| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| 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 application with standard normal distribution. Part (i) requires routine calculation of a z-test statistic and comparison to critical values. Part (ii) asks for standard definitions of Type I error, which is direct recall. The question involves no novel problem-solving or conceptual challenges beyond applying a standard procedure taught in S2. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(H_0: \mu = 21.2\) | B1 | For \(H_0\) and \(H_1\) correct must be \(\neq\) |
| \(H_1: \mu \neq 21.2\) | ||
| Test statistic \(z = \frac{19.4 - 21.2}{(7.3/\sqrt{90})} = -2.34\) | M1 | For standardising must have \(\sqrt{90}\) |
| \(CV = z \pm 1.96\) | A1 | For correct \(z\) accept \(+/-\) |
| In CR, reject \(H_0\). Sig evidence to say not the same author | M1 A1ft (5) | For correct comparison with correct critical value, ft from their \(H_1\) / For correct conclusion ft on their \(z\) and their CV |
| or \(\Phi(-2.339) = 1 - 0.9903 = 0.0097/0.0096\) | M1 | For correct \(\Phi\) and correct comparison (consistent with \(H_1\)) |
| Compare with 0.025 say sig evidence to say not the same sentence length or author | A1ft | For correct conclusion ft on their \(\Phi\) and 0.025 |
| or \(x = 21.2 \pm 1.96x(7.3/\sqrt{90}) = 19.2(22.7)\) | M1 A1 | For expression for \(x\) with correct (consistent) \(z\) / For correct comparison and conclusion(ft) |
| Compare with 19.4 etc. | M1 A1ft | |
| (ii) Say it is not the same sentence length or author when it is \(P(\text{Type I error}) = 5\%\) | B1 | For correct statement |
| B1 (2) | For correct answer |
**(i)** $H_0: \mu = 21.2$ | B1 | For $H_0$ and $H_1$ correct must be $\neq$
$H_1: \mu \neq 21.2$ | |
Test statistic $z = \frac{19.4 - 21.2}{(7.3/\sqrt{90})} = -2.34$ | M1 | For standardising must have $\sqrt{90}$
$CV = z \pm 1.96$ | A1 | For correct $z$ accept $+/-$
In CR, reject $H_0$. Sig evidence to say not the same author | M1 A1ft (5) | For correct comparison with correct critical value, ft from their $H_1$ / For correct conclusion ft on their $z$ and their CV
or $\Phi(-2.339) = 1 - 0.9903 = 0.0097/0.0096$ | M1 | For correct $\Phi$ and correct comparison (consistent with $H_1$)
Compare with 0.025 say sig evidence to say not the same sentence length or author | A1ft | For correct conclusion ft on their $\Phi$ and 0.025
or $x = 21.2 \pm 1.96x(7.3/\sqrt{90}) = 19.2(22.7)$ | M1 A1 | For expression for $x$ with correct (consistent) $z$ / For correct comparison and conclusion(ft)
Compare with 19.4 etc. | M1 A1ft |
**(ii)** Say it is not the same sentence length or author when it is $P(\text{Type I error}) = 5\%$ | B1 | For correct statement
| B1 (2) | For correct answer4 A study of a large sample of books by a particular author shows that the number of words per sentence can be modelled by a normal distribution with mean 21.2 and standard deviation 7.3. A researcher claims to have discovered a previously unknown book by this author. The mean length of 90 sentences chosen at random in this book is found to be 19.4 words.\\
(i) Assuming the population standard deviation of sentence lengths in this book is also 7.3, test at the $5 \%$ level of significance whether the mean sentence length is the same as the author's. State your null and alternative hypotheses.\\
(ii) State in words relating to the context of the test what is meant by a Type I error and state the probability of a Type I error in the test in part (i).
\hfill \mbox{\textit{CAIE S2 2005 Q4 [7]}}