| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type II error |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question covering standard S2 content: one-tailed z-test with known variance, calculating Type II error probability using normal distribution, and stating a definition. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): Pop mean time (or \(\mu\)) \(= 20.5\) | ||
| \(H_1\): Pop mean time (or \(\mu\)) \(< 20.5\) | B1 | Not just "mean" |
| \(\frac{20.3 - 20.5}{1.2 \div \sqrt{100}}\) | M1 | Allow without \(\sqrt{}\) sign (accept \(\pm 1.667/1.67\)) |
| \(= -1.667\) or \(0.0478/0.952\) if areas compared | A1 | |
| \(1.667 < 1.751\) (or \(-1.667 > -1.751\)) | M1 | Correct comparison of their \(z_\text{calc}\) with \(1.751/1.75\) oe valid comparison of areas \((0.0478 > 0.04)\) |
| No evidence that (pop) mean time has decreased | A1ft [5] | No contradictions (ft their \(z\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{cv - 20.5}{1.2 \div \sqrt{100}} = -1.751\) | M1* | |
| \(cv = 20.29\) or \(20.3\) | A1 | |
| \(\frac{20.29 - 20.1}{1.2 \div \sqrt{100}}\) \((= 1.583\) or \(1.582)\) | DM1 | Allow \(\frac{20.3 - 20.1}{1.2 \div \sqrt{100}}\) \((= 1.667)\) — M1 |
| \(1 - \Phi(1.583)\) | M1 | \(1 - \Phi(1.667)\) — M1 |
| \(= 0.0567 - 0.0569\) (3 sf) | A1 [5] | \(= 0.0478\) (3 sf) — A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Concluding (mean) time not decreased when in fact it has. | B1 [1] | Must be in context, oe |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: Pop mean time (or $\mu$) $= 20.5$ | | |
| $H_1$: Pop mean time (or $\mu$) $< 20.5$ | B1 | Not just "mean" |
| $\frac{20.3 - 20.5}{1.2 \div \sqrt{100}}$ | M1 | Allow without $\sqrt{}$ sign (accept $\pm 1.667/1.67$) |
| $= -1.667$ or $0.0478/0.952$ if areas compared | A1 | |
| $1.667 < 1.751$ (or $-1.667 > -1.751$) | M1 | Correct comparison of their $z_\text{calc}$ with $1.751/1.75$ oe valid comparison of areas $(0.0478 > 0.04)$ |
| No evidence that (pop) mean time has decreased | A1ft [5] | No contradictions (ft their $z$) |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{cv - 20.5}{1.2 \div \sqrt{100}} = -1.751$ | M1* | |
| $cv = 20.29$ or $20.3$ | A1 | |
| $\frac{20.29 - 20.1}{1.2 \div \sqrt{100}}$ $(= 1.583$ or $1.582)$ | DM1 | Allow $\frac{20.3 - 20.1}{1.2 \div \sqrt{100}}$ $(= 1.667)$ — M1 |
| $1 - \Phi(1.583)$ | M1 | $1 - \Phi(1.667)$ — M1 |
| $= 0.0567 - 0.0569$ (3 sf) | A1 [5] | $= 0.0478$ (3 sf) — A1 |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Concluding (mean) time not decreased when in fact it has. | B1 [1] | Must be in context, oe |7 In the past the time, in minutes, taken for a particular rail journey has been found to have mean 20.5 and standard deviation 1.2. Some new railway signals are installed. In order to test whether the mean time has decreased, a random sample of 100 times for this journey are noted. The sample mean is found to be 20.3 minutes. You should assume that the standard deviation is unchanged.\\
(i) Carry out a significance test, at the $4 \%$ level, of whether the population mean time has decreased.
Later another significance test of the same hypotheses, using another random sample of size 100 , is carried out at the $4 \%$ level.\\
(ii) Given that the population mean is now 20.1, find the probability of a Type II error.\\
(iii) State what is meant by a Type II error in this context.
\hfill \mbox{\textit{CAIE S2 2016 Q7 [11]}}