OCR MEI Paper 2 2022 June — Question 12 8 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample t-test
DifficultyModerate -0.8 This is a straightforward hypothesis testing question requiring students to read software output and apply standard procedures. Part (a) is routine recall of null/alternative hypotheses for a one-tailed test about a mean. Part (b) requires identifying the t-distribution from output (standard for small samples). Part (c) involves comparing a p-value to significance level—a mechanical process with no problem-solving. The question tests basic statistical literacy rather than mathematical reasoning, making it easier than average A-level content.
Spec2.05e Hypothesis test for normal mean: known variance

A retailer sells bags of flour which are advertised as containing 1.5 kg of flour. A trading standards officer is investigating whether there is enough flour in each bag. He collects a random sample and uses software to carry out a hypothesis test at the 5\% level. The analysis is shown in the software printout below. \includegraphics{figure_12}
  1. State the hypotheses the officer uses in the test, defining any parameters used. [2]
  2. State the distribution used in the analysis. [3]
  3. Carry out the hypothesis test, giving your conclusion in context. [3]

Question 12:
AnswerMarks Guidance
12(a) H : μ = 1.5
0
H : μ < 1.5
1
μ is the population mean weight of flour in
AnswerMarks
a bagB1
B11.1
2.5both hypotheses in terms of μ
allow μ is the population mean weight of a bag of flour
[2]
Alternatively
32 1
11𝑝 = 1− may be implied by 𝑝 =
120 15
(from (𝑃(𝑋 ≠ 12))
k = 4
𝑘𝑓 32
or =
120 120
(from 11f = 120 ‒ 32 = 88 so f = 8 and so 𝑘𝑝 = ⋯)
k = 4
AnswerMarks Guidance
12(b) N(𝑎,𝑏)
0.242
a = 1.5 or b = or 0.0018
32
0.242
N(1.5, ) isw or N(1.5,0.0018) isw
AnswerMarks
32M1
A1
AnswerMarks
A13.3
2.2a
AnswerMarks
3.1aa and b are numerical values
allow 0.0424² for variance
A0 if answer spoiled by wrong variable quoted eg
0.242 0.242
𝜇~N(1.5, ) or 𝑋~N(1.5, );
32 32
𝑋̅
allow only oe if variable included
[3]
AnswerMarks Guidance
12(c) 0.0786 > 0.05 or ‒1.4142 > ‒1.645
do not reject H
0
there is insufficient evidence at the 5% level
to suggest that the mean weight of the flour
AnswerMarks
in the bags is less than 1.5 kgM1
A1
AnswerMarks
A13.4
1.1
AnswerMarks
2.2bor 1.44 > 1.43(02586…)
NB 1.43(02586…) is from InvNorm(0.05, 1.5,0.0424)
allow accept H or not significant or reject H
0 1
do not allow eg conclude / prove / indicate or other assertive
statement instead of suggest
if calculated values are used full marks may be awarded for
awrt 0.07865 or 0.0786, or ‒ 1.415 ≤ z ≤ ‒ 1.414 ;
otherwise award a maximum of M1A1 for 0.07…or ‒1.4…
other calculated values score M0
[3]
Question 12:
12 | (a) | H : μ = 1.5
0
H : μ < 1.5
1
μ is the population mean weight of flour in
a bag | B1
B1 | 1.1
2.5 | both hypotheses in terms of μ
allow μ is the population mean weight of a bag of flour
[2]
Alternatively
32 1
11𝑝 = 1− may be implied by 𝑝 =
120 15
(from (𝑃(𝑋 ≠ 12))
k = 4
𝑘𝑓 32
or =
120 120
(from 11f = 120 ‒ 32 = 88 so f = 8 and so 𝑘𝑝 = ⋯)
k = 4
12 | (b) | N(𝑎,𝑏)
0.242
a = 1.5 or b = or 0.0018
32
0.242
N(1.5, ) isw or N(1.5,0.0018) isw
32 | M1
A1
A1 | 3.3
2.2a
3.1a | a and b are numerical values
allow 0.0424² for variance
A0 if answer spoiled by wrong variable quoted eg
0.242 0.242
𝜇~N(1.5, ) or 𝑋~N(1.5, );
32 32
𝑋̅
allow only oe if variable included
[3]
12 | (c) | 0.0786 > 0.05 or ‒1.4142 > ‒1.645
do not reject H
0
there is insufficient evidence at the 5% level
to suggest that the mean weight of the flour
in the bags is less than 1.5 kg | M1
A1
A1 | 3.4
1.1
2.2b | or 1.44 > 1.43(02586…)
NB 1.43(02586…) is from InvNorm(0.05, 1.5,0.0424)
allow accept H or not significant or reject H
0 1
do not allow eg conclude / prove / indicate or other assertive
statement instead of suggest
if calculated values are used full marks may be awarded for
awrt 0.07865 or 0.0786, or ‒ 1.415 ≤ z ≤ ‒ 1.414 ;
otherwise award a maximum of M1A1 for 0.07…or ‒1.4…
other calculated values score M0
[3]
A retailer sells bags of flour which are advertised as containing 1.5 kg of flour. A trading standards officer is investigating whether there is enough flour in each bag. He collects a random sample and uses software to carry out a hypothesis test at the 5\% level. The analysis is shown in the software printout below.

\includegraphics{figure_12}

\begin{enumerate}[label=(\alph*)]
\item State the hypotheses the officer uses in the test, defining any parameters used. [2]
\item State the distribution used in the analysis. [3]
\item Carry out the hypothesis test, giving your conclusion in context. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Paper 2 2022 Q12 [8]}}