Standard +0.3 This is a straightforward two-tail hypothesis test with all standard steps clearly signposted. Students must calculate sample mean and variance from summations, then perform a z-test using the normal distribution. While it requires multiple computational steps and understanding of hypothesis testing framework, it follows a completely standard template with no conceptual surprises or novel problem-solving required—slightly easier than average due to its routine nature.
4 The areas, \(X \mathrm {~cm} ^ { 2 }\), of petals of a certain kind of flower have mean \(\mu \mathrm { cm } ^ { 2 }\). In the past it has been found that \(\mu = 8.9\). Following a change in the climate, a botanist claims that the mean is no longer 8.9. The areas of a random sample of 200 petals from this kind of flower are measured, and the results are summarized by
$$\Sigma x = 1850 , \quad \Sigma x ^ { 2 } = 17850 .$$
Test the botanist's claim at the \(2.5 \%\) significance level.
Use of biased variance (\(3.6875\)) still scores M1
\(= 2.57\) (3sf) (or using areas \(0.00507 - 0.0051\))
A1
Accept \(2.58\) (3sf) or using areas \(0.0049\)–\(0.005\) where biased variance used
\(2.24 < 2.57\) or \(0.00507 < 0.0125\)
M1
For valid comparison with \(2.240\) or \(2.241\) or valid comparison with \(0.0125\). Accept \(2.24 < 2.58\) or \(0.00496 < 0.0125\) where biased variance used
(Reject \(H_0\)) There is evidence that \(\mu\) is not \(8.9\)
A1 FT
Not definite, e.g. NOT '\(\mu \neq 8.9\)'. Must be in context. No contradictions. (Accept cv method). Note: Use of 1 tail test scores Max B1M1A1B0M1A1M1A0, max 6 out of 8
Total: 8
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{est}(\mu) = \frac{1850}{200}$ or $9.25$ | B1 | |
| $\text{est}(\sigma^2) = \frac{200}{199}\left(\frac{17850}{200} - \left(\frac{1850}{200}\right)^2\right)$ or $\frac{1}{199}\left(17850 - \frac{1850^2}{200}\right)$ | M1 | |
| $= 3.71$ or $3.7060$ or $\frac{1475}{398}$ | A1 | |
| $H_0: \mu = 8.9$, $H_1: \mu \neq 8.9$ | B1 | Accept "population mean" (not just mean) |
| $\dfrac{\frac{1850}{200} - 8.9}{\sqrt{\frac{"3.706"}{200}}}$ | M1 | Use of biased variance ($3.6875$) still scores M1 |
| $= 2.57$ (3sf) (or using areas $0.00507 - 0.0051$) | A1 | Accept $2.58$ (3sf) or using areas $0.0049$–$0.005$ where biased variance used |
| $2.24 < 2.57$ or $0.00507 < 0.0125$ | M1 | For valid comparison with $2.240$ or $2.241$ or valid comparison with $0.0125$. Accept $2.24 < 2.58$ or $0.00496 < 0.0125$ where biased variance used |
| (Reject $H_0$) There is evidence that $\mu$ is not $8.9$ | A1 FT | Not definite, e.g. NOT '$\mu \neq 8.9$'. Must be in context. No contradictions. (Accept cv method). **Note:** Use of 1 tail test scores Max B1M1A1B0M1A1M1A0, max 6 out of 8 |
| Total: 8 | | |
4 The areas, $X \mathrm {~cm} ^ { 2 }$, of petals of a certain kind of flower have mean $\mu \mathrm { cm } ^ { 2 }$. In the past it has been found that $\mu = 8.9$. Following a change in the climate, a botanist claims that the mean is no longer 8.9. The areas of a random sample of 200 petals from this kind of flower are measured, and the results are summarized by
$$\Sigma x = 1850 , \quad \Sigma x ^ { 2 } = 17850 .$$
Test the botanist's claim at the $2.5 \%$ significance level.\\
\hfill \mbox{\textit{CAIE S2 2020 Q4 [8]}}