| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 14 |
| 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 | Challenging +1.8 This is a sophisticated hypothesis testing question requiring understanding of Type II errors and power calculations. Part (i) is routine, but part (ii) demands simultaneous manipulation of two conditions (significance level and Type II error probability) to derive and solve a system of equations involving sample size and critical value—well beyond standard S2 exercises and requiring genuine statistical insight. |
| Spec | 2.05c Significance levels: one-tail and two-tail2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\alpha:\) | M1 | Standardise 100.7 with \(\frac{-2.576}{\sqrt{80}}\) or −2.076 |
| \(\frac{100.7-102}{5.6/\sqrt{80}} = -2.076\) | A1 | 3 |
| Compare with \(-2.576\) | B1 | Compare correct tail with 0.005 or 0.995 |
| or | \(\beta:\) | M1 |
| \(\Phi(-2.076) = 0.0189\) | A1 | a.r.t. 0.019, allow 0.981 only if compared with 0.995 |
| and compare with 0.005 [or 0.995] | B1 | (3) |
| or | \(\gamma:\) | M1 |
| \(102 - \frac{k×5.6}{\sqrt{80}}\) | B1 | \(k = 2.576/2.58\), – sign, and compare 100.7 with CV |
| \(k = 2.576\), compare 100.7 | B1 | (3) |
| Do not reject \(H_0\) | M1 | Reject/Do not reject, √, needs minimum 80 or √80, \(\Phi^{-1}\) or equivalent, correct comparison, not if clearly μ = 100.7 |
| A1 | 2 | Correct contextualised conclusion |
| M1 | One equation for c and n, equated to \(\Phi^{-1}\), allow cc, wrong sign, σ²; 2.326 or 2.33 | |
| B1 | 3 | Correctly obtain given equation, needs in principle to have started from \(c - 102, -2.326\) |
| (ii) (a) \(c - 102 = -2.326\) | A1 | |
| \(\frac{5.6/\sqrt{n}}{13.0256}\) | AG | |
| \(c - 100 = 1.645\) or \(c - 100 = \frac{9.212}{\sqrt{n}}\) | M1 | Second equation, as before |
| A1 | 2 | Completely correct, aef |
| (b) Solve simultaneous equations | M1 | Correct method for simultaneous equations, find c or \(\sqrt{n}\) |
| \(\sqrt{n} = 11.12\) | A1 | \(\sqrt{n}\) correct to 3 SF |
| \(n_{\min} = 124\) | A1 | \(n_{\min} = 124\) only |
| \(c = 100.83\) | A1 | 4 |
(i) $\alpha:$ | M1 | Standardise 100.7 with $\frac{-2.576}{\sqrt{80}}$ or −2.076
$\frac{100.7-102}{5.6/\sqrt{80}} = -2.076$ | A1 | 3 | a.r.t. −2.08 obtained, must be −, not from $\mu = 100.7$; −2.576 or −2.58 seen and compare $z$, allow both +
Compare with $-2.576$ | B1 | Compare correct tail with 0.005 or 0.995
or | $\beta:$ | M1 | Standardise 100.7 with $\sqrt{80}$ or 80
$\Phi(-2.076) = 0.0189$ | A1 | a.r.t. 0.019, allow 0.981 only if compared with 0.995
and compare with 0.005 [or 0.995] | B1 | (3) | Compare correct tail with 0.005 or 0.995
or | $\gamma:$ | M1 | This formula, allow +, 80, wrong SD, any $k$ from $\Phi^{-1}$
$102 - \frac{k×5.6}{\sqrt{80}}$ | B1 | $k = 2.576/2.58$, – sign, and compare 100.7 with CV
$k = 2.576$, compare 100.7 | B1 | (3) | CV a.r.t. 100.4
**Do not reject $H_0$** | M1 | Reject/Do not reject, √, needs minimum 80 or √80, $\Phi^{-1}$ or equivalent, correct comparison, not if clearly μ = 100.7
| A1 | 2 | Correct contextualised conclusion
| M1 | One equation for c and n, equated to $\Phi^{-1}$, allow cc, wrong sign, σ²; 2.326 or 2.33
| B1 | 3 | Correctly obtain given equation, needs in principle to have started from $c - 102, -2.326$
(ii) (a) $c - 102 = -2.326$ | A1 | |
$\frac{5.6/\sqrt{n}}{13.0256}$ | AG | |
$c - 100 = 1.645$ or $c - 100 = \frac{9.212}{\sqrt{n}}$ | M1 | Second equation, as before
| A1 | 2 | Completely correct, aef
(b) Solve simultaneous equations | M1 | Correct method for simultaneous equations, find c or $\sqrt{n}$
$\sqrt{n} = 11.12$ | A1 | $\sqrt{n}$ correct to 3 SF
$n_{\min} = 124$ | A1 | $n_{\min} = 124$ only
$c = 100.83$ | A1 | 4 | Critical value correct, 100.8 or better8 The quantity, $X$ milligrams per litre, of silicon dioxide in a certain brand of mineral water is a random variable with distribution $\mathrm { N } \left( \mu , 5.6 ^ { 2 } \right)$.\\
(i) A random sample of 80 observations of $X$ has sample mean 100.7. Test, at the $1 \%$ significance level, the null hypothesis $\mathrm { H } _ { 0 } : \mu = 102$ against the alternative hypothesis $\mathrm { H } _ { 1 } : \mu \neq 102$.\\
(ii) The test is redesigned so as to meet the following conditions.
\begin{itemize}
\item The hypotheses are $\mathrm { H } _ { 0 } : \mu = 102$ and $\mathrm { H } _ { 1 } : \mu < 102$.
\item The significance level is $1 \%$.
\item The probability of making a Type II error when $\mu = 100$ is to be (approximately) 0.05 .
\end{itemize}
The sample size is $n$, and the critical region is $\bar { X } < c$, where $\bar { X }$ denotes the sample mean.
\begin{enumerate}[label=(\alph*)]
\item Show that $n$ and $c$ satisfy (approximately) the equation $102 - c = \frac { 13.0256 } { \sqrt { n } }$.
\item Find another equation satisfied by $n$ and $c$.
\item Hence find the values of $n$ and $c$.
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2007 Q8 [14]}}