| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (lower tail) |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard procedures: calculating sample statistics, performing a one-sample z-test with known variance, and justifying use of CLT. The calculations are routine with clear structure, though it requires careful execution across multiple parts. Slightly easier than average due to the mechanical nature of the steps. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Est}(\mu) = 495.9\) | B1 | Accept 496 |
| \(\text{Est}(\sigma^2) = \frac{10}{9}\left(\frac{2459283}{10} - 495.9^2\right)\) | M1 | Attempt \(\Sigma x^2\) and subst in correct formula \((1/9(``2459283" - ``4959"^2/10))\). May be implied by correct answer |
| \(= 12.8\) (3 sf) or \(383/30\) | A1 | (Note: Biased var "11.49" scores M0 A0) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu = 505\), \(H_1: \mu < 505\) | B1 | Allow 'Pop mean' but not just 'mean' |
| \(\frac{75660/150 - 505}{3.6 \div \sqrt{150}}\) | M1 | Correct stand'n; must have \(\sqrt{150}\). No sd/var mixes. Condone sample SD (3.58/3.39). Accept standardisation of totals \(((75660-75750)/44.091)\). Accept CV method |
| \(= -2.04\) | A1 | Accept +2.04 (Note: if valid area comparison done 0.0207/0.0206 or 0.979 needed for A1) |
| comp \(z = -2.054\) | M1 | Valid comparison of z's or area \((0.0207/6 > 0.02; 0.979(3) < 0.98)\) |
| No evidence (at 2%) that machine pkts mean mass \(< 505\) | A1ft | oe No contradictions. SC Two tail test can score B0 M1 A1 M1 for comparison with 2.326 A0 (max 3/5) |
| Answer | Marks | Guidance |
|---|---|---|
| Large sample, so sample mean approx normally distributed | B1 | Allow just 'Sample is large' or '\(n\) is large' \(n > 30\) |
## Question 4(i):
| $\text{Est}(\mu) = 495.9$ | B1 | Accept 496 |
| $\text{Est}(\sigma^2) = \frac{10}{9}\left(\frac{2459283}{10} - 495.9^2\right)$ | M1 | Attempt $\Sigma x^2$ and subst in correct formula $(1/9(``2459283" - ``4959"^2/10))$. May be implied by correct answer |
| $= 12.8$ (3 sf) or $383/30$ | A1 | (Note: Biased var "11.49" scores M0 A0) |
---
## Question 4(ii):
| $H_0: \mu = 505$, $H_1: \mu < 505$ | B1 | Allow 'Pop mean' but not just 'mean' |
| $\frac{75660/150 - 505}{3.6 \div \sqrt{150}}$ | M1 | Correct stand'n; must have $\sqrt{150}$. No sd/var mixes. Condone sample SD (3.58/3.39). Accept standardisation of totals $((75660-75750)/44.091)$. Accept CV method |
| $= -2.04$ | A1 | Accept +2.04 (Note: if valid area comparison done 0.0207/0.0206 or 0.979 needed for A1) |
| comp $z = -2.054$ | M1 | Valid comparison of z's or area $(0.0207/6 > 0.02; 0.979(3) < 0.98)$ |
| No evidence (at 2%) that machine pkts mean mass $< 505$ | A1ft | oe No contradictions. SC Two tail test can score B0 M1 A1 M1 for comparison with 2.326 A0 (max 3/5) |
---
## Question 4(iii):
| Large sample, so sample mean approx normally distributed | B1 | Allow just 'Sample is large' or '$n$ is large' $n > 30$ |
---4 The mean mass of packets of sugar is supposed to be 505 g . A random sample of 10 packets filled by a certain machine was taken and the masses, in grams, were found to be as follows.
$$\begin{array} { l l l l l l l l l l }
500 & 499 & 496 & 495 & 498 & 490 & 492 & 501 & 494 & 494
\end{array}$$
(i) Find unbiased estimates of the population mean and variance.\\
The mean mass of packets produced by this machine was found to be less than 505 g , so the machine was adjusted. Following the adjustment, the masses of a random sample of 150 packets from the machine were measured and the total mass was found to be 75660 g .\\
(ii) Given that the population standard deviation is 3.6 g , test at the $2 \%$ significance level whether the machine is still producing packets with mean mass less than 505 g .\\
(iii) Explain why the use of the normal distribution is justified in carrying out the test in part (ii). [1]\\
\hfill \mbox{\textit{CAIE S2 2018 Q4 [9]}}