| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| 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 covering standard S2 content: calculating sample statistics, conducting a one-tail z-test with known variance, and interpreting Type I errors. All steps are routine applications of formulas with no novel problem-solving required, making it slightly easier than average for A-level statistics. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Number of green sweets | 0 | 1 | 2 | 3 | \(> 3\) |
| Number of packets | 32 | 50 | 97 | 21 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\hat{\mu} = \frac{307}{200}\) or \(1.535\) | B1 | Accept 3sf if nothing better seen (1.53 or 1.54) |
| \(\Sigma x^2 f = 627\), \(\text{Est}(\sigma^2) = \frac{200}{199}\left(\frac{627}{200} - 1.535^2\right)\) or \(\frac{1}{199}\left(627 - \frac{307^2}{200}\right)\) | M1 | Use of a correct formula with *their* values |
| \(= 0.783\) | A1 | AG. Correctly obtained with no errors seen. |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 1.65\); \(H_1: \mu < 1.65\) | B1 | Accept 'population mean' but not just 'mean' |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1.535 - 1.65}{\sqrt{0.783 \div 200}}\) | M1 | Standardising with *their* mean |
| \(= -1.838\) or \(-1.84\) | A1* | |
| \(\Phi(0.05)\) and \(\Phi(0.01)\) attempted | M1 | Or \(P(z < -1.838)\) attempted. SC: Condone \(\Phi(0.025) = 2.807\) and \(\Phi(0.005) = 3.291\) following two-tailed test in (b) |
| \(-1.645 > -1.838 > -2.326\) [Hence significant at 5% but not 1% level] | DA1 | AG. \(= 0.033\) and \(0.05 > 0.033 > 0.01\). SC: use of 1.54 or 1.53 for the mean leading to \(-1.645 > -1.758 > -2.326\) or \(-1.645 > -1.918 > -2.326\); or \(0.95 < 0.9606\) or \(0.9724 < 0.99\) scores M1 M1 A1. Accept critical value method \(1.535 < 1.547\) or accept \(1.65 > 1.638\). |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At the 1% level \(H_0\) is not rejected. Or a Type I error can only occur if \(H_0\) is rejected. | B1 | OE |
| Total: 1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\hat{\mu} = \frac{307}{200}$ or $1.535$ | B1 | Accept 3sf if nothing better seen (1.53 or 1.54) |
| $\Sigma x^2 f = 627$, $\text{Est}(\sigma^2) = \frac{200}{199}\left(\frac{627}{200} - 1.535^2\right)$ or $\frac{1}{199}\left(627 - \frac{307^2}{200}\right)$ | M1 | Use of a correct formula with *their* values |
| $= 0.783$ | A1 | AG. Correctly obtained with no errors seen. |
| **Total: 3** | | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 1.65$; $H_1: \mu < 1.65$ | B1 | Accept 'population mean' but not just 'mean' |
| **Total: 1** | | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1.535 - 1.65}{\sqrt{0.783 \div 200}}$ | M1 | Standardising with *their* mean |
| $= -1.838$ or $-1.84$ | A1* | |
| $\Phi(0.05)$ and $\Phi(0.01)$ attempted | M1 | Or $P(z < -1.838)$ attempted. SC: Condone $\Phi(0.025) = 2.807$ and $\Phi(0.005) = 3.291$ following two-tailed test in (b) |
| $-1.645 > -1.838 > -2.326$ [Hence significant at 5% but not 1% level] | DA1 | AG. $= 0.033$ and $0.05 > 0.033 > 0.01$. SC: use of 1.54 or 1.53 for the mean leading to $-1.645 > -1.758 > -2.326$ or $-1.645 > -1.918 > -2.326$; or $0.95 < 0.9606$ or $0.9724 < 0.99$ scores **M1 M1 A1**. Accept critical value method $1.535 < 1.547$ or accept $1.65 > 1.638$. |
| **Total: 4** | | |
---
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At the 1% level $H_0$ is not rejected. Or a Type I error can only occur if $H_0$ is rejected. | B1 | OE |
| **Total: 1** | | |
---6 The numbers of green sweets in 200 randomly chosen packets of Frutos are summarised in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Number of green sweets & 0 & 1 & 2 & 3 & $> 3$ \\
\hline
Number of packets & 32 & 50 & 97 & 21 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate an unbiased estimate for the population mean of the number of green sweets in a packet of Frutos, and show that an unbiased estimate of the population variance is 0.783 correct to 3 significant figures.\\
The manufacturers of Frutos claim that the mean number of green sweets in a packet is 1.65 .\\
Anji believes that the true value of the mean, $\mu$, is less than 1.65 . She uses the results from the 200 randomly chosen packets to test the manufacturers' claim.
\item State suitable null and alternative hypotheses for the test.\\
\includegraphics[max width=\textwidth, alt={}, center]{7c078a14-98f9-4292-ae76-a2642238176f-08_2714_37_143_2008}
\item Show that the result of Anji's test is significant at the $5 \%$ level but not at the $1 \%$ level.
\item It is given that Anji made a Type I error.
Explain how this shows that the significance level that Anji used in her test was not $1 \%$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q6 [9]}}