| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2014 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Type I/II errors and power of test |
| Type | State meaning of Type I error |
| Difficulty | Challenging +1.8 This is a challenging S4 question requiring students to work backwards from Type I and Type II error constraints to find both sample size and critical value simultaneously, then calculate power. It demands strong understanding of hypothesis testing theory, manipulation of normal distribution equations, and solving a system involving z-scores. While the individual techniques are standard for Further Maths Statistics, the reverse-engineering aspect and multi-constraint optimization make it substantially harder than routine hypothesis test questions. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Type I error: rejecting null hypothesis | B1 | Allow B1 out of 2 for P(...) |
| when it is true | B1 | |
| Type II error: accepting null hypothesis | B1 | Allow B1 out of 2 for P(...) |
| when it is false | B1 | |
| OC: P(accepting null hypothesis | B1 | P(Type II error |
| as a function of the parameter under investigation) | B1 | B1+B1 |
| Power: P(rejecting null hypothesis | B1 | P(Type I error |
| as a function of the parameter under investigation) | B1 | B1+B1. Definition of power as \(1 - \text{OC}\) scores zero |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X \sim N(\mu, \sigma = 3.7)\), \(H_0: \mu = 63\), \(H_1: \mu < 63\) | Note: candidates might not exhibit work in style of mark scheme | |
| \(0.01 = P\text{ reject } H_0 \mid \mu = 63 = P(\bar{X} < c \mid \mu = 63)\) | M1 | May be implied |
| \(= P\left(N\ 63, 3.7^2/n < c\right) = P\left(N\ 0,1 < \frac{c-63}{3.7/\sqrt{n}}\right)\) | M1 | M1 for distribution of \(\bar{X}\) (may be implied) |
| M1 | M1 for standardising (ignore "c") | |
| \(\therefore \frac{c-63}{3.7/\sqrt{n}} = -2.326\) | B1 | B1 for \(-2.326\) used |
| \(0.90 = P\text{ reject } H_0 \mid \mu = 60\) | ||
| \(= P\left(N\ 60, 3.7^2/n < c\right) = P\left(N\ 0,1 < \frac{c-60}{3.7/\sqrt{n}}\right)\) | M1* | M1 for distribution of \(\bar{X}\) |
| M1dep* | M1 for standardising (ignore "c") | |
| \(\therefore \frac{c-60}{3.7/\sqrt{n}} = 1.282\) | B1 | B1 for 1.282 used |
| \(\therefore c = 63 - \frac{8.6062}{\sqrt{n}}\) and \(c = 60 + \frac{4.7434}{\sqrt{n}}\) | M1 | M1 for two equations for \(c\) and \(n\) |
| A1 | A1 if both correct (f.t. from any previous errors) | |
| Attempt to solve | M1 | |
| \(c = 61.0658\) [allow 61.1 or awrt] | B1 | cao (can allow if st error is incorrect) |
| \(\sqrt{n} = 4.45(053)\), \(n = 19.8(07)\) | A1 | cao (awrt 19.8) |
| Take \(n\) as "next integer up" from candidate's value | B1 | |
| \(P\text{ reject } H_0 \mid \mu = 58.5\) | M1 | Correct "translation" of power of test for \(\mu = 58.5\) |
| \(= P\left(N\ 58.5, 3.7^2/20 < 61.0658\right)\) | M1 | Use of candidate's values of \(c\) and \(n\) (NB \(n\) must be an integer here) |
| \(= P\left(N\ 0,1 < \frac{2.5658}{3.7/\sqrt{20}} = 3.10\right) = 0.9990\) | A1 | cao (awrt 0.999) |
| [16] |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Type I error: rejecting null hypothesis | B1 | Allow B1 out of 2 for P(...) |
| when it is true | B1 | |
| Type II error: accepting null hypothesis | B1 | Allow B1 out of 2 for P(...) |
| when it is false | B1 | |
| OC: P(accepting null hypothesis | B1 | P(Type II error | the true value of the parameter) scores |
| as a function of the parameter under investigation) | B1 | B1+B1 |
