| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (upper tail) |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard procedures: stating hypotheses, finding a critical value from normal tables (z = 1.645), and identifying common misconceptions about hypothesis tests. The calculations are routine (critical region is mean > 161.76), and part (c) tests basic understanding of hypothesis test language rather than deep statistical insight. Slightly easier than average due to its procedural nature and explicit guidance. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu = 161.6\), \(H_1: \mu > 161.6\) | B1 | AO1.1 |
| \(\mu\) is the population mean height of adult females (in 2020) | B1 | AO2.5 |
| 1-tailed test because it is suspected that the mean height has increased | B1 | AO2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(\bar{X} \sim N\!\left(161.6, \frac{1.96}{200}\right)\) | M1 | AO3.3 |
| awrt \(\bar{X} > 161.8\) BC | A1 | AO1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1st statement is correct because \(\bar{X} > 161.8\), or \(\bar{X} > 161.8\) so the sample mean is in the critical region | M1 | AO2.3 |
| 2nd statement is incorrect because the sample mean is in the critical region, or because the result is significant, or '...so/hence the null hypothesis is rejected.' following from the first statement | A1 | AO2.4 |
| 3rd statement is incorrect; because it is only possible to infer, not prove, using a hypothesis test | B1 | AO2.2b |
# Question 11:
## Part (a):
$H_0: \mu = 161.6$, $H_1: \mu > 161.6$ | **B1** | AO1.1 | Both hypotheses; allow other parameters (but not $X$ or $\bar{X}$) if defined as mean
$\mu$ is the **population** mean height of adult females (in 2020) | **B1** | AO2.5 |
1-tailed test because it is suspected that the mean height has increased | **B1** | AO2.4 |
**[3 marks]**
## Part (b):
Use of $\bar{X} \sim N\!\left(161.6, \frac{1.96}{200}\right)$ | **M1** | AO3.3 | May see $1.645 = \frac{\bar{X}-161.6}{\sqrt{\frac{1.96}{200}}}$; $\bar{X} = 161.76...$ implies **M1** only
awrt $\bar{X} > 161.8$ BC | **A1** | AO1.1 | Allow if inequality not strict; NB 161.76283...; allow e.g. sample mean $> 161.8$
**[2 marks]**
## Part (c):
1st statement is correct because $\bar{X} > 161.8$, or $\bar{X} > 161.8$ so the sample mean is in the critical region | **M1** | AO2.3 | FT their 161.8 (must be greater than 161.6) or allow if $P(\bar{X} > 161.9)[= 0.00122] < 0.05$ or their $z > 1.645$; critical region must be upper tail; probability must be less than 0.5
2nd statement is incorrect because the sample mean is in the critical region, or because the result is significant, or '...so/hence the null hypothesis is rejected.' following from the first statement | **A1** | AO2.4 | FT their calculated critical region or their calculated probability
3rd statement is incorrect; because it is only possible to infer, not prove, using a hypothesis test | **B1** | AO2.2b | Ignore comments about rejecting the null hypothesis
**[3 marks]**
---11 In 2010 the heights of adult women in the UK were found to have mean $\mu = 161.6 \mathrm {~cm}$ and variance $\sigma ^ { 2 } = 1.96 \mathrm {~cm} ^ { 2 }$.
It is believed that the mean height of adult women in 2020 in the UK is greater than in 2010.
In 2020 a researcher collected a random sample of the heights of 200 adult women in the UK.
The researcher calculated the sample mean height and carried out a hypothesis test at the $5 \%$ level to investigate whether there was any evidence to suggest that the mean height of adult women in the UK had increased.
The researcher assumed that the variance was unaltered.
\begin{enumerate}[label=(\alph*)]
\item - State suitable hypotheses for the test, defining any variables you use.
\begin{itemize}
\item Explain whether the researcher conducted a 1-tail or a 2-tail test.
\item Determine the critical region for the test.
\end{itemize}
The researcher found that the sample mean was 161.9 cm and made the following statements.
\begin{itemize}
\item The sample mean is in the critical region.
\item The null hypothesis is accepted.
\item This proves that the mean height of adult women in the UK is unaltered at 161.6 cm .
\item Explain whether each of these statements is correct.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2021 Q11 [8]}}