| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Standard +0.3 This is a straightforward two-part hypothesis testing question requiring calculation of unbiased estimates followed by a standard one-sample z-test. The calculations are routine (sample mean, variance formula, z-statistic) with clear instructions and no conceptual surprises, making it slightly easier than average for an S2 question. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Est}(\mu) = 3.25 = 13/4\) or \(1625/500\) | B1 | |
| \(\text{Est}(\sigma^2) = \frac{500}{499}\left(\frac{5663.5}{500} - \text{"3.25"}^2\right)\) or \(\frac{1}{499}\left(5663.5 - \frac{1625^2}{500}\right)\) | M1 | Expression of correct form |
| \(= 0.766\) (3 sf) or \(1529/1996\) | A1 | Biased variance of 0.7645 scores M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean (or \(\mu\)) \(=\) '3.25'; \(H_1\): Pop mean (or \(\mu\)) \(\neq\) '3.25' | B1FT | Not just 'mean'. FT their 3.25 |
| \(\frac{2.95 - \text{"3.25"}}{\sqrt{\text{"0.766"} \div 60}}\) | M1 | Standardising with their values. Must have \(\sqrt{60}\) |
| \(= -2.655\) | A1 | Or \(P(\bar{X} < 2.95) = 0.0039\) or \(0.00396\) or \(0.00397\). SC FT their biased \(\text{est}(\sigma^2)\), i.e. 0.7645 to give \(z = 2.658\) A1 |
| \(\text{'2.655'} > 2.576\) or \(\text{'}-2.655\text{'} < -2.576\) | M1 | For valid comparison, e.g. 0.0039 or 0.00396 or 0.00397 \(< 0.005\), or \(0.0078 < 0.01\), or \(0.00792 < 0.01\) |
| [Reject \(H_0\)] There is evidence that (mean) mass in (country B) is different (from country A) | A1FT | OE. Must be in context and not definite. Context needs either 'mass' or 'countries' OE. SC: Use of one-tail test: '2.655' \(> 2.326\) or \(0.0039 < 0.01\) M1A0 (Max B0M1A1M1A0 3/5). Accept critical value method: \(X_\text{crit}=2.959\) M1A1 \(2.95<2.959\) M1A1FT with correct conclusion, or \(X_\text{crit}=3.241\) M1A1 \(3.25>3.241\) M1A1FT with correct conclusion |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Est}(\mu) = 3.25 = 13/4$ or $1625/500$ | B1 | |
| $\text{Est}(\sigma^2) = \frac{500}{499}\left(\frac{5663.5}{500} - \text{"3.25"}^2\right)$ or $\frac{1}{499}\left(5663.5 - \frac{1625^2}{500}\right)$ | M1 | Expression of correct form |
| $= 0.766$ (3 sf) or $1529/1996$ | A1 | Biased variance of 0.7645 scores M0A0 |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean (or $\mu$) $=$ '3.25'; $H_1$: Pop mean (or $\mu$) $\neq$ '3.25' | B1FT | Not just 'mean'. FT their 3.25 |
| $\frac{2.95 - \text{"3.25"}}{\sqrt{\text{"0.766"} \div 60}}$ | M1 | Standardising with their values. Must have $\sqrt{60}$ |
| $= -2.655$ | A1 | Or $P(\bar{X} < 2.95) = 0.0039$ or $0.00396$ or $0.00397$. SC FT their biased $\text{est}(\sigma^2)$, i.e. 0.7645 to give $z = 2.658$ A1 |
| $\text{'2.655'} > 2.576$ or $\text{'}-2.655\text{'} < -2.576$ | M1 | For valid comparison, e.g. 0.0039 or 0.00396 or 0.00397 $< 0.005$, or $0.0078 < 0.01$, or $0.00792 < 0.01$ |
| [Reject $H_0$] There is evidence that (mean) mass in (country B) is different (from country A) | A1FT | OE. Must be in context and not definite. Context needs either 'mass' or 'countries' OE. SC: Use of one-tail test: '2.655' $> 2.326$ or $0.0039 < 0.01$ M1A0 (Max B0M1A1M1A0 3/5). Accept critical value method: $X_\text{crit}=2.959$ M1A1 $2.95<2.959$ M1A1FT with correct conclusion, or $X_\text{crit}=3.241$ M1A1 $3.25>3.241$ M1A1FT with correct conclusion |
---3 The masses, in kilograms, of newborn babies in country $A$ are represented by the random variable $X$, with mean $\mu$ and variance $\sigma ^ { 2 }$. The masses of a random sample of 500 newborn babies in this country were found and the results are summarised below.
$$n = 500 \quad \Sigma x = 1625 \quad \Sigma x ^ { 2 } = 5663.5$$
\begin{enumerate}[label=(\alph*)]
\item Calculate unbiased estimates of $\mu$ and $\sigma ^ { 2 }$.\\
A researcher wishes to test whether the mean mass of newborn babies in a neighbouring country, $B$, is different from that in country $A$. He chooses a random sample of 60 newborn babies in country $B$ and finds that their sample mean mass is 2.95 kg .
Assume that your unbiased estimates in part (a) are the correct values for $\mu$ and $\sigma ^ { 2 }$. Assume also that the variance of the masses of newborn babies in country $B$ is the same as in country $A$.
\item Carry out the test at the $1 \%$ significance level.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q3 [8]}}