| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Challenging +1.2 Part (a) is a standard t-distribution confidence interval calculation with given summary statistics. Part (b) requires understanding that when the true mean shifts by 1.6, we need to find P(interval captures μ+1.6), which involves recognizing this as a normal distribution probability calculation with known variance. While multi-step, both parts follow established procedures with no novel insight required, making it moderately above average difficulty for Further Maths Statistics. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | ˆ x 9 . 9 = = |
| Answer | Marks |
|---|---|
| 9.9 1.22 = (8.68, 11.12) | B1 |
| Answer | Marks |
|---|---|
| [7] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | 9.9 seen anywhere |
| Answer | Marks |
|---|---|
| (b) | X 2 .5 7 6 1 0 / 5 0 ( = X 1 .1 5 2 ) |
| Answer | Marks |
|---|---|
| = (6.15) – (1.002) = 0.158(2) | M1* |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | Find new confidence interval in terms of X (not ), needn’t use 50 here |
Question 7:
7 | (a) | ˆ x 9 . 9 = =
4271.40
s2 = −9.92 [= 8.775]
40
4 0
2 8 .7 7 5 =
3 9
= 9
9
9 .9 z
4 0
z = 2.576, allow 2.58
9.9 1.22 = (8.68, 11.12) | B1
M1
M1
A1
M1
B1
A1
[7] | 1.1
2.1
1.2
1.1
3.1b
1.1
1.1 | 9.9 seen anywhere
Correct method for biased or unbiased estimate used
40/39, or divisor 39, seen anywhere
9 only (or 3 if clear)
Their 9.9 and 9. 40 and numerical z must be seen or implied. Allow
errors
Explicit, or implied by correct answer only (not for just “Z ”)
0.995
Both, needs 40/39 oe. Allow awrt 8.68, and awrt 11.1.
Use of 8.775 gives 8.69: B1M1 M0A0 M1B1 A0
Use of 2.236 gives (8.80, 11.0): 5/7.
(b) | X 2 .5 7 6 1 0 / 5 0 ( = X 1 .1 5 2 )
P ( X 1 .1 5 2 1 .6 X 1 .1 5 2 ) − + +
0.448 (X −) 2.752
=P
0.2 0.2 0.2
= (6.15) – (1.002) = 0.158(2) | M1*
depM1
A1
[3] | 3.1b
1.1
1.1 | Find new confidence interval in terms of X (not ), needn’t use 50 here
9.9 1.152 or (8.75, 11.05) on their own are M0, but see SCs below
Obtain at least LH inequality, 50 needed here, allow errors
Awrt 0.158 (No marks just for finding new interval, or 10.348 & 12.472)
If RH inequality not considered at all, withhold final A1
SC: Explicitly take specific value of , e.g. 0 or 9.9 (must be notX , and not just ~ N(9.9, 0.2): first M1* as above;
e.g. P(0.448X 2.752) from = 0 or P (1 0 .3 4 8 X 1 2 .4 7 2 ) from = 9.9 depM1;
0.158(2) A1
SC: 9.9 1.152 = (8.748, 11.05); ~ N(9.9, 0.2) M0 so far.
P(8.748 – 1.6 < < 11.052 – 1.6) M1
= 0.158 A1 max 2/37 A club secretary collects data about the time, $T$ minutes, needed to process the details of a new member. The mean of $T$ is denoted by $\mu$. The variance of $T$ is denoted by $\sigma ^ { 2 }$. The results of a random sample of 40 observations of $T$ are summarised as follows.\\
$\mathrm { n } = 40 \quad \Sigma \mathrm { t } = 396.0 \quad \Sigma \mathrm { t } ^ { 2 } = 4271.40$
\begin{enumerate}[label=(\alph*)]
\item Determine a 99\% confidence interval for $\mu$.
\item The secretary discovers that over a long period the value of $\sigma ^ { 2 }$ is in fact 10.0 . The secretary collects an independent random sample of 50 observations of $T$ and constructs a new 99\% confidence interval for $\mu$ based on this sample of size 50 , but using $\sigma ^ { 2 } = 10.0$.
Find the probability that this new confidence interval contains the value $\mu + 1.6$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2023 Q7 [10]}}