| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2022 |
| Session | June |
| 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: applying CLT, interpreting software output, verifying a z-statistic calculation, and finding a critical value. All steps are routine A-level statistics techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Mean | 3.565 |
| s | 1.05 |
| N | 40 |
| Z Test of a Mean | |
| Mean | 3.565 |
| S | 1.05 |
| SE | 0.1660 |
| N | 40 |
| Z | 0.994 |
| p | 0.160 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | The sample is large |
| Answer | Marks |
|---|---|
| large n, sample means are Normally distributed | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| 2.4 | Allow ‘n > 30’ in place of ‘large’ |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | p > 0.05 so do not reject H |
| Answer | Marks |
|---|---|
| amount of kitchen waste per household has increased | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 2.2b | Or Z < 1.645 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (c) | S a m p le m e a n − 3.565−3.4 3.565−3.4 |
| Answer | Marks |
|---|---|
| = 0.994 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1a | |
| 1.1 | AG | |
| 3 | (d) | invNorm(0.95, 3.4, 0.1660) |
| x =3.673 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | =3.4+0.1660−1(0.95) |
Question 3:
3 | (a) | The sample is large
The central limit theorem states that, for sufficiently
large n, sample means are Normally distributed | E1
E1
[2] | 2.4
2.4 | Allow ‘n > 30’ in place of ‘large’
oe
3 | (b) | p > 0.05 so do not reject H
0
There is insufficient evidence to suggest that the mean
amount of kitchen waste per household has increased | M1
A1
[2] | 3.4
2.2b | Or Z < 1.645
Must include ‘mean’ or ‘average’
Allow’ insufficient evidence to suggest that the campaign
has been successful’
Condone ‘suggests the campaign has not been successful’
3 | (c) | S a m p le m e a n − 3.565−3.4 3.565−3.4
Z = or or
S E 0.1660 1.05
√40
= 0.994 | M1
A1
[2] | 1.1a
1.1 | AG
3 | (d) | invNorm(0.95, 3.4, 0.1660)
x =3.673 | M1
A1
[2] | 3.1a
1.1 | =3.4+0.1660−1(0.95)
For distribution Or x
Allow M1 for incorrect z-value EG 1.282 or 1.963 A local council collects domestic kitchen waste for composting. Householders place their kitchen waste in a 'compost bin' and this is emptied weekly by the council.
The average weight of kitchen waste collected per household each week is known to be 3.4 kg . The council runs a campaign to try to increase the amount of kitchen waste per household which is put in the compost bin. After the campaign, a random sample of 40 households is selected and the weights in kg of kitchen waste in their compost bins are measured.
A hypothesis test is carried out in order to investigate whether the campaign has been successful, using software to analyse the sample. The output from the software is shown below.\\
□\\
Z Test of a Mean\\
Null Hypothesis $\mu = 3.4$\\
Alternative Hypothesis $\bigcirc < 0 > 0 \neq$\\
Sample
\begin{center}
\begin{tabular}{ r l }
Mean & 3.565 \\
s & 1.05 \\
N & 40 \\
\end{tabular}
\end{center}
Result
\begin{center}
\begin{tabular}{ l l }
\multicolumn{2}{|l|}{Z Test of a Mean} \\
& \\
Mean & 3.565 \\
S & 1.05 \\
SE & 0.1660 \\
N & 40 \\
Z & 0.994 \\
p & 0.160 \\
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Explain why the test is based on the Normal distribution even though the distribution of the population of amounts of kitchen waste per household is not known.
\item Using the output from the software, complete the test at the $5 \%$ significance level.
\item Show how the value of $Z$ in the software output was calculated.
\item Calculate the least value of the sample mean which would have resulted in the conclusion of the test in part (b) being different. You should assume that the standard error is unchanged.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2022 Q3 [8]}}