| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | Simultaneous critical region and Type II error |
| Difficulty | Standard +0.3 This is a standard hypothesis testing question requiring routine application of normal distribution theory. Part (i)(a) involves finding critical values using z-scores (±1.96) and standard error, part (i)(b) requires calculating Type II error probability using a shifted distribution, and part (ii) tests conceptual understanding. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(50.0 \pm 1.96\sqrt{\frac{20.25}{81}} = 50.0 \pm 0.98\) | M1, B1 | \(50.0 \pm z\sqrt{(1.96/81)}\), allow one sign only, allow \(\sqrt{}\) errors; \(z = 1.96\) in equation (*not* just stated) |
| \(= 49.02, 50.98\) | A1A1 | Both critical values, min 4 SF at some stage (if both 3SF, A1) |
| \(\bar{W} < 49.02\) and \(\bar{W} > 50.98\) | A1\(\sqrt{}\) 5 | CR, allow \(\leq / \geq\), don't need \(\bar{W}\), \(\sqrt{}\) on their CVs, can't recover. [Ans \(50 \pm 0.98\): A1 only] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{50.98 - 50.2}{0.5} = 1.56\) | M1, A1 | Standardise one limit with same SD as in (i); A.r.t. 1.56, allow \(-\); can allow \(\sqrt{}\) here |
| \(\frac{49.02 - 50.2}{0.5} = -2.36\) | A1, M1 | A.r.t. \(-2.36\), allow \(+\); if very unfair; Correct handling of tails for Type II error |
| \(\Phi(1.56) - \Phi(-2.36) = \mathbf{0.9315}\) | A1 5 | Answer in range [0.931, 0.932] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| It would get smaller | B1 1 | No reason needed, but withhold if definitely wrong reason seen. Allow from 1-tail |
# Question 6:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $50.0 \pm 1.96\sqrt{\frac{20.25}{81}} = 50.0 \pm 0.98$ | M1, B1 | $50.0 \pm z\sqrt{(1.96/81)}$, allow one sign only, allow $\sqrt{}$ errors; $z = 1.96$ in equation (*not* just stated) |
| $= 49.02, 50.98$ | A1A1 | Both critical values, min 4 SF at some stage (if both 3SF, A1) |
| $\bar{W} < 49.02$ and $\bar{W} > 50.98$ | A1$\sqrt{}$ **5** | CR, allow $\leq / \geq$, don't need $\bar{W}$, $\sqrt{}$ on their CVs, can't recover. [Ans $50 \pm 0.98$: A1 only] |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{50.98 - 50.2}{0.5} = 1.56$ | M1, A1 | Standardise one limit with same SD as in (i); A.r.t. 1.56, allow $-$; can allow $\sqrt{}$ here |
| $\frac{49.02 - 50.2}{0.5} = -2.36$ | A1, M1 | A.r.t. $-2.36$, allow $+$; if very unfair; Correct handling of tails for Type II error |
| $\Phi(1.56) - \Phi(-2.36) = \mathbf{0.9315}$ | A1 **5** | Answer in range [0.931, 0.932] |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| It would get smaller | B1 **1** | No reason needed, but withhold if definitely wrong reason seen. Allow from 1-tail |
---6 The weight of a plastic box manufactured by a company is $W$ grams, where $W \sim \mathrm {~N} ( \mu , 20.25 )$. A significance test of the null hypothesis $\mathrm { H } _ { 0 } : \mu = 50.0$, against the alternative hypothesis $\mathrm { H } _ { 1 } : \mu \neq 50.0$, is carried out at the $5 \%$ significance level, based on a sample of size $n$.\\
(i) Given that $n = 81$,
\begin{enumerate}[label=(\alph*)]
\item find the critical region for the test, in terms of the sample mean $\bar { W }$,
\item find the probability that the test results in a Type II error when $\mu = 50.2$.\\
(ii) State how the probability of this Type II error would change if $n$ were greater than 81 .
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2009 Q6 [11]}}