| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| 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 standard procedures: calculating sample size from a given z-statistic using the formula z = (x̄ - μ)/(s/√n), then comparing to critical value. Both parts involve direct application of learned formulas with no conceptual challenges or novel problem-solving required. Slightly above average difficulty only due to the algebraic manipulation in part (i). |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{85.7 - 85}{4.8/\sqrt{n}} = 1.786\) | M1 | |
| \(n = \left(\frac{1.786 \times 4.8}{0.7}\right)^2\) | A1 | Correct equation in \(n\) |
| \(= 150\) | A1 | |
| [3] | ||
| (ii) \(H_0: \mu = 85.0\) \(H_1: \mu > 85.0\) | B1 | |
| \(z = 1.645\) | M1 | Comparison 1.786 and 1.645 |
| Allow 1.96 if \(H_1: \mu \neq 85.0\) | ||
| Evidence that \(\mu\) increased | A1f | Correct conc. No contradictions, fl \(H_1\) |
| [3] |
**(i)** $\frac{85.7 - 85}{4.8/\sqrt{n}} = 1.786$ | M1 |
$n = \left(\frac{1.786 \times 4.8}{0.7}\right)^2$ | A1 | Correct equation in $n$
$= 150$ | A1 |
| [3] |
**(ii)** $H_0: \mu = 85.0$ $H_1: \mu > 85.0$ | B1 |
$z = 1.645$ | M1 | Comparison 1.786 and 1.645
| | Allow 1.96 if $H_1: \mu \neq 85.0$
Evidence that $\mu$ increased | A1f | Correct conc. No contradictions, fl $H_1$
| [3] |3 Past experience has shown that the heights of a certain variety of rose bush have been normally distributed with mean 85.0 cm . A new fertiliser is used and it is hoped that this will increase the heights. In order to test whether this is the case, a botanist records the heights, $x \mathrm {~cm}$, of a large random sample of $n$ rose bushes and calculates that $\bar { x } = 85.7$ and $s = 4.8$, where $\bar { x }$ is the sample mean and $s ^ { 2 }$ is an unbiased estimate of the population variance. The botanist then carries out an appropriate hypothesis test.\\
(i) The test statistic, $z$, has a value of 1.786 correct to 3 decimal places. Calculate the value of $n$.\\
(ii) Using this value of the test statistic, carry out the test at the $5 \%$ significance level.
\hfill \mbox{\textit{CAIE S2 2011 Q3 [6]}}