| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Paired sample confidence interval |
| Difficulty | Standard +0.3 This is a straightforward application of paired sample confidence intervals using the CLT. Part (i) requires standard formula application with n=40, parts (ii-iv) test interpretation and understanding of assumptions. The CLT justification (n≥30) and basic statistical reasoning are routine for Further Statistics, making this slightly above average difficulty due to the multi-part nature but not requiring novel insight. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | C.I. is given by |
| -0.218 ≤ μ ≤ 0.798 or 0.29 ± 0.5082 | M1 |
| Answer | Marks |
|---|---|
| [4] | For general form |
| Answer | Marks |
|---|---|
| everything else | If full working not shown allow SC2 |
| Answer | Marks |
|---|---|
| (ii) | Confidence interval does not suggest that |
| Answer | Marks |
|---|---|
| …since the interval contains zero. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | Condone ‘it does not’ or similar | Do not allow ‘suggests that the |
| Answer | Marks |
|---|---|
| (iii) | The sample is large and mention of Central Limit |
| Answer | Marks |
|---|---|
| distributed | E1 |
| Answer | Marks |
|---|---|
| [2] | For full answer |
| (iv) | A random sample enables proper inference about |
| the population to be undertaken. | B2 |
| [2] | B2 for correct explanation as |
Question 6:
6 | (i) | C.I. is given by
-0.218 ≤ μ ≤ 0.798 or 0.29 ± 0.5082 | M1
B1
M1
A1
[4] | For general form
For 1.96
For can get this indep of
everything else | If full working not shown allow SC2
for correct answer
At least 2sf required but not just 0.8
(ii) | Confidence interval does not suggest that
the additive reduces fuel consumption …
…since the interval contains zero. | B1
B1
[2] | Condone ‘it does not’ or similar | Do not allow ‘suggests that the
additive does not affect fuel
consumption’ (but second mark still
available)
(iii) | The sample is large and mention of Central Limit
Theorem or Normal distribution
The Central Limit Theorem states that sample
means are therefore approximately Normally
distributed | E1
E1
[2] | For full answer
(iv) | A random sample enables proper inference about
the population to be undertaken. | B2
[2] | B2 for correct explanation as
shown in answer column
OR
B1 for partially correct
explanation e.g. ‘a random
sample is less likely to be
biased’ or ‘so that the sample
is more likely to be
representative of the
population’
PPMMTT
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© OCR 20186 A company has a large fleet of cars. It is claimed that use of a fuel additive will reduce fuel consumption. In order to test this claim a researcher at the company randomly selects 40 of the cars. The fuel consumption of each of the cars is measured, both with and without the fuel additive. The researcher then calculates the difference $d$ litres per kilometre between the two figures for each car, where $d$ is the fuel consumption without the additive minus the fuel consumption with the additive. The sample mean of $d$ is 0.29 and the sample standard deviation is 1.64 .\\
(i) Showing your working, find a 95\% confidence interval for the population mean difference.\\
(ii) Explain whether the confidence interval suggests that, on average, the fuel additive does reduce fuel consumption.\\
(iii) Explain why you can construct the interval in part (i) despite not having any information about the distribution of the population of differences.\\
(iv) Explain why the sample used was random.
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2018 Q6 [10]}}