| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a straightforward application of a one-sample t-test with standard bookwork components: calculating test statistic from given data, constructing a confidence interval, and identifying the normality assumption. The calculations are routine (12 data points, basic summary statistics), and part (iii) simply requires recalling that a sign test is the non-parametric alternative. Slightly easier than average due to clear structure and standard procedures. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution5.07b Sign test: and Wilcoxon signed-rank |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu = 0.6\) | B1 | Allow absence of "population" if correct notation \(\mu\) is used |
| \(H_1: \mu < 0.6\) | B1 | Do NOT allow "\(\bar{X} = ...\)" or similar unless \(\bar{X}\) clearly stated as population mean |
| Where \(\mu\) is the (population) mean height of the saplings | B1 | Hypotheses in words only must include "population" |
| \(\bar{x} = 0.5883\), \(s_{n-1} = 0.03664\) (\(s_{n-1}^2 = 0.00134\)) | B1 | Do not allow \(s_n = 0.03507\) (\(s_n^2 = 0.00123\)) |
| Test statistic is \(\dfrac{0.5883 - 0.6}{\left(\dfrac{0.03664}{\sqrt{12}}\right)}\) | M1 | Allow c's \(\bar{x}\) and/or \(s_{n-1}\). Allow alternative: \(0.6 \pm (\text{c's} - 1.796) \times \frac{0.03664}{\sqrt{12}}\) |
| \(= -1.103\) | A1 | c.a.o. Use of \(0.6 - \bar{x}\) scores M1A0 |
| Refer to \(t_{11}\). Lower 5% point is \(-1.796\) | M1, A1 | Must be \(-1.796\) unless clear absolute values used |
| \(-1.103 > -1.796\), \(\therefore\) Result is not significant | E1 | ft only c's test statistic |
| Seems mean height of saplings meets the manager's requirements | E1 | ft only c's test statistic |
| Underlying population is Normal | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CI given by \(0.5883 \pm 2.201 \times \dfrac{0.03664}{\sqrt{12}}\) | M1, B1 | ft c's \(\bar{x} \pm\); ft c's \(s_{n-1}\) |
| \(= 0.5883 \pm 0.0233 = (0.565(0),\ 0.611(6))\) | A1 | c.a.o. Must be expressed as interval. ZERO if not same distribution as test |
| In repeated sampling, 95% of intervals constructed in this way will contain the true population mean | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Could use the Wilcoxon test | E1 | |
| Null hypothesis is "Median \(= 0.6\)" | E1 |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 0.6$ | B1 | Allow absence of "population" if correct notation $\mu$ is used |
| $H_1: \mu < 0.6$ | B1 | Do NOT allow "$\bar{X} = ...$" or similar unless $\bar{X}$ clearly stated as population mean |
| Where $\mu$ is the (population) mean height of the saplings | B1 | Hypotheses in words only must include "population" |
| $\bar{x} = 0.5883$, $s_{n-1} = 0.03664$ ($s_{n-1}^2 = 0.00134$) | B1 | Do not allow $s_n = 0.03507$ ($s_n^2 = 0.00123$) |
| Test statistic is $\dfrac{0.5883 - 0.6}{\left(\dfrac{0.03664}{\sqrt{12}}\right)}$ | M1 | Allow c's $\bar{x}$ and/or $s_{n-1}$. Allow alternative: $0.6 \pm (\text{c's} - 1.796) \times \frac{0.03664}{\sqrt{12}}$ |
| $= -1.103$ | A1 | c.a.o. Use of $0.6 - \bar{x}$ scores M1A0 |
| Refer to $t_{11}$. Lower 5% point is $-1.796$ | M1, A1 | Must be $-1.796$ unless clear absolute values used |
| $-1.103 > -1.796$, $\therefore$ Result is not significant | E1 | ft only c's test statistic |
| Seems mean height of saplings meets the manager's requirements | E1 | ft only c's test statistic | [11 marks] |
| Underlying population is Normal | B1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| CI given by $0.5883 \pm 2.201 \times \dfrac{0.03664}{\sqrt{12}}$ | M1, B1 | ft c's $\bar{x} \pm$; ft c's $s_{n-1}$ |
| $= 0.5883 \pm 0.0233 = (0.565(0),\ 0.611(6))$ | A1 | c.a.o. Must be expressed as interval. ZERO if not same distribution as test | [5 marks] |
| In repeated sampling, 95% of intervals constructed in this way will contain the true population mean | E1 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Could use the Wilcoxon test | E1 | |
| Null hypothesis is "Median $= 0.6$" | E1 | | [2 marks] |
---2 The manager of a large country estate is preparing to plant an area of woodland. He orders a large number of saplings (young trees) from a nursery. He selects a random sample of 12 of the saplings and measures their heights, which are as follows (in metres).
$$\begin{array} { l l l l l l l l l l l l }
0.63 & 0.62 & 0.58 & 0.56 & 0.59 & 0.62 & 0.64 & 0.58 & 0.55 & 0.61 & 0.56 & 0.52
\end{array}$$
(i) The manager requires that the mean height of saplings at planting is at least 0.6 metres. Carry out the usual $t$ test to examine this, using a $5 \%$ significance level. State your hypotheses and conclusion carefully. What assumption is needed for the test to be valid?\\
(ii) Find a $95 \%$ confidence interval for the true mean height of saplings. Explain carefully what is meant by a $95 \%$ confidence interval.\\
(iii) Suppose the assumption needed in part (i) cannot be justified. Identify an alternative test that the manager could carry out in order to check that the saplings meet his requirements, and state the null hypothesis for this test.
\hfill \mbox{\textit{OCR MEI S3 2007 Q2 [18]}}