| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| 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 two-part hypothesis testing question requiring standard formulas for unbiased estimates and a routine one-tail z-test. Part (i) involves direct substitution into memorized formulas, while part (ii) follows a standard hypothesis test procedure with clearly stated parameters. The question requires no novel insight or complex multi-step reasoning, making it slightly easier than average for A-level statistics. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Est}(\mu) = 923/400\) or \(2.3075\) or \(2.31\) (3 sf) | B1 | |
| \(\text{Est}(\sigma^2) = \frac{400}{399}\left(\frac{3170}{400} - \text{"2.3075"}^2\right)\) OE | M1 | |
| \(= 2.60696\) or \(2.61\) (3 sf) | A1 | Note: Biased Var \(= 2.600\) scores M0 |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean (or \(\mu\)) \(=\) "2.31" or "2310"; \(H_1\): Pop mean (or \(\mu\)) \(>\) "2.31" or "2310" | B1 FT | |
| \(\pm\frac{2.6 - \text{"2.310"}}{\sqrt{2.60696 \div 50}} = 1.27\) | M1 A1 | Standardising using their values. Accept 1.28 |
| Comp \(1.645\) (OE) | M1 | Valid comparison \(z\) values or areas |
| No evidence that incomes in the region greater | A1 FT | OE FT their \(z\). No contradictions. (No FT for 2 tail test – max score B0 M1 A1 M1 for comp 1.96 A0). Note: Accept alternative CV method |
| Total: | 5 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Est}(\mu) = 923/400$ or $2.3075$ or $2.31$ (3 sf) | B1 | |
| $\text{Est}(\sigma^2) = \frac{400}{399}\left(\frac{3170}{400} - \text{"2.3075"}^2\right)$ OE | M1 | |
| $= 2.60696$ or $2.61$ (3 sf) | A1 | Note: Biased Var $= 2.600$ scores **M0** |
| **Total:** | **3** | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean (or $\mu$) $=$ "2.31" or "2310"; $H_1$: Pop mean (or $\mu$) $>$ "2.31" or "2310" | B1 FT | |
| $\pm\frac{2.6 - \text{"2.310"}}{\sqrt{2.60696 \div 50}} = 1.27$ | M1 A1 | Standardising using their values. Accept 1.28 |
| Comp $1.645$ (OE) | M1 | Valid comparison $z$ values or areas |
| No evidence that incomes in the region greater | A1 FT | OE FT their $z$. No contradictions. (No FT for 2 tail test – max score **B0 M1 A1 M1** for comp 1.96 **A0**). Note: Accept alternative CV method |
| **Total:** | **5** | |3 Household incomes, in thousands of dollars, in a certain country are represented by the random variable $X$ with mean $\mu$ and standard deviation $\sigma$. The incomes of a random sample of 400 households are found and the results are summarised below.
$$n = 400 \quad \Sigma x = 923 \quad \Sigma x ^ { 2 } = 3170$$
(i) Calculate unbiased estimates of $\mu$ and $\sigma ^ { 2 }$.\\
(ii) A random sample of 50 households in one particular region of the country is taken and the sample mean income, in thousands of dollars, is found to be 2.6 . Using your values from part (i), test at the $5 \%$ significance level whether household incomes in this region are greater, on average, than in the country as a whole.\\
\hfill \mbox{\textit{CAIE S2 2017 Q3 [8]}}