| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Confidence interval interpretation |
| Difficulty | Standard +0.3 This is a straightforward application of standard confidence interval procedures with given summary statistics. Parts (i)-(iii) are routine calculations requiring sample variance formula and CI construction. Parts (iv)-(v) test basic understanding of sampling variability and confidence level interpretation—conceptually important but not mathematically demanding. Slightly above average difficulty due to the multi-part nature and requiring conceptual understanding in later parts, but all techniques are standard A-level statistics. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (i) | Estimate of population variance = |
| Answer | Marks |
|---|---|
| = 0.00008515 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (ii) | 1.49597 |
| Answer | Marks |
|---|---|
| = 1.49597 ± 0.00233 or (1.4936,1.4983) | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 3.4 | N |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (iii) | It appears that the (population) mean content |
| Answer | Marks |
|---|---|
| not contain 1.5. | E1 |
| [1] | I |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (iv) | Each time a sample is taken it will be |
| Answer | Marks |
|---|---|
| interval. | E |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| 2.4 | Samples vary |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (v) | 300(cid:117)0.95(cid:32)285 |
| [1] | 1.1 |
Question 10:
10 | (i) | Estimate of population variance =
89.7582
134.280(cid:16)
60
59
= 0.00008515 | M1
A1
[2] | 1.1
1.1
10 | (ii) | 1.49597
(cid:114)1.96
0.00008515
(cid:117)
60
= 1.49597 ± 0.00233 or (1.4936,1.4983) | B1
M1
M1
A1
[4] | 1.1
3.3
1.1
3.4 | N
E
M
Allow (1.494, 1.498)
10 | (iii) | It appears that the (population) mean content
is not 1.5 litres as the calculated interval does
not contain 1.5. | E1
[1] | I
C3.5a
10 | (iv) | Each time a sample is taken it will be
different, so e.g. will have a different mean
P
hence different midpoint for confidence
interval. | E
E1
E1
[2] | 2.4
2.4 | Samples vary
…so confidence intervals vary
10 | (v) | 300(cid:117)0.95(cid:32)285 | B1
[1] | 1.110 The label on a particular size of milk carton states that it contains 1.5 litres of milk. In an investigation at the packaging plant the contents, $x$ litres, of each of 60 randomly selected cartons are measured. The data are summarised as follows.
$$\Sigma x = 89.758 \quad \Sigma x ^ { 2 } = 134.280$$
(i) Estimate the variance of the underlying population.\\
(ii) Find a 95\% confidence interval for the mean of the underlying population.\\
(iii) What does the confidence interval which you have calculated suggest about the statement on the carton?
Each day for 300 days a random sample of 60 cartons is selected and for each sample a $95 \%$ confidence interval is constructed.\\
(iv) Explain why the confidence intervals will not be identical.\\
(v) What is the expected number of confidence intervals to contain the population mean?
\hfill \mbox{\textit{OCR MEI Further Statistics Major Q10 [10]}}