3 A scientist is investigating the masses of birds of a certain species in country \(X\) and country \(Y\). She takes a random sample of 50 birds of this species from country \(X\) and a random sample of 80 birds of this species from country \(Y\). She records their masses in \(\mathrm { kg } , x\) and \(y\), respectively. Her results are summarised as follows. $$\sum x = 75.5 \quad \sum x ^ { 2 } = 115.2 \quad \sum y = 116.8 \quad \sum y ^ { 2 } = 172.6$$ The population mean masses of these birds in countries \(X\) and \(Y\) are \(\mu _ { x } \mathrm {~kg}\) and \(\mu _ { y } \mathrm {~kg}\) respectively.
Test, at the \(5 \%\) significance level, the null hypothesis \(\mu _ { \mathrm { x } } = \mu _ { \mathrm { y } }\) against the alternative hypothesis \(\mu _ { \mathrm { x } } > \mu _ { \mathrm { y } }\). State your conclusion in the context of the question.

## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_x^2 = \frac{1}{49}\left(115.2 - \frac{75.5^2}{50}\right) = 0.02439$ | M1 | $\frac{239}{9800}$, $\frac{259}{9875}$ |
| $s_y^2 = \frac{1}{79}\left(172.6 - \frac{116.8^2}{80}\right) = 0.02623$ | | |
| | A1 | Both correct |
| $s^2 = \frac{0.02439}{50} + \frac{0.02623}{80}\ [= 0.00081565]$ | M1 A1 | May be implied by 1.75 for $z$ |
| $z = \dfrac{\frac{75.5}{50} - \frac{116.8}{80}}{s}$ | M1 | |
| 1.75 | A1 | |
| Compare with 1.645: $1.75 > 1.645$; Reject null hypothesis | M1 | Using areas, $0.04 < 0.05$ |
| Sufficient evidence to suggest that population mean in country $X$ is greater than population mean in country $Y$ | A1 | Correct conclusion in context, following correct work, level of uncertainty in language. 'Prove' gives A0 |
| | **8** | Pooled variance: $z = 1.74$; M1A1 M0A0M0A0 M1A0 max 3/8 |

---

3 A scientist is investigating the masses of birds of a certain species in country $X$ and country $Y$. She takes a random sample of 50 birds of this species from country $X$ and a random sample of 80 birds of this species from country $Y$. She records their masses in $\mathrm { kg } , x$ and $y$, respectively. Her results are summarised as follows.

$$\sum x = 75.5 \quad \sum x ^ { 2 } = 115.2 \quad \sum y = 116.8 \quad \sum y ^ { 2 } = 172.6$$

The population mean masses of these birds in countries $X$ and $Y$ are $\mu _ { x } \mathrm {~kg}$ and $\mu _ { y } \mathrm {~kg}$ respectively.\\
Test, at the $5 \%$ significance level, the null hypothesis $\mu _ { \mathrm { x } } = \mu _ { \mathrm { y } }$ against the alternative hypothesis $\mu _ { \mathrm { x } } > \mu _ { \mathrm { y } }$. State your conclusion in the context of the question.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q3 [8]}}