| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2024 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a standard one-sample t-test question with routine calculations (sample mean, standard deviation) and straightforward hypothesis testing procedure. Part (a) requires basic statistical calculations, parts (b)-(c) test understanding of normality testing, part (d) is a textbook t-test application, and part (e) checks knowledge of non-parametric alternatives. While it's a Further Maths topic, the question follows a completely standard template with no novel problem-solving required, making it slightly easier than average overall. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.07e Test medians |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | Estimate of population mean = 11.07 (11.069) |
| Estimate of population standard deviation = 3.33 (3.3299...) | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (b) | Normal probability plot is roughly straight |
| Answer | Marks |
|---|---|
| Both suggest that the data may be from a Normal distribution | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.2b | No marks if Wilcoxon suggested |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (c) | H : the Normal distribution fits the data |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | B1 | |
| [1] | 1.2 | Do not allow ‘The data is Normal’ |
| 7 | (d) | DR |
| Answer | Marks |
|---|---|
| different (from 9.4 ng/g) | B1 |
| Answer | Marks |
|---|---|
| [8] | 1.1a |
| Answer | Marks |
|---|---|
| 3.5a | No marks for Wilcoxon |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (e) | A Wilcoxon test (or A Wilcoxon signed rank test) |
| Answer | Marks |
|---|---|
| distributed. | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.2 |
| 2.2b | Do not allow ‘bivariate Normally distributed’ |
Question 7:
7 | (a) | Estimate of population mean = 11.07 (11.069)
Estimate of population standard deviation = 3.33 (3.3299...) | B1
B1
[2] | 1.1
1.1
7 | (b) | Normal probability plot is roughly straight
Very high p-value
Both suggest that the data may be from a Normal distribution | B1
B1
B1
[3] | 1.1
1.1
2.2b | No marks if Wilcoxon suggested
Allow ‘not too low’
Dep on at least one previous B mark
Do not allow ‘from bivariate Normal’ oe
Do not allow ‘the data is Normally distributed’
7 | (c) | H : the Normal distribution fits the data
0
Condone H : the data comes from or follows a Normal
0
distribution
Condone H : Population is Normally distributed
0 | B1
[1] | 1.2 | Do not allow ‘The data is Normal’
7 | (d) | DR
H : μ = 9.4 H : μ ≠ 9.4
0 1
Where μ is the population mean selenium level in carrots from
these fields
1 1 .0 6 9 − 9 .4
Test statistic is
3 .3 2 9 9 / 1 0
= 1.585
Refer to t
9
Critical value (2-tailed) at 5% level is 2.262
1.585 < 2.262 so not significant (do not reject H )
0
Insufficient evidence to suggest that the selenium level is
different (from 9.4 ng/g) | B1
B1
M1
A1
M1
A1
M1
A1
[8] | 1.1a
1.2
3.3
1.1
3.4
1.1
2.2b
3.5a | No marks for Wilcoxon
Hypotheses in words only must include “population”.
For definition in context including “population”
FT their mean and/or sd
3.3299
For confidence interval method 99.4±Crit value×
√10
BC For confidence interval method 7.018 to 11.782
No FT if not t . Can be implied by 2.262 or 1.833
9
Or p-value = 1 − 0.926 = 0.074. M1 A1
Or using p-value 7.4% > 2.5% Or compare 11.069 with
confidence limit
FT their sensibly obtained test statistic and cv (2.262 or
1.833) for M1 only
Answer must be in context.
7 | (e) | A Wilcoxon test (or A Wilcoxon signed rank test)
Since the population would be unlikely to be Normally
distributed. | B1
B1
[2] | 1.2
2.2b | Do not allow ‘bivariate Normally distributed’7 An environmental investigator wants to check whether the level of selenium in carrots in fields near a mine is different from the usual level in the country, which is $9.4 \mathrm { ng } / \mathrm { g }$ (nanograms per gram). She takes a random sample of 10 carrots from fields near the mine and measures the selenium level of each of them in $\mathrm { ng } / \mathrm { g }$, with results as follows.\\
$\begin{array} { l l l l l l l l l l } 6.20 & 10.72 & 11.42 & 16.32 & 15.33 & 10.56 & 8.83 & 9.21 & 7.78 & 14.32 \end{array}$
\begin{enumerate}[label=(\alph*)]
\item Find estimates of each of the following.
\begin{itemize}
\item The population mean
\item The population standard deviation
\end{itemize}
The investigator produces a Normal probability plot and carries out a Kolmogorov-Smirnov test for these data as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{bab116b3-6e5f-44db-ac86-670e4040d649-06_583_1499_959_242}
\item Comment on what the Normal probability plot and the $p$-value of the test suggest about the data.
\item State the null hypothesis for the Kolmogorov-Smirnov test for Normality.
\item In this question you must show detailed reasoning.
Carry out a test at the $5 \%$ significance level to investigate whether the mean selenium level in carrots from fields near the mine is different from $9.4 \mathrm { ng } / \mathrm { g }$.
\item If the $p$-value of the Kolmogorov-Smirnov test for Normality had been 0.007, explain what procedure you could have used to investigate the selenium level in carrots from fields near the mine.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2024 Q7 [16]}}