| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Find minimum sample size for Type II error constraint |
| Difficulty | Standard +0.8 This is a two-part hypothesis testing question requiring (i) a standard one-tailed z-test with given significance level, and (ii) a reverse probability calculation to find sample size given a power requirement. Part (i) is routine, but part (ii) requires understanding of sampling distributions and working backwards from a probability condition, which is less standard and requires careful setup of the inequality involving the standard error formula. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(H_0: \mu = 750\) \(H_1: \mu < 750\) Test statistic \(z = \frac{746 - 750}{(11/\sqrt{20})} = +/-1.626\) CV \(= -1.751\) or \(-1.752\) Not enough evidence to say mean is less. | B1, B1, M1, M1*dep, A1 ft [4] | \(H_0\) and \(H_1\) correct. Correct test statistic used in comparison. oe. Correct CV and valid comparison. oe. [0.0520 or 745.7] [2 tail 2.054/5]. Correct conclusion ft their test statistic. No contradictions. Condone cc's. |
| (ii) \(z = +/-1.881\) or \(+/-1.882\) \(-1.881 \leq \frac{745 - 750}{(11/\sqrt{n})} \leq -1.881\) \(n = 18\) | B1, M1*, M1*dep, A1 [4] | Accept 1.881 or 1.882. Equation or inequality relating their z to standardised value. Condone cc's. Solving attempt for \(n\). Correct answer. cwo. 3 sf of accuracy. |
(i) $H_0: \mu = 750$ $H_1: \mu < 750$ Test statistic $z = \frac{746 - 750}{(11/\sqrt{20})} = +/-1.626$ CV $= -1.751$ or $-1.752$ Not enough evidence to say mean is less. | B1, B1, M1, M1*dep, A1 ft [4] | $H_0$ and $H_1$ correct. Correct test statistic used in comparison. oe. Correct CV and valid comparison. oe. [0.0520 or 745.7] [2 tail 2.054/5]. Correct conclusion ft their test statistic. No contradictions. Condone cc's.
(ii) $z = +/-1.881$ or $+/-1.882$ $-1.881 \leq \frac{745 - 750}{(11/\sqrt{n})} \leq -1.881$ $n = 18$ | B1, M1*, M1*dep, A1 [4] | Accept 1.881 or 1.882. Equation or inequality relating their z to standardised value. Condone cc's. Solving attempt for $n$. Correct answer. cwo. 3 sf of accuracy.5 The masses of packets of cornflakes are normally distributed with standard deviation 11 g . A random sample of 20 packets was weighed and found to have a mean mass of 746 g .\\
(i) Test at the $4 \%$ significance level whether there is enough evidence to conclude that the population mean mass is less than 750 g .\\
(ii) Given that the population mean mass actually is 750 g , find the smallest possible sample size, $n$, for which it is at least $97 \%$ certain that the mean mass of the sample exceeds 745 g .
\hfill \mbox{\textit{CAIE S2 2009 Q5 [8]}}