| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Standard +0.3 This is a straightforward confidence interval interpretation question requiring standard calculations and basic critical thinking. Part (a) is routine CI construction, parts (b) and (e) require simple observations about CI position/width relative to the target value, part (c) tests understanding of CLT assumptions, and part (d) is trivial reformatting. No novel insight or complex reasoning required—slightly easier than average due to its guided structure. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Mean | 19.68 |
| s | 1.38 |
| N | 40 |
| Mean | 19.68 |
| s | 1.38 |
| SE | 0.2182 |
| N | 40 |
| Interval | \(19.68 \pm 0.4277\) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 7652 |
| Answer | Marks |
|---|---|
| = 19.125 ± 1.034 or (18.09, 20.16) | M1 |
| Answer | Marks |
|---|---|
| [6] | 1.1 |
| Answer | Marks |
|---|---|
| 3.4 | Accept denominator of 40 rather than 39 for M1 leading to |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | It does to some extent because the confidence interval |
| Answer | Marks |
|---|---|
| accurate. | E1 |
| Answer | Marks |
|---|---|
| [2] | 3.5a |
| 2.2b | Must be unassertive FT their interval |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | If the population from which the sample was drawn was |
| Answer | Marks |
|---|---|
| sample size | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 2.4 | Allow ‘don’t know if population is Normally distributed so |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | 19.25 < µ < 20.11 |
| [1] | 1.1 | |
| 5 | (e) | It suggests that he may be correct because the interval |
| Answer | Marks |
|---|---|
| And the variance is much lower this time | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.2b |
| 3.5a | Must be unassertive. Do not allow ‘Amari is correct’ |
Question 5:
5 | (a) | 7652
15065−
Est of pop variance = 40
39
3475
= 11.138 =[ ]
312
Confidence interval is
19.125
± 1.96
11.138
×√
40
= 19.125 ± 1.034 or (18.09, 20.16) | 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 leading to
est of var = 10.859 or sd =3.295
Allow 11.1 or sd = 3.34 (3.33736…)
765
Or seen anywhere
40
Accept t-value of 2.02 to 2.03
𝑡ℎ𝑒𝑖𝑟 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
For √
40
Accept based on t-distribution from correct working
t = 2.02 leads to (18.06, 20.19) t = 2.0227 leads to (18.044,
20.206) t = 2.03 leads to (18.0422, 20.1961)
5 | (b) | It does to some extent because the confidence interval
does contain 20
But there is a lot of variation so estimates are not very
accurate. | E1
E1
[2] | 3.5a
2.2b | Must be unassertive FT their interval
Allow other suitable comments.
Allow valid comment on variation relating to the claim
EG Might not be representative of the population.
Lower confidence level could result in 20 being outside.
Near the end of interval so not very reliable.
The interval is wide.
5 | (c) | If the population from which the sample was drawn was
Normally distributed then you could have formed it using
the t distribution.
If not then you could not have formed it, due to the small
sample size | E1
E1
[2] | 2.2a
2.4 | Allow ‘don’t know if population is Normally distributed so
cannot use the t distribution.
Max 1 mark if t distribution not mentioned. Marks are
independent so can get mark for second comment only
5 | (d) | 19.25 < µ < 20.11 | B1
[1] | 1.1
5 | (e) | It suggests that he may be correct because the interval
again contains 20
And the variance is much lower this time | E1
E1
[2] | 2.2b
3.5a | Must be unassertive. Do not allow ‘Amari is correct’
Allow ‘The centre of the interval is nearer to 20 this time’
Allow ‘This interval is narrower.5 Amari is investigating how accurately people can estimate a short time period. He asks each of a random sample of 40 people to estimate a period of 20 seconds. For each person, he starts a stopwatch and then stops it when they tell him that they think that 20 s has elapsed. The times which he records are denoted by $x \mathrm {~s}$. You are given that\\
$\sum x = 765 , \quad \sum x ^ { 2 } = 15065$.
\begin{enumerate}[label=(\alph*)]
\item Determine a 95\% confidence interval for the mean estimated time.
\item Amari says that the confidence interval supports the suggestion that people can estimate 20 s accurately.
Make two comments about Amari's statement.
\item Discuss whether you could have constructed the confidence interval if there had only been 10 people involved in the experiment.
Amari thinks that people would be able to estimate more accurately if he gave them a second attempt. He repeats the experiment with each person and again records the times. Software is used to produce a $95 \%$ confidence interval for the mean estimated time. The output from the software is shown below.
Z Estimate of a Mean
Confidence level 0.95
Sample
\begin{center}
\begin{tabular}{ r l }
Mean & 19.68 \\
s & 1.38 \\
N & 40 \\
\end{tabular}
\end{center}
Result\\
Z Estimate of a Mean
\begin{center}
\begin{tabular}{ l l }
Mean & 19.68 \\
s & 1.38 \\
SE & 0.2182 \\
N & 40 \\
Interval & $19.68 \pm 0.4277$ \\
\end{tabular}
\end{center}
\item State the confidence interval in the form $\mathrm { a } < \mu < \mathrm { b }$.
\item Make two comments based on this confidence interval about Amari's opinion that second attempts result in more accurate estimates.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2023 Q5 [13]}}