| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Unbiased estimates calculation |
| Difficulty | Moderate -0.3 This is a straightforward hypothesis testing question with standard procedures. Part (a) requires routine calculation of sample mean and variance from given data. Part (b)(i) is a one-sample z-test with known population standard deviation—a textbook application requiring only substitution into formulas. Part (b)(ii) tests understanding of the Central Limit Theorem with large samples (n=100), which is standard bookwork. The question involves multiple parts but each is procedural with no novel insight required, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Est}(\mu) = \frac{2044}{8}\ [=255.5]\) | B1 | Accept 3sf if nothing better seen. |
| \(\text{Est}(\sigma^2) = \frac{8}{7}\!\left(\frac{522348}{8} - \text{"255.5"}^2\right)\) or \(\frac{1}{7}\!\left(\text{'522348'} - \frac{\text{'2044'}^2}{8}\right)\) | M1 | Attempt to find \(\Sigma x^2\) and substitute in correct formula. May be implied by correct answer. Biased \(13.25\) scores M0. |
| \(= 15.1\) (3sf) or \(\frac{106}{7}\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 253\), \(H_1: \mu > 253\) | B1 | Allow 'Population mean' but not just 'mean'. |
| \(\frac{\frac{25360}{100} - 253}{3.5 \div \sqrt{100}}\) | M1 | Standardising must have \(\sqrt{100}\). |
| \(= 1.714\) | A1 | |
| \(1.714 > 1.645\) or \(0.0432 < 0.05\) | M1 | OE |
| [Reject \(H_0\)] There is sufficient evidence (at 5% level) to suggest [mean] mass is greater than 253 | A1FT | OE. FT *their* '1.714' in context, not definite, no contradictions. Accept critical value method of \(253.57 < 253.60\) or \(253.02 > 253\). Use of a two-tailed test scores B0 M1 A1 M1 A0 (comp with \(\frac{0.025}{1.96}\)). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Not true. Large sample, [so sample mean is approx normally distributed]. | B1 | OE. Allow 'Not true. Large sample' or 'Not true. \(n\) is large' or 'Not true. CLT used'. |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Est}(\mu) = \frac{2044}{8}\ [=255.5]$ | B1 | Accept 3sf if nothing better seen. |
| $\text{Est}(\sigma^2) = \frac{8}{7}\!\left(\frac{522348}{8} - \text{"255.5"}^2\right)$ or $\frac{1}{7}\!\left(\text{'522348'} - \frac{\text{'2044'}^2}{8}\right)$ | M1 | Attempt to find $\Sigma x^2$ and substitute in correct formula. May be implied by correct answer. Biased $13.25$ scores M0. |
| $= 15.1$ (3sf) or $\frac{106}{7}$ | A1 | OE |
## Question 4(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 253$, $H_1: \mu > 253$ | B1 | Allow 'Population mean' but not just 'mean'. |
| $\frac{\frac{25360}{100} - 253}{3.5 \div \sqrt{100}}$ | M1 | Standardising must have $\sqrt{100}$. |
| $= 1.714$ | A1 | |
| $1.714 > 1.645$ or $0.0432 < 0.05$ | M1 | OE |
| [Reject $H_0$] There is sufficient evidence (at 5% level) to suggest [mean] mass is greater than 253 | A1FT | OE. FT *their* '1.714' in context, not definite, no contradictions. Accept critical value method of $253.57 < 253.60$ or $253.02 > 253$. Use of a two-tailed test scores B0 M1 A1 M1 A0 (comp with $\frac{0.025}{1.96}$). |
## Question 4(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Not true. Large sample, [so sample mean is approx normally distributed]. | B1 | OE. Allow 'Not true. Large sample' or 'Not true. $n$ is large' or 'Not true. CLT used'. |4
\begin{enumerate}[label=(\alph*)]
\item A random sample of 8 boxes of cereal from a certain supplier was taken. Each box was weighed and the masses in grams were as follows.
$$\begin{array} { l l l l l l l l }
261 & 249 & 259 & 252 & 255 & 256 & 258 & 254
\end{array}$$
Find unbiased estimates of the population mean and variance.
\item The supplier claims that the mean mass of boxes of cereal is 253 g . A quality control officer suspects that the mean mass is actually more than 253 g . In order to test this claim, he weighs a random sample of 100 boxes of cereal and finds that the total mass is 25360 g .
\begin{enumerate}[label=(\roman*)]
\item Given that the population standard deviation of the masses is 3.5 g , test at the $5 \%$ significance level whether the population mean mass is more than 253 g .\\
An employee says, 'This test is invalid because it uses the normal distribution, but we do not know whether the masses of the boxes are normally distributed.'
\item Explain briefly whether this statement is true or not.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q4 [9]}}