| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type II error |
| Difficulty | Standard +0.8 This is a multi-part hypothesis testing question requiring calculation of test statistic from summary statistics, understanding of Type II error definition, and most challengingly, computing the power function for a specific alternative hypothesis value. Part (ii)(b) requires working backwards from critical values through the sampling distribution under the alternative hypothesis—a conceptually demanding task that goes beyond routine hypothesis testing and requires deep understanding of the relationship between Type I/II errors. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu = 2.0\); \(H_1: \mu \neq 2.0\) | B1 | |
| \(\bar{x} = \frac{430}{200} = 2.15\) | B1 | For \(\bar{x}\) |
| \(s^2 = \frac{200}{199}\left(\frac{1290}{200} - \left(\frac{430}{200}\right)^2\right) = 1.8366834\) | B1 | Correct subst in \(s^2\) formula; for \(s^2\) correct (or \(s = 1.35524\)) |
| \(\frac{2.15 - 2.0}{\sqrt{\frac{1.8366834}{200}}}\ (= 1.565)\) | M1 | For standardising (need 200), accept sd/var mixes |
| \(z = 1.645\) | M1 | For correct comparison of \(z\) values or areas |
| No evidence that \(\mu \neq 2.0\) | A1 [6] | Cwo (condone biased variance for last 3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Concluding \(\mu = 2.0\) although not true | B1 [1] | Not concluding \(\mu \neq 2.0\) although this is true |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\bar{x} - 2.0}{\sqrt{\frac{1.85}{200}}} = 1.645\) | M1 | Attempt at finding rejection region |
| Answer | Marks | Guidance |
|---|---|---|
| Rejection region is \(\bar{x} < 1.8418\) and \(\bar{x} > 2.1582\) | A1 | |
| \(\frac{2.1582 - 2.12}{\sqrt{\frac{1.85}{200}}}\ (= 0.397)\) | M1 | |
| \(P(\bar{x} < 2.1582 \mid \mu = 2.12) = \Phi(0.397)\) | M1 | |
| \(= 0.6543\) | ||
| \(\frac{1.8418 - 2.12}{\sqrt{\frac{1.85}{200}}}\ (= -2.893)\) | M1 | |
| \(P(\bar{x} < -2.893 \mid \mu = 2.12) = 1 - \Phi(2.893)\) | M1 | |
| \((= 0.0019)\) | ||
| \(\Rightarrow P(1.8418 \leq \bar{x} \leq 2.1582 \mid \mu = 2.12) = 0.6543 - 0.0019 = 0.6524\) | ||
| \(P(\text{Type II error}) = 0.652\) (3 sfs) | A1 [7] | Using only RH tail (ans 0.654) scores max M1A0M1M1M0M0A0; SR if zero scored allow SC M1 for one standardisation attempt with den \(\sqrt{(1.85/200)}\) |
## Question 7:
**(i)**
$H_0: \mu = 2.0$; $H_1: \mu \neq 2.0$ | B1 |
$\bar{x} = \frac{430}{200} = 2.15$ | B1 | For $\bar{x}$
$s^2 = \frac{200}{199}\left(\frac{1290}{200} - \left(\frac{430}{200}\right)^2\right) = 1.8366834$ | B1 | Correct subst in $s^2$ formula; for $s^2$ correct (or $s = 1.35524$)
$\frac{2.15 - 2.0}{\sqrt{\frac{1.8366834}{200}}}\ (= 1.565)$ | M1 | For standardising (need 200), accept sd/var mixes
$z = 1.645$ | M1 | For correct comparison of $z$ values or areas
No evidence that $\mu \neq 2.0$ | A1 [6] | Cwo (condone biased variance for last 3 marks)
**(ii)(a)**
Concluding $\mu = 2.0$ although not true | B1 [1] | Not concluding $\mu \neq 2.0$ although this is true
**(ii)(b)**
$\frac{\bar{x} - 2.0}{\sqrt{\frac{1.85}{200}}} = 1.645$ | M1 | Attempt at finding rejection region
$\bar{x} = 2 + 0.1582$
Rejection region is $\bar{x} < 1.8418$ and $\bar{x} > 2.1582$ | A1 |
$\frac{2.1582 - 2.12}{\sqrt{\frac{1.85}{200}}}\ (= 0.397)$ | M1 |
$P(\bar{x} < 2.1582 \mid \mu = 2.12) = \Phi(0.397)$ | M1 |
$= 0.6543$ | |
$\frac{1.8418 - 2.12}{\sqrt{\frac{1.85}{200}}}\ (= -2.893)$ | M1 |
$P(\bar{x} < -2.893 \mid \mu = 2.12) = 1 - \Phi(2.893)$ | M1 |
$(= 0.0019)$ | |
$\Rightarrow P(1.8418 \leq \bar{x} \leq 2.1582 \mid \mu = 2.12) = 0.6543 - 0.0019 = 0.6524$ | |
$P(\text{Type II error}) = 0.652$ (3 sfs) | A1 [7] | Using only RH tail (ans 0.654) scores max M1A0M1M1M0M0A0; SR if zero scored allow SC M1 for one standardisation attempt with den $\sqrt{(1.85/200)}$7 The weights, $X$ kilograms, of bags of carrots are normally distributed. The mean of $X$ is $\mu$. An inspector wishes to test whether $\mu = 2.0$. He weighs a random sample of 200 bags and his results are summarised as follows.
$$\Sigma x = 430 \quad \Sigma x ^ { 2 } = 1290$$
(i) Carry out the test, at the $10 \%$ significance level.\\
(ii) You may now assume that the population variance of $X$ is 1.85 . The inspector weighs another random sample of 200 bags and carries out the same test at the $10 \%$ significance level.
\begin{enumerate}[label=(\alph*)]
\item State the meaning of a Type II error in this context.
\item Given that $\mu = 2.12$, show that the probability of a Type II error is 0.652 , correct to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2012 Q7 [14]}}