OCR S2 2011 June — Question 3 7 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2011
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicType I/II errors and power of test
TypeCalculate probability of Type II error
DifficultyChallenging +1.2 This question requires understanding of Type II errors, working with normal distributions under both null and alternative hypotheses, and calculating probabilities using the standard normal distribution. While it involves multiple conceptual steps (finding critical region, then probability under true distribution), it's a standard S2 textbook exercise with clear setup and straightforward calculation once the method is understood.
Spec2.05c Significance levels: one-tail and two-tail2.05e Hypothesis test for normal mean: known variance

3 The random variable \(X\) has the distribution \(\mathrm { N } \left( \mu , 5 ^ { 2 } \right)\). A hypothesis test is carried out of \(\mathrm { H } _ { 0 } : \mu = 20.0\) against \(\mathrm { H } _ { 1 } : \mu < 20.0\), at the \(1 \%\) level of significance, based on the mean of a sample of size 16. Given that in fact \(\mu = 15.0\), find the probability that the test results in a Type II error.

Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\text{CV} = 20 - \frac{5}{\sqrt{16}} \times 2.326 = 17.0925\)M1, B1, A1 Attempt \(20 - 5z/\sqrt{16}\), allow \(\text{SD} \leftrightarrow\) var errors, allow \(20 \pm 5z/\sqrt{16}\), *not* \(20 + 5z/\sqrt{16}\); 2.326 seen; CV a.r.t. 17.1 [*NB: not 17.9075*]
\(P(X > 17.0925)\)M1* Standardise any attempt at a CV (from \(\mu = 20\)) with 15 and any SD that would have got first M1
\(= \Phi\left(\frac{17.0925-15}{5/\sqrt{16}}\right) = \Phi(1.674)\)A1, dep M1 \(z = 1.674\) seen or implied; probability \(< 0.5\), or \(> 0.5\) if their CV is \(< 15\)
Answer \(\mathbf{0.0471}\)A1 7
# Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{CV} = 20 - \frac{5}{\sqrt{16}} \times 2.326 = 17.0925$ | M1, B1, A1 | Attempt $20 - 5z/\sqrt{16}$, allow $\text{SD} \leftrightarrow$ var errors, allow $20 \pm 5z/\sqrt{16}$, *not* $20 + 5z/\sqrt{16}$; 2.326 seen; CV a.r.t. 17.1 [*NB: not 17.9075*] |
| $P(X > 17.0925)$ | M1* | Standardise any attempt at a CV (from $\mu = 20$) with 15 and any SD that would have got first M1 |
| $= \Phi\left(\frac{17.0925-15}{5/\sqrt{16}}\right) = \Phi(1.674)$ | A1, dep M1 | $z = 1.674$ seen or implied; probability $< 0.5$, or $> 0.5$ if their CV is $< 15$ |
| Answer $\mathbf{0.0471}$ | A1 | **7** | Answer a.r.t. 0.047 [including 0.0465 from CV 17.1] |

---
3 The random variable $X$ has the distribution $\mathrm { N } \left( \mu , 5 ^ { 2 } \right)$. A hypothesis test is carried out of $\mathrm { H } _ { 0 } : \mu = 20.0$ against $\mathrm { H } _ { 1 } : \mu < 20.0$, at the $1 \%$ level of significance, based on the mean of a sample of size 16. Given that in fact $\mu = 15.0$, find the probability that the test results in a Type II error.

\hfill \mbox{\textit{OCR S2 2011 Q3 [7]}}