| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type I error |
| Difficulty | Standard +0.3 This is a straightforward hypothesis test question requiring standard procedures: identifying test type, stating Type I error probability (simply 0.05), explaining Type I error contextually, and performing a routine z-test with known variance. The calculations are simple (sample mean = 568.5/6 = 94.75) and the test follows a standard template. Slightly above average difficulty only because it requires understanding of error types and assumptions, but no novel problem-solving is needed. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Test is for "difference" | B1 | Test is not for 'increase' or 'decrease' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.05\) | B1 | |
| Conclude mean time is different when it is not | B1 | oe, in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assume \(\sigma = 6.4\) | B1 | |
| \(H_0\): pop mean \(= 91.4\), \(H_1\): pop mean \(\neq 91.4\) | B1 | Allow \(\mu\), but not 'mean' |
| \(\bar{x} = \frac{568.5}{6}\) \((= 94.75)\) | B1 | |
| \(\frac{'94.75' - 91.4}{\frac{6.4}{\sqrt{6}}}\) | M1 | Must have \(\sqrt{6}\) |
| \(= 1.282\), cv of \(z = 1.96\) | A1 | |
| \('1.282' < 1.96\) | M1 | Valid comparison or comp \(\Phi\)("1.282") with \(0.975\); \(0.9(001) < 0.975\) or \(0.0999\) (or \(0.1) > 0.025\); consistent use of one tail test can score M1 for comparison with 1.645 oe but not A1ft |
| No evidence mean time different | A1ft | CV method \(x = 96.52\) M1 A1 \(94.75 < 96.52\) M1 Conc A1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Test is for "difference" | B1 | Test is not for 'increase' or 'decrease' |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.05$ | B1 | |
| Conclude mean time is different when it is not | B1 | oe, in context |
---
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume $\sigma = 6.4$ | B1 | |
| $H_0$: pop mean $= 91.4$, $H_1$: pop mean $\neq 91.4$ | B1 | Allow $\mu$, but not 'mean' |
| $\bar{x} = \frac{568.5}{6}$ $(= 94.75)$ | B1 | |
| $\frac{'94.75' - 91.4}{\frac{6.4}{\sqrt{6}}}$ | M1 | Must have $\sqrt{6}$ |
| $= 1.282$, cv of $z = 1.96$ | A1 | |
| $'1.282' < 1.96$ | M1 | Valid comparison or comp $\Phi$("1.282") with $0.975$; $0.9(001) < 0.975$ or $0.0999$ (or $0.1) > 0.025$; consistent use of one tail test can score M1 for comparison with 1.645 oe but not A1ft |
| No evidence mean time different | A1ft | CV method $x = 96.52$ M1 A1 $94.75 < 96.52$ M1 Conc A1 |
---6 The time taken by volunteers to complete a certain task is normally distributed. In the past the time, in minutes, has had mean 91.4 and standard deviation 6.4. A new, similar task is introduced and the times, $t$ minutes, taken by a random sample of 6 volunteers to complete the new task are summarised by $\Sigma t = 568.5$. Andrea plans to carry out a test, at the $5 \%$ significance level, of whether the mean time for the new task is different from the mean time for the old task.\\
(i) Give a reason why Andrea should use a two-tail test.\\
(ii) State the probability that a Type I error is made, and explain the meaning of a Type I error in this context.\\
You may assume that the times taken for the new task are normally distributed.\\
(iii) Stating another necessary assumption, carry out the test.\\
\hfill \mbox{\textit{CAIE S2 2019 Q6 [10]}}