| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| 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 | Standard +0.3 This is a straightforward hypothesis testing question requiring standard normal distribution calculations. Part (a) is routine one-tailed z-test application, and part (b) requires calculating Type II error probability given a specific alternative mean—a standard textbook exercise with clear steps and no novel insight required. Slightly easier than average due to being computational rather than conceptual. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): Population mean mass \(= 510\) g; \(H_1\): Population mean mass \(< 510\) g | B1 | Allow '\(\mu\)' but not just 'mean'. |
| \(\pm\frac{508 - 510}{10 \div \sqrt{120}}\) | M1 | Standardising must have \(\sqrt{120}\). |
| \(= \pm 2.191\) or \(-2.190\) | A1 | |
| \(-2.191 < -1.96\) or \(2.191 > 1.96\); Area comparison: \(0.0143\) or \(0.0142 < 0.025\) | M1 | OE. For valid comparison. Inequality sign the wrong way round scores M1 A0. |
| [Reject \(H_0\)] There is sufficient evidence to suggest that the [mean] mass has decreased | A1FT | OE. In context (must be 'decreased' OE, not 'changed'); not definite. No contradictions. NB: Accept alternative method using critical value (\(= 508.21\)) and comparison with 508. Condone 509.79 compared with 510. Two tail test scores maximum B0 M1 A1 M1 A0; must have comparison with 0.0125 or 2.24/2.241. |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{cv - 510}{10 \div \sqrt{120}} = -1.96\) | M1 | Standardising to find critical value (must use 510 and \(10 \div \sqrt{120}\)). Accept \(\pm 1.96\). |
| \(cv = 508.21\) | A1 | Accept 3 sf if nothing better seen. Note: cv could be found in (a). |
| \(z = \pm\frac{508.21 - 506}{10 \div \sqrt{120}}\ [= 2.421]\) | M1 | Standardising with their 508.21 and 506 (must use \(10 \div \sqrt{120}\)). |
| \(P(\bar{X} > 508.21 \mid \mu = 506) = 1 - \Phi(2.421)\) | M1 | For area consistent with their working. |
| \(= 0.0077\) to \(0.0080\) (2sf) | A1 | Note: \(\frac{510-506}{10 \div \sqrt{120}}\) scores max M0 A0 M1 M1 A0. |
| 5 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Population mean mass $= 510$ g; $H_1$: Population mean mass $< 510$ g | B1 | Allow '$\mu$' but not just 'mean'. |
| $\pm\frac{508 - 510}{10 \div \sqrt{120}}$ | M1 | Standardising must have $\sqrt{120}$. |
| $= \pm 2.191$ or $-2.190$ | A1 | |
| $-2.191 < -1.96$ or $2.191 > 1.96$; Area comparison: $0.0143$ or $0.0142 < 0.025$ | M1 | OE. For valid comparison. Inequality sign the wrong way round scores M1 A0. |
| [Reject $H_0$] There is sufficient evidence to suggest that the [mean] mass has decreased | A1FT | OE. In context (must be 'decreased' OE, not 'changed'); not definite. No contradictions. NB: Accept alternative method using critical value ($= 508.21$) and comparison with 508. Condone 509.79 compared with 510. Two tail test scores maximum B0 M1 A1 M1 A0; must have comparison with 0.0125 or 2.24/2.241. |
| | **5** | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{cv - 510}{10 \div \sqrt{120}} = -1.96$ | M1 | Standardising to find critical value (must use 510 and $10 \div \sqrt{120}$). Accept $\pm 1.96$. |
| $cv = 508.21$ | A1 | Accept 3 sf if nothing better seen. Note: cv could be found in (a). |
| $z = \pm\frac{508.21 - 506}{10 \div \sqrt{120}}\ [= 2.421]$ | M1 | Standardising with their 508.21 and 506 (must use $10 \div \sqrt{120}$). |
| $P(\bar{X} > 508.21 \mid \mu = 506) = 1 - \Phi(2.421)$ | M1 | For area consistent with their working. |
| $= 0.0077$ to $0.0080$ (2sf) | A1 | Note: $\frac{510-506}{10 \div \sqrt{120}}$ scores max M0 A0 M1 M1 A0. |
| | **5** | |6 The masses of cereal boxes filled by a certain machine have mean 510 grams. An adjustment is made to the machine and an inspector wishes to test whether the mean mass of cereal boxes filled by the machine has decreased.
After the adjustment is made, he chooses a random sample of 120 cereal boxes. The mean mass of these boxes is found to be 508 grams.
Assume that the standard deviation of the masses is 10 grams.
\begin{enumerate}[label=(\alph*)]
\item Test at the $2.5 \%$ significance level whether the mean mass of cereal boxes filled by the machine has decreased.\\
Later the inspector carries out a similar test at the $2.5 \%$ significance level, using the same hypotheses and another 120 randomly chosen cereal boxes.\\[0pt]
\item Given that the mean mass is now actually 506 grams, find the probability of a Type II error. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q6 [10]}}