| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (upper tail) |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring calculation of sample statistics followed by a standard one-tail z-test. The steps are routine (calculate mean/variance, set up hypotheses, compute test statistic, compare to critical value) with no conceptual challenges beyond applying the standard procedure. Slightly above average difficulty due to the two-part structure and need for careful calculation, but well within typical A-level statistics content. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Est}(\mu) = 2.3\) | B1 | |
| \(\text{Est}(\sigma^2) = \dfrac{200}{199}\!\left(\dfrac{1636}{200} - \left(\dfrac{460}{200}\right)^2\right)\) | M1 | Allow \(\sqrt{\dfrac{200}{199}\!\left(\dfrac{1636}{200}-\left(\dfrac{460}{200}\right)^2\right)}\) or \(1.7043\) for M1; or \(\frac{1}{199}(1636 - 460^2/200)\) |
| \(= 2.90\) (3 sf) or \(2.91\) or \(578/199\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): Pop mean wt loss \(= 2\) kg; \(H_1\): Pop mean wt loss \(> 2\) kg | B1 | Allow '\(\mu\)' but not just 'mean' |
| \(\dfrac{2.3-2}{\sqrt{\dfrac{2.9045}{200}}}\) | M1 | Standardise with \(\sqrt{200}\); accept sd/var mixes; or \(x_\text{crit} = 2 + 2.326\sqrt{2.9045/200}\) |
| \(= 2.489\) or \(\pm 2.49\); or \(0.0064/0.9936\) for area comparison; or \(x_\text{crit} = 2.28(03)\) | A1 | |
| comp \(z = 2.326\) | M1 | For valid comparison (\(z\) or area or \(x_\text{crit}\)) |
| Evidence that mean wt loss \(> 2\) kg | A1ft [5] | No contradictions; reject \(H_0\)/accept \(H_1\) only if \(H_0/H_1\) correctly defined; if \(\frac{200}{199}\) not used in (i): var \(= 2.89\), sd \(= 1.7\), cr \(z = 2.496\) can score all marks |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Est}(\mu) = 2.3$ | B1 | |
| $\text{Est}(\sigma^2) = \dfrac{200}{199}\!\left(\dfrac{1636}{200} - \left(\dfrac{460}{200}\right)^2\right)$ | M1 | Allow $\sqrt{\dfrac{200}{199}\!\left(\dfrac{1636}{200}-\left(\dfrac{460}{200}\right)^2\right)}$ or $1.7043$ for M1; or $\frac{1}{199}(1636 - 460^2/200)$ |
| $= 2.90$ (3 sf) or $2.91$ or $578/199$ | A1 [3] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: Pop mean wt loss $= 2$ kg; $H_1$: Pop mean wt loss $> 2$ kg | B1 | Allow '$\mu$' but not just 'mean' |
| $\dfrac{2.3-2}{\sqrt{\dfrac{2.9045}{200}}}$ | M1 | Standardise with $\sqrt{200}$; accept sd/var mixes; or $x_\text{crit} = 2 + 2.326\sqrt{2.9045/200}$ |
| $= 2.489$ or $\pm 2.49$; or $0.0064/0.9936$ for area comparison; or $x_\text{crit} = 2.28(03)$ | A1 | |
| comp $z = 2.326$ | M1 | For valid comparison ($z$ or area or $x_\text{crit}$) |
| Evidence that mean wt loss $> 2$ kg | A1ft [5] | No contradictions; reject $H_0$/accept $H_1$ only if $H_0/H_1$ correctly defined; if $\frac{200}{199}$ not used in (i): var $= 2.89$, sd $= 1.7$, cr $z = 2.496$ can score all marks |
---5 It is claimed that, on average, people following the Losefast diet will lose more than 2 kg per month. The weight losses, $x$ kilograms per month, of a random sample of 200 people following the Losefast diet were recorded and summarised as follows.
$$n = 200 \quad \Sigma x = 460 \quad \Sigma x ^ { 2 } = 1636$$
(i) Calculate unbiased estimates of the population mean and variance.\\
(ii) Test the claim at the $1 \%$ significance level.
\hfill \mbox{\textit{CAIE S2 2012 Q5 [8]}}