| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type I error |
| Difficulty | Standard +0.3 This is a straightforward one-sample z-test with standard steps: calculate sample statistics, state hypotheses, perform the test at a given significance level, and interpret Type I error. All techniques are routine for S2 level with no novel problem-solving required, though it involves multiple computational steps and understanding of hypothesis testing concepts, making it slightly above average difficulty. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\overline{x} = 14.8 (890/60 \text{ oe})\) | B1 | Correct answer |
| \(s^2 = \frac{1}{59}\left(13780 - \frac{890^2}{60}\right)\) | M1 | Substituting in formula from book, o.e. |
| \(= 9.80\) | A1 [3] | Correct answer |
| (ii) \(H_0: \mu = 15.2\) | B1 | Correct \(H_1\) and \(H_0\) |
| \(H_1: \mu < 15.2\) | B1 | Correct answer |
| \(P(\text{Type I error}) = 0.1 (10\%)\) | B1ft [3] | o.e. must be related to question. No contradictions. ft their \(H_1\) |
| Say the photographer has fewer discards when she doesn't | ||
| (iii) Test statistic \(z = \frac{14.83 - 15.2}{\sqrt{\frac{9.802}{60}}}\) | M1 | Standardising must have \(\sqrt{60}\) |
| \(= -0.915\) | A1 | Correct \(z\) (\(\pm 0.91\) to \(0.92\)) or correct area 0.18 |
| \(CV: z = \pm 1.282\) | M1 | Valid comparison with correct CV must be \(+\) with \(+\) or \(-\) with \(-\) and consistent with their \(H_1\) oe comparison of areas |
| Not enough evidence to support photographer's claim | A1ft [4] | Correct conclusion ft their \(z\) and their CV. No contradictions |
**(i)** $\overline{x} = 14.8 (890/60 \text{ oe})$ | B1 | Correct answer
$s^2 = \frac{1}{59}\left(13780 - \frac{890^2}{60}\right)$ | M1 | Substituting in formula from book, o.e.
$= 9.80$ | A1 [3] | Correct answer
**(ii)** $H_0: \mu = 15.2$ | B1 | Correct $H_1$ and $H_0$
$H_1: \mu < 15.2$ | B1 | Correct answer
$P(\text{Type I error}) = 0.1 (10\%)$ | B1ft [3] | o.e. must be related to question. No contradictions. ft their $H_1$
Say the photographer has fewer discards when she doesn't |
**(iii)** Test statistic $z = \frac{14.83 - 15.2}{\sqrt{\frac{9.802}{60}}}$ | M1 | Standardising must have $\sqrt{60}$
$= -0.915$ | A1 | Correct $z$ ($\pm 0.91$ to $0.92$) or correct area 0.18
$CV: z = \pm 1.282$ | M1 | Valid comparison with correct CV must be $+$ with $+$ or $-$ with $-$ and consistent with their $H_1$ oe comparison of areas
Not enough evidence to support photographer's claim | A1ft [4] | Correct conclusion ft their $z$ and their CV. No contradictions6 Photographers often need to take many photographs of families until they find a photograph which everyone in the family likes. The number of photographs taken until obtaining one which everybody likes has mean 15.2. A new photographer claims that she can obtain a photograph which everybody likes with fewer photographs taken. To test at the $10 \%$ level of significance whether this claim is justified, the numbers of photographs, $x$, taken by the new photographer with a random sample of 60 families are recorded. The results are summarised by $\Sigma x = 890$ and $\Sigma x ^ { 2 } = 13780$.\\
(i) Calculate unbiased estimates of the population mean and variance of the number of photographs taken by the new photographer.\\
(ii) State null and alternative hypotheses for the test, and state also the probability that the test results in a Type I error. Say what a Type I error means in the context of the question.\\
(iii) Carry out the test.
\hfill \mbox{\textit{CAIE S2 2009 Q6 [10]}}