| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.3 This is a straightforward one-sample t-test question with standard interpretation of normality checks. Part (a) requires routine interpretation of a normal probability plot and K-S test p-value (both clearly support normality), while part (b) is a standard t-test calculation with given data. The question is slightly easier than average because the normality assumption is clearly satisfied, the hypotheses are straightforward, and the calculations follow a standard template that students practice extensively. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | The Normal probability plot is roughly straight and the p- |
| Answer | Marks |
|---|---|
| population to be Normally distributed) | E1 |
| Answer | Marks |
|---|---|
| B1 | 1.1 |
| Answer | Marks |
|---|---|
| 3.3 | No marks if Wilcoxon suggested |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (b) | DR |
| Answer | Marks |
|---|---|
| concentration exceeds 1.0% | B1 |
| Answer | Marks |
|---|---|
| [10] | 1.1a |
| Answer | Marks |
|---|---|
| 3.5a | No marks for Wilcoxon except for mean and sd if found |
Question 7:
7 | (a) | The Normal probability plot is roughly straight and the p-
value is not too low
which are both consistent with the data coming from a
Normal distribution
A t test should be carried out (since this test requires the
population to be Normally distributed) | E1
E1
B1 | 1.1
2.2b
3.3 | No marks if Wilcoxon suggested
Dependent on at least 1 of the previous marks
[3]
7 | (b) | DR
H : μ = 1.0 H : μ > 1.0
0 1
Where μ is the population mean concentration
Sample mean = 1.015
Sample SD = 0.0981
1.015−1.0
Test statistic is
0.0981/√12
= 0.530
Use of t
11
Critical value (1-tailed) at 5% level is 1.796
0.530 < 1.796 not significant (do not reject H )
0
Insufficient evidence to suggest that the mean
concentration exceeds 1.0% | B1
B1
B1
B1
M1
A1
M1
A1
M1
E1
[10] | 1.1a
1.2
1.1
1.1
3.3
1.1
3.4
1.1
2.2b
3.5a | No marks for Wilcoxon except for mean and sd if found
Hypotheses in words only must include “population”.
Allow H : μ = 0.01 H : μ > 0.01
0 1
For definition in context. Must include population.
FT their mean and/or sd
No FT if not t Can be implied by 1.796 or 2.201
11
OR p-value = 0.3034 and compare with 5%
Or confidence interval method [0.9641, 1.0659]
Answer must be in context
FT their sensibly obtained test statistic and cv (provided
from t )for M1 but not for A1
117 An analyst routinely examines bottles of hair shampoo in order to check that the average percentage of a particular chemical which the shampoo contains does not exceed the value of $1.0 \%$ specified by the manufacturer. The percentages of the chemical in a random sample of 12 bottles of the shampoo are as follows.\\
$\begin{array} { l l l l l l l l l l l } 1.087 & 1.171 & 1.047 & 0.846 & 0.909 & 1.052 & 1.042 & 0.893 & 1.021 & 1.085 & 1.096 \end{array} 0.931$\\
The analyst uses software to draw a Normal probability plot for these data, and to carry out a Normality test as shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{c692fb20-436f-4bc1-89bd-10fdba41ceba-08_524_1539_694_264}
\begin{enumerate}[label=(\alph*)]
\item The analyst is going to carry out a hypothesis test to check whether the average percentage exceeds 1.0\%.
Explain which test the analyst should use, referring to each of the following.
\begin{itemize}
\item The Normal probability plot
\item The $p$-value of the Kolmogorov-Smirnov test
\item In this question you must show detailed reasoning.
\end{itemize}
Carry out the test at the 5\% significance level.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2023 Q7 [13]}}