| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring standard formulas (t-distribution CI, sample size calculation) and interpretation. While it's a multi-part question worth several marks, each part follows routine procedures taught in Further Statistics with no novel problem-solving required. The CLT justification in part (c) is standard bookwork, making this slightly easier than average for Further Maths material. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | 0.1442 |
| Answer | Marks |
|---|---|
| = 0.1442 ± 0.0800 or (0.0642, 0.2242) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [4] | Allow 0.064 to 0.224 | |
| 7 | (b) | It seems that the coach’s belief may be correct, |
| as the confidence interval contains 0.2 | E1 |
| Answer | Marks |
|---|---|
| [2] | FT their interval |
| Answer | Marks |
|---|---|
| coach may be correct | Condone ‘The |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (c) | By the CLT, for large samples the distribution of |
| the sample mean is approximately Normal | B1 |
| Answer | Marks |
|---|---|
| [2] | For mention of central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (d) | For halving 0.12 |
| Answer | Marks |
|---|---|
| n = 71.03 so minimum sample size is 72 | M1 |
| Answer | Marks |
|---|---|
| [3] | If 0.12 not halved allow M0M1A0 |
Question 7:
7 | (a) | 0.1442
± 1.96
0.2580
40
= 0.1442 ± 0.0800 or (0.0642, 0.2242) | B1
M1
M1
A1
[4] | Allow 0.064 to 0.224
7 | (b) | It seems that the coach’s belief may be correct,
as the confidence interval contains 0.2 | E1
E1
[2] | FT their interval
Allow E1E0 for whole interval is
above zero so evidence to suggest
coach may be correct | Condone ‘The
coaches’ belief is
correct’
7 | (c) | By the CLT, for large samples the distribution of
the sample mean is approximately Normal | B1
B1
[2] | For mention of central limit theorem
For full statement (including CLT) and
including sample mean
7 | (d) | For halving 0.12
0.2580
1.96 0.06
n
n = 71.03 so minimum sample size is 72 | M1
M1
A1
[3] | If 0.12 not halved allow M0M1A07 A swimming coach believes that times recorded by people using stopwatches are on average 0.2 seconds faster than those recorded by an electronic timing system.
In order to test this, the coach takes a random sample of 40 competitors' times recorded by both methods, and finds the differences between the times recorded by the two methods. The mean difference in the times (electronic time minus stopwatch time) is 0.1442 s and the standard deviation of the differences is 0.2580 s .
\begin{enumerate}[label=(\alph*)]
\item Find a 95\% confidence interval for the mean difference between electronic and stopwatch times.
\item Explain whether there is evidence to suggest that the coach's belief is correct.
\item Explain how you can calculate the confidence interval in part (a) even though you do not know the distribution of the parent population of differences.
\item If the coach wanted to produce a $95 \%$ confidence interval of width no more than 0.12 s , what is the minimum sample size that would be needed, assuming that the standard deviation remains the same?
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2019 Q7 [11]}}