| Power: P(rejecting null hypothesis | B1 | P(Type I error | the true value of the parameter) scores |
| as a function of the parameter under investigation) | B1 | B1+B1. Definition of power as $1 - \text{OC}$ scores zero |
| **[8]** | | |
---
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim N(\mu, \sigma = 3.7)$, $H_0: \mu = 63$, $H_1: \mu < 63$ | | Note: candidates might not exhibit work in style of mark scheme |
| $0.01 = P\text{ reject } H_0 \mid \mu = 63 = P(\bar{X} < c \mid \mu = 63)$ | M1 | May be implied |
| $= P\left(N\ 63, 3.7^2/n < c\right) = P\left(N\ 0,1 < \frac{c-63}{3.7/\sqrt{n}}\right)$ | M1 | M1 for distribution of $\bar{X}$ (may be implied) |
| | M1 | M1 for standardising (ignore "c") |
| $\therefore \frac{c-63}{3.7/\sqrt{n}} = -2.326$ | B1 | B1 for $-2.326$ used |
| $0.90 = P\text{ reject } H_0 \mid \mu = 60$ | | |
| $= P\left(N\ 60, 3.7^2/n < c\right) = P\left(N\ 0,1 < \frac{c-60}{3.7/\sqrt{n}}\right)$ | M1* | M1 for distribution of $\bar{X}$ |
| | M1dep* | M1 for standardising (ignore "c") |
| $\therefore \frac{c-60}{3.7/\sqrt{n}} = 1.282$ | B1 | B1 for 1.282 used |
| $\therefore c = 63 - \frac{8.6062}{\sqrt{n}}$ and $c = 60 + \frac{4.7434}{\sqrt{n}}$ | M1 | M1 for two equations for $c$ and $n$ |
| | A1 | A1 if both correct (f.t. from any previous errors) |
| Attempt to solve | M1 | |
| $c = 61.0658$ [allow 61.1 or awrt] | B1 | cao (can allow if st error is incorrect) |
| $\sqrt{n} = 4.45(053)$, $n = 19.8(07)$ | A1 | cao (awrt 19.8) |
| Take $n$ as "next integer up" from candidate's value | B1 | |
| $P\text{ reject } H_0 \mid \mu = 58.5$ | M1 | Correct "translation" of power of test for $\mu = 58.5$ |
| $= P\left(N\ 58.5, 3.7^2/20 < 61.0658\right)$ | M1 | Use of candidate's values of $c$ and $n$ (NB $n$ must be an integer here) |
| $= P\left(N\ 0,1 < \frac{2.5658}{3.7/\sqrt{20}} = 3.10\right) = 0.9990$ | A1 | cao (awrt 0.999) |
| **[16]** | | |
---3 (i) Explain the meaning of the following terms in the context of hypothesis testing: Type I error, Type II error, operating characteristic, power.\\
(ii) A chemical manufacturer is endeavouring to reduce the amount of a certain impurity in one of its bulk products by improving the production process. The amount of impurity is measured in a convenient unit of concentration, and this is modelled by the Normally distributed random variable $X$. In the old production process, the mean of $X$, denoted by $\mu$, was 63 and the standard deviation of $X$ was 3.7. Experimental batches of the product are to be made using the new process, and it is desired to examine the hypotheses $\mathrm { H } _ { 0 } : \mu = 63$ and $\mathrm { H } _ { 1 } : \mu < 63$ for the new process. Investigation of the variability in the new process has established that the standard deviation may be assumed unchanged.
The usual Normal test based on $\bar { X }$ is to be used, where $\bar { X }$ is the mean of $X$ over $n$ experimental batches (regarded as a random sample), with a critical value $c$ such that $\mathrm { H } _ { 0 }$ is rejected if the value of $\bar { X }$ is less than $c$. The following criteria are set out.
\begin{itemize}
\item If in fact $\mu = 63$, the probability of concluding that $\mu < 63$ must be only $1 \%$.
\item If in fact $\mu = 60$, the probability of concluding that $\mu < 63$ must be $90 \%$.
\end{itemize}
Find $c$ and the smallest value of $n$ that is required. With these values, what is the power of the test if in fact $\mu = 58.5$ ?
\hfill \mbox{\textit{OCR MEI S4 2014 Q3 [24]}}