| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Confidence interval interpretation |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question testing standard procedures: calculating a CI from data, interpreting it, explaining the random sampling assumption, and working backwards from a given CI to find sample statistics. All parts involve routine application of formulas with no novel problem-solving or proof required. Slightly easier than average due to mechanical nature. |
| Spec | 2.01a Population and sample: terminology5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | DR |
| Answer | Marks |
|---|---|
| ± 0.081 or | M1 |
| Answer | Marks |
|---|---|
| [6] | 1.1 |
| Answer | Marks |
|---|---|
| 3.4 | Accept denominator of 40 rather than 39 for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | AIt ldloowes support± t h0e.0 b8e0li e of rb ecause the confidence |
| interval does not contain 12.2 | E1 | |
| [1] | 3.5a | Must be unassertive EG do not allow ‘the researcher is correct’ |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (c) | A random sample enables proper inference about the |
| population to be undertaken | E1 | |
| [1] | 2.4 | Do NOT allow ‘so that the distribution can be modelled by a |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (d) | Sample mean = 1.3 |
| Answer | Marks |
|---|---|
| Sample size = 100 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Do not allow M1 for 0.25 rather than |
Question 6:
6 | (a) | DR
491.842
6050.3−
40
Est of pop variance =
39
2.63536
= =0.06757
39
Confidence interval is
12.296
1.96
0.06757 0.25994
× or ×
±
40 40
= 12.296 (12.215, 12.377)
12.296 (12.216, 12.376)
± 0.081 or | M1
A1
B1
M1
M1
A1
[6] | 1.1
1.1
1.1
3.3
1.1
3.4 | Accept denominator of 40 rather than 39 for M1
Allow 0.676 Or sd = 0.25994
or seen anywhere
491.84
Accept t-value of 2.02
40
Accept based on t-distribution (12.213, 12.379)
Allow 12.22 to 12.38
6 | (b) | AIt ldloowes support± t h0e.0 b8e0li e of rb ecause the confidence
interval does not contain 12.2 | E1
[1] | 3.5a | Must be unassertive EG do not allow ‘the researcher is correct’
FT their interval
6 | (c) | A random sample enables proper inference about the
population to be undertaken | E1
[1] | 2.4 | Do NOT allow ‘so that the distribution can be modelled by a
Normal…’ oe
6 | (d) | Sample mean = 1.3
0.25
0.098=1.96×
n
Sample size = 100 | B1
M1
A1
[3] | 1.1
3.1b
1.1 | Do not allow M1 for 0.25 rather than
√0.25\begin{enumerate}[label=(\alph*)]
\item Determine a 95\% confidence interval for the mean weight of liquid paraffin in a tub.
\item Explain whether the confidence interval supports the researcher's belief.
\item Explain why the sample has to be random in order to construct the confidence interval.\\[0pt]
\item A 95\% confidence interval for the mean weight in grams of another ingredient in the skin cream is [1.202, 1.398]. This confidence interval is based on a large sample and the unbiased estimate of the population variance calculated from the sample is 0.25 .
Find each of the following.
\begin{itemize}
\item The mean of the sample
\item The size of the sample
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2022 Q6 [11]}}