| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State what extra info needed for Type II |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question covering standard procedures: conducting a one-tailed z-test (part i), defining Type II error in context (part ii), and identifying what's needed to calculate its probability (part iii). All parts require recall and application of standard techniques with no novel problem-solving or insight 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 mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): pop mean (or \(\mu\)) \(= 12.4\); \(H_1\): pop mean (or \(\mu\)) \(> 12.4\) | B1 | not just "mean" |
| \(\frac{12.9 - 12.4}{2.1 \div \sqrt{50}}\) | M1 | Allow with 50 instead of \(\sqrt{50}\) |
| \(1.684\) | A1 | |
| comp cv \(z = 1.96\); No evidence that pop mean time has increased | B1f [4] | or \(P(z > 1.684) = 0.0461 > 0.025\). Allow accept \(H_0\) if correctly defined. Ft their test statistic. No contradictions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Not reject (or accept) that mean time is unchanged (or is 12.4) oe | B1 | |
| although mean time has increased (or is more than 12.4) oe | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| True (or new) mean | B1 [1] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: pop mean (or $\mu$) $= 12.4$; $H_1$: pop mean (or $\mu$) $> 12.4$ | B1 | not just "mean" |
| $\frac{12.9 - 12.4}{2.1 \div \sqrt{50}}$ | M1 | Allow with 50 instead of $\sqrt{50}$ |
| $1.684$ | A1 | |
| comp cv $z = 1.96$; No evidence that pop mean time has increased | B1f [4] | or $P(z > 1.684) = 0.0461 > 0.025$. Allow accept $H_0$ if correctly defined. Ft their test statistic. No contradictions |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Not reject (or accept) that mean time is unchanged (or is 12.4) oe | B1 | |
| although mean time has increased (or is more than 12.4) oe | B1 [2] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| True (or new) mean | B1 [1] | |
---4 In the past, the time taken by vehicles to drive along a particular stretch of road has had mean 12.4 minutes and standard deviation 2.1 minutes. Some new signs are installed and it is expected that the mean time will increase. In order to test whether this is the case, the mean time for a random sample of 50 vehicles is found. You may assume that the standard deviation is unchanged.\\
(i) The mean time for the sample of 50 vehicles is found to be 12.9 minutes. Test at the $2.5 \%$ significance level whether the population mean time has increased.\\
(ii) State what is meant by a Type II error in this context.\\
(iii) State what extra piece of information would be needed in order to find the probability of a Type II error.
\hfill \mbox{\textit{CAIE S2 2015 Q4 [7]}}