| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Challenging +1.2 This is a multi-part hypothesis testing question requiring understanding of normal distribution properties, finding parameters from percentiles, and determining sample size for significance. Part (i) is routine z-test application, part (ii) requires solving simultaneous equations from percentiles (moderately challenging), and part (iii) involves working backwards from significance level to find minimum n (requires careful reasoning but uses standard formulas). The question is above average difficulty due to the multi-step nature and part (ii)'s requirement to reconcile two percentile conditions, but remains within standard Further Maths Statistics scope. |
| Spec | 2.05d Sample mean as random variable2.05e Hypothesis test for normal mean: known variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Calculation of \(P(X<14)\) and \(P(X>18)\) | M1 (AO3.4) | Or solves \(-1.476=\frac{14-\mu}{\sigma}\) and \(0.496=\frac{18-\mu}{\sigma}\) simultaneously; or solves \(-1.476=\frac{x-15}{2}\) and \(0.496=\frac{x-15}{2}\) |
| \(0.3085\) and \(0.0668\) to 1 sf or better | A1 (AO1.1) | \(\mu\approx17\) and \(\sigma\approx2.02\); \(x=12.048\) and \(15.992\) to nearest whole number or better, which are not close to 14 and 18 |
| These figures do not support the model | A1 (AO3.5a) | 17 is (relatively) far from 15 so not a good fit; the second A1 is only available if the first A1 is awarded; allow SC2 for showing the model is not a good fit for either value with all working correct; or for a complete argument based on symmetry which refers to both tails; or \(\frac{14-15}{2}\) and \(\frac{18-15}{2}\) evaluated giving \(-0.5\) and \(1.5\), which are not close to \(-1.476\) and \(0.496\) respectively |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Phi^{-1}(0.07) = -1.476 = \frac{14-\mu}{2}\) giving \([\mu = 16.95]\) OR \(\Phi^{-1}(0.69) = 0.496 = \frac{18-\mu}{2}\) giving \([\mu = 17.008]\), \([\mu =]17\) | M1 | AO 3.5c |
| \(16 + 1 = 17\) | A1 [2] | AO 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = \pm 1.96\) used | B1 | AO 1.1a |
| \(\frac{16-\mu}{2/\sqrt{n}} < -1.96\) or \(\frac{\mu-16}{2/\sqrt{n}} > 1.96\) | M1 | AO 3.1b |
| \(\sqrt{n}\) isolated from their \(\frac{16-\mu}{\sigma/\sqrt{n}} < -1.96\) | M1 | AO 2.1 |
| \([n>]\ 15.3664 - 15.4\), \(n = 16\) cao | A1 | AO 3.4 |
| previous A1 must be awarded for award of final A1 | A1 [5] | AO 2.2b |
## Question 13(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Calculation of $P(X<14)$ and $P(X>18)$ | M1 (AO3.4) | Or solves $-1.476=\frac{14-\mu}{\sigma}$ and $0.496=\frac{18-\mu}{\sigma}$ simultaneously; or solves $-1.476=\frac{x-15}{2}$ and $0.496=\frac{x-15}{2}$ |
| $0.3085$ and $0.0668$ to 1 sf or better | A1 (AO1.1) | $\mu\approx17$ and $\sigma\approx2.02$; $x=12.048$ and $15.992$ to nearest whole number or better, which are not close to 14 and 18 |
| These figures do not support the model | A1 (AO3.5a) | 17 is (relatively) far from 15 so not a good fit; the second A1 is only available if the first A1 is awarded; allow SC2 for showing the model is not a good fit for either value with all working correct; or for a complete argument based on symmetry which refers to both tails; or $\frac{14-15}{2}$ and $\frac{18-15}{2}$ evaluated giving $-0.5$ and $1.5$, which are not close to $-1.476$ and $0.496$ respectively |
| **[3]** | | |
## Question 13(ii):
$\Phi^{-1}(0.07) = -1.476 = \frac{14-\mu}{2}$ giving $[\mu = 16.95]$ **OR** $\Phi^{-1}(0.69) = 0.496 = \frac{18-\mu}{2}$ giving $[\mu = 17.008]$, $[\mu =]17$ | M1 | AO 3.5c | alternatively since the variance is assumed to be correct, the mean must be as far above the midpoint as it was previously below it
$16 + 1 = 17$ | A1 [2] | AO 2.4 |
---
## Question 13(iii):
$z = \pm 1.96$ used | B1 | AO 1.1a | NB 1.959963985...rounded to 3 or more sf; **M0** if other value for $\sigma$ used
$\frac{16-\mu}{2/\sqrt{n}} < -1.96$ or $\frac{\mu-16}{2/\sqrt{n}} > 1.96$ | M1 | AO 3.1b | allow method marks only if other $z$ value, eg $-1.645$ used; FT $\mu$
$\sqrt{n}$ isolated from their $\frac{16-\mu}{\sigma/\sqrt{n}} < -1.96$ | M1 | AO 2.1 | eg $\sqrt{n} > 2 \times 1.96$; all marks available if works with $=$ instead of $<$ or $>$ throughout, but withhold final **A1** if works with $<$ instead of $>$ or $>$ instead of $<$ throughout
$[n>]\ 15.3664 - 15.4$, $n = 16$ cao | A1 | AO 3.4 |
previous A1 must be awarded for award of final A1 | A1 [5] | AO 2.2b |
---13 Each weekday Keira drives to work with her son Kaito. She always sets off at 8.00 a.m. She models her journey time, $x$ minutes, by the distribution $X \sim \mathrm {~N} ( 15,4 )$.
Over a long period of time she notes that her journey takes less than 14 minutes on $7 \%$ of the journeys, and takes more than 18 minutes on $31 \%$ of the journeys.\\
(i) Investigate whether Keira's model is a good fit for the data.
Kaito believes that Keira's value for the variance is correct, but realises that the mean is not correct.\\
(ii) Find, correct to two significant figures, the value of the mean that Keira should use in a refined model which does fit the data.
Keira buys a new car. After driving to work in it each day for several weeks, she randomly selects the journey times for $n$ of these days. Her mean journey time for these $n$ days is 16 minutes. Using the refined model she conducts a hypothesis test to see if her mean journey time has changed, and finds that the result is significant at the $5 \%$ level.\\
(iii) Determine the smallest possible value of $n$.
\hfill \mbox{\textit{OCR MEI Paper 2 2018 Q13 [10]}}