| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Hypothesis test then Type II error probability |
| Difficulty | Challenging +1.2 This is a standard hypothesis testing question with a Type II error calculation. Part (a) is routine z-test application requiring calculation of sample mean and standard deviation. Part (b) requires understanding of Type II errors and computing probability under the alternative hypothesis, which is a step beyond basic testing but follows a standard procedure taught in S2. The multi-step nature and Type II error component elevate it slightly above average difficulty. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Est}(\mu) = 23/50 = 0.46\) | B1 | |
| \(\text{Est}(\sigma) = \sqrt{\frac{50}{49}} \times \sqrt{\frac{13.02}{50} - 0.46^2}\) or \(\text{Est}(\sigma^2) = \frac{50}{49} \times \left(\frac{13.02}{50} - 0.46^2\right)\) or estimated unbiased variance \(= \frac{1}{49}\left(13.02 - \frac{(23.0)^2}{50}\right)\) | M1 | For an expression of the correct form for unbiased standard deviation or variance |
| \(\text{Est}(\sigma) = 0.22315\) or \(\text{Est}(\sigma^2) = 0.0497959 = \left(\frac{61}{1225}\right)\) or \(0.0498\) | A1 | |
| \(\dfrac{0.46 - 0.5}{\frac{0.22315}{\sqrt{50}}}\) | M1 | Standardising with their values |
| \(= -1.268\) or \(-1.267\) or \(-1.27\) (3sf) | A1 | |
| \(-1.268 > -1.645\) or \(0.102\) to \(0.103 > 0.05\) | M1 | For a valid comparison |
| [Do not reject \(H_0\)] There is insufficient evidence [at 5% level] that the mean concentration is less than 0.5 | A1 FT | In context, not definite. E.g., not 'Mean concentration is not less than 0.5'. No contradictions |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{cv - 0.5}{\frac{0.22315}{\sqrt{50}}} = -1.645\) | M1 | |
| \(cv = 0.448(1)\) or \(0.448\) (3 sf) | A1 | |
| \(\dfrac{0.448 - 0.4}{\frac{0.22315}{\sqrt{50}}}\) \([= 1.521 \text{ to } 1.524]\) | M1 | |
| \(1 - \phi(1.524)\) | M1 | For area consistent with their working |
| \(= 0.0638\) to \(0.0642\) | A1 | |
| Total: 5 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Est}(\mu) = 23/50 = 0.46$ | **B1** | |
| $\text{Est}(\sigma) = \sqrt{\frac{50}{49}} \times \sqrt{\frac{13.02}{50} - 0.46^2}$ or $\text{Est}(\sigma^2) = \frac{50}{49} \times \left(\frac{13.02}{50} - 0.46^2\right)$ or estimated unbiased variance $= \frac{1}{49}\left(13.02 - \frac{(23.0)^2}{50}\right)$ | **M1** | For an expression of the correct form for unbiased standard deviation or variance |
| $\text{Est}(\sigma) = 0.22315$ or $\text{Est}(\sigma^2) = 0.0497959 = \left(\frac{61}{1225}\right)$ or $0.0498$ | **A1** | |
| $\dfrac{0.46 - 0.5}{\frac{0.22315}{\sqrt{50}}}$ | **M1** | Standardising with their values |
| $= -1.268$ or $-1.267$ or $-1.27$ (3sf) | **A1** | |
| $-1.268 > -1.645$ or $0.102$ to $0.103 > 0.05$ | **M1** | For a valid comparison |
| [Do not reject $H_0$] There is insufficient evidence [at 5% level] that the mean concentration is less than 0.5 | **A1 FT** | In context, not definite. E.g., not 'Mean concentration is not less than 0.5'. No contradictions |
| **Total: 7** | | |
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## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{cv - 0.5}{\frac{0.22315}{\sqrt{50}}} = -1.645$ | **M1** | |
| $cv = 0.448(1)$ or $0.448$ (3 sf) | **A1** | |
| $\dfrac{0.448 - 0.4}{\frac{0.22315}{\sqrt{50}}}$ $[= 1.521 \text{ to } 1.524]$ | **M1** | |
| $1 - \phi(1.524)$ | **M1** | For area consistent with their working |
| $= 0.0638$ to $0.0642$ | **A1** | |
| **Total: 5** | | |7 A biologist wishes to test whether the mean concentration $\mu$, in suitable units, of a certain pollutant in a river is below the permitted level of 0.5 . She measures the concentration, $x$, of the pollutant at 50 randomly chosen locations in the river. The results are summarised below.
$$n = 50 \quad \Sigma x = 23.0 \quad \Sigma x ^ { 2 } = 13.02$$
\begin{enumerate}[label=(\alph*)]
\item Carry out a test at the $5 \%$ significance level of the null hypothesis $\mu = 0.5$ against the alternative hypothesis $\mu < 0.5$.\\
Later, a similar test is carried out at the $5 \%$ significance level using another sample of size 50 and the same hypotheses as before. You should assume that the standard deviation is unchanged.
\item Given that, in fact, the value of $\mu$ is 0.4 , find the probability of a Type II error.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q7 [12]}}