Standard +0.3 This is a standard two-sample t-test with summary statistics requiring calculation of sample means, variances, pooled variance, test statistic, and comparison with critical value. While it involves multiple computational steps (5-6 marks typical), it follows a completely routine procedure taught directly in Further Statistics with no conceptual challenges or novel elements—making it slightly easier than an average A-level question overall.
4 A scientist is investigating the lengths of the leaves of birch trees in different regions. He takes a random sample of 50 leaves from birch trees in region \(A\) and a random sample of 60 leaves from birch trees in region \(B\). He records their lengths in \(\mathrm { cm } , x\) and \(y\), respectively. His results are summarised as follows.
$$\sum x = 282 \quad \sum x ^ { 2 } = 1596 \quad \sum y = 328 \quad \sum y ^ { 2 } = 1808$$
The population mean lengths of leaves from birch trees in regions \(A\) and \(B\) are \(\mu _ { A } \mathrm {~cm}\) and \(\mu _ { B } \mathrm {~cm}\) respectively.
Carry out a test at the \(5 \%\) significance level to test the null hypothesis \(\mu _ { \mathrm { A } } = \mu _ { \mathrm { B } }\) against the alternative hypothesis \(\mu _ { \mathrm { A } } \neq \mu _ { \mathrm { B } }\).
Comparison with tabular value \(1.96\) OE; \(2.155 > 1.96\); Reject \(H_0\)
M1
Comparison with 1.96 OE and conclusion
Sufficient evidence to reject population means are equal OR sufficient evidence to accept population means are not equal OR sufficient evidence to suggest \(\mu_A \neq \mu_B\) OE
A1
Correct conclusion following correct work. Level of uncertainty in language is used
Total
8
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_x^2 = \frac{1}{49}\left(1596 - \frac{282^2}{50}\right) = 0.11265 \left[= \frac{138}{1225}\right]$ | M1 | One correct, unsimplified |
| $s_y^2 = \frac{1}{59}\left(1808 - \frac{328^2}{60}\right) = 0.25311 \left[= \frac{224}{885}\right]$ | A1 | Both correct to 3sf |
| $s^2 = \frac{0.11265}{50} + \frac{0.25312}{60} = 0.006471\ldots$ | M1 A1 | May be implied |
| $z = \dfrac{\frac{282}{50} - \frac{328}{60}}{\text{their } s}$ | M1 | |
| $[z] = 2.155$ | A1 | Tolerance $2.15 - 2.16$ |
| Comparison with tabular value $1.96$ OE; $2.155 > 1.96$; Reject $H_0$ | M1 | Comparison with 1.96 OE and conclusion |
| Sufficient evidence to reject population means are equal OR sufficient evidence to accept population means are not equal OR sufficient evidence to suggest $\mu_A \neq \mu_B$ OE | A1 | Correct conclusion following correct work. Level of uncertainty in language is used |
| **Total** | **8** | |
---
4 A scientist is investigating the lengths of the leaves of birch trees in different regions. He takes a random sample of 50 leaves from birch trees in region $A$ and a random sample of 60 leaves from birch trees in region $B$. He records their lengths in $\mathrm { cm } , x$ and $y$, respectively. His results are summarised as follows.
$$\sum x = 282 \quad \sum x ^ { 2 } = 1596 \quad \sum y = 328 \quad \sum y ^ { 2 } = 1808$$
The population mean lengths of leaves from birch trees in regions $A$ and $B$ are $\mu _ { A } \mathrm {~cm}$ and $\mu _ { B } \mathrm {~cm}$ respectively.
Carry out a test at the $5 \%$ significance level to test the null hypothesis $\mu _ { \mathrm { A } } = \mu _ { \mathrm { B } }$ against the alternative hypothesis $\mu _ { \mathrm { A } } \neq \mu _ { \mathrm { B } }$.\\
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q4 [8]}}