < μ < 301.3 or 285.2 ± 16.1 | B1 |
| Answer | Marks |
|---|---|
| [7] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | DR |
Question 7:
7 | (a) | The population from which the sample is drawn must
be Normally distributed | E1
[1] | 1.2 | Do not allow ‘The sample is Normally distributed’
7 | (b) | Because the Normal probability plot appears to be
linear. | E1
[1] | 3.5a | Do not allow ‘linear correlation’
7 | (c) | Sample mean = 285.2
Sample SD = 15.30
Confidence interval is given by
15.30
285.2±2.571×
√6
269.1 < μ < 301.3 or 285.2 ± 16.1 | B1
B1
M1
M1
A1
M1
A1
[7] | 1.1
1.1
3.4
1.1a
1.1
1.1
1.1 | DR
BC Allow 285
BC
For general form
For 5 degrees of freedom
For 2.571
For
15.30
OE √6
Condone 269 < μ < 301 provided working seen, but must be from
285.2, not from 2857 The lengths in mm of a random sample of 6 one-year-old fish of a particular species are as follows.\\
$\begin{array} { l l l l l l } 271 & 293 & 306 & 287 & 264 & 290 \end{array}$
\begin{enumerate}[label=(\alph*)]
\item State an assumption required in order to find a confidence interval for the mean length of one-year-old fish of this species.
Fig. 7 shows a Normal probability plot for these data.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d36bc92-07ac-40c3-9e75-26f2bc9d2fcc-07_599_753_646_246}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
\item Explain why the Normal probability plot suggests that the assumption in part (a) may be valid.
\item In this question you must show detailed reasoning.
Assuming that this assumption is true, find a 95\% confidence interval for the mean length of one-year-old fish of this species.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2020 Q7 [9]}}