| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Paired sample confidence interval |
| Difficulty | Moderate -0.3 This is a straightforward application of CLT with standard confidence interval interpretation. Part (a) tests basic CLT knowledge (n=60 is large), parts (b-c) are routine calculations from given output, and part (d) requires simple interpretation of whether 0 is in the interval. All parts are textbook-standard with no novel problem-solving required. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Z Estimate of a Mean | |||
| Confidence Level |
| ||
| Sample | |||
| Mean 0.1173 | |||
| □ | |||
| Result | |||
| Z Estimate of a Mean | |||
| Mean | 0.1173 | ||
| \(\sigma\) | 0.5766 | ||
| SE | 0.07444 | ||
| N | 60 | ||
| Lower Limit | -0.0286 | ||
| Upper Limit | 0.2632 | ||
| Interval | \(0.1173 \pm 0.1459\) | ||
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | Because the sample size is large |
| Answer | Marks |
|---|---|
| approximately Normally distributed | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 2.4 | |
| 2.4 | Must mention sample mean for second mark | |
| 4 | (b) | –0.0286 < µ < 0.2632 |
| [1] | 1.1 | |
| 4 | (c) | This is σ/√60 |
| [1] | 3.4 | |
| 4 | (d) | The confidence interval contains zero |
| Answer | Marks |
|---|---|
| agree. | E1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 2.4 | Allow valid alternative answers such |
Question 4:
4 | (a) | Because the sample size is large
so by the central limit theorem the sample mean is
approximately Normally distributed | E1
E1
[2] | 2.4
2.4 | Must mention sample mean for second mark
4 | (b) | –0.0286 < µ < 0.2632 | B1
[1] | 1.1
4 | (c) | This is σ/√60 | B1
[1] | 3.4
4 | (d) | The confidence interval contains zero
Which suggests that the two measurements broadly
agree. | E1
E1
[2] | 3.4
2.4 | Allow valid alternative answers such
as ‘Although the interval contains
zero almost all of the interval is
above zero, which suggests that on
average the digital gauge may be
reading lower than the traditional
gauge.’4 An amateur meteorologist records the total rainfall at her home each day using a traditional rain gauge. This means that she has to go out each day at 9 am to read the rain gauge and then to empty it. She wants to save time by using a digital rain gauge, but she also wants to ensure that the readings from the digital gauge are similar to those of her traditional gauge. Over a period of 100 days, she uses both gauges to measure the rainfall.
The meteorologist uses software to produce a 95\% confidence interval for the difference between the two readings (the traditional gauge reading minus the digital gauge reading). The output from the software is shown in Fig. 4. Although rainfall was measured over a period of 100 days, there was no rain on 40 of those days and so the sample size in the software output is 60 rather than 100.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|}
\hline
\multicolumn{2}{|c|}{Z Estimate of a Mean} \\
\hline
Confidence Level & \begin{tabular}{l}
□ \\
0.95 \\
\end{tabular} \\
\hline
\multicolumn{2}{|l|}{Sample} \\
\hline
\multicolumn{2}{|c|}{Mean 0.1173} \\
\hline
\multicolumn{2}{|c|}{□} \\
\hline
& \\
\hline
\multicolumn{2}{|l|}{Result} \\
\hline
\multicolumn{2}{|l|}{Z Estimate of a Mean} \\
\hline
Mean & 0.1173 \\
\hline
$\sigma$ & 0.5766 \\
\hline
SE & 0.07444 \\
\hline
N & 60 \\
\hline
Lower Limit & -0.0286 \\
\hline
Upper Limit & 0.2632 \\
\hline
Interval & $0.1173 \pm 0.1459$ \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Explain why this confidence interval can be calculated even though nothing is known about the distribution of the population of differences.
\item State the confidence interval which the software gives in the form $a < \mu < b$.
\item Show how the value 0.07444 (labelled SE) was calculated.
\item Comment on whether you think that the confidence interval suggests that the two different methods of measurement are broadly in agreement.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2020 Q4 [6]}}