OCR Further Statistics 2019 June — Question 4 9 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2019
SessionJune
Marks9
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Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeOne-tail z-test (lower tail)
DifficultyStandard +0.3 This is a straightforward one-sample z-test with known variance following a standard template: calculate test statistic, compare to critical value, and state conclusion. Parts (b) and (c) test basic understanding of assumptions and CLT, which are standard bookwork. The question requires minimal problem-solving beyond applying the learned procedure, making it slightly easier than average.
Spec5.05c Hypothesis test: normal distribution for population mean
4 The greatest weight \(W N\) that can be supported by a shelving bracket of traditional design is a normally distributed random variable with mean 500 and standard deviation 80 . A sample of 40 shelving brackets of a new design are tested and it is found that the mean of the greatest weights that the brackets in the sample can support is 473.0 N .
  1. Test at the \(1 \%\) significance level whether the mean of the greatest weight that a bracket of the new design can support is less than the mean of the greatest weight that a bracket of the traditional design can support.
  2. State an assumption needed in carrying out the test in part (a).
  3. Explain whether it is necessary to use the central limit theorem in carrying out the test.
Question 4:
AnswerMarks Guidance
4(a) H :  = 500, H :  < 500
0 1
where  is the mean of the greatest weight (that
the new design can support)
2
80
X ~ N(500, ) = N(500, 160) and X = 473
40
P(X < 473) = 0.01640 or z = –2.13(45)
or CV = 470.6
p > 0.01 or z > –2.326 or 473 > 470.6
Do not reject H . Insufficient evidence that
0
greatest weight that new design can support is
less than the greatest weight that the traditional
AnswerMarks
design can support.B1
B1
M1
A1
A1
M1ft
A1ft
AnswerMarks
[7]1.1
2.5
3.3
3.4
1.1
1.1
AnswerMarks
2.2bOne error, e.g. H :   500, or  not
1
defined, or all in words: B1
40 needed but allow  errors, e.g.
variance 80/40 etc. If CV found, not
centred on 473.
p or z correct to 3 sf.
Compare p with 0.01 or z with –
2.326, or 2.326 used in CV
Correct first conclusion, needs
correct method and like-with-like, ft
on test statistic if method correct
AnswerMarks
Contextualised, not too definitex or x : 0 unless defined as
population mean (then B1)
Can be implied by 0.0164,
0.9836, 0.433, 0.198, 0.000
but not 0.3679 or 0.00127
NOT 0.9836
Must be like-with-like,
Not e.g. 0.9836 > 0.01
or p < 2.326
But BOD if no explicit
comparison of p with 0.01
Not “the new design does
not have a smaller greatest
weight …”
AnswerMarks Guidance
(b)Standard deviation/variance remains unchanged,
or sample must be randomB1
[1]1.2 No extras. Not “same distribution”.
not needed
AnswerMarks
(c)Either: Yes as we do not know that the
distribution of weights for the new
design is normal
Or: No as the population distribution known
AnswerMarks Guidance
to be normalB1
[1]2.1 Allow “population distribution
assumed to be normal”. No extras,
AnswerMarks
e.g. “and sample size is large”.Allow “yes as we do not
know that the distribution
for the new design is
normal” only if clearly refers
to the new design only
Question 4:
4 | (a) | H :  = 500, H :  < 500
0 1
where  is the mean of the greatest weight (that
the new design can support)
2
80
X ~ N(500, ) = N(500, 160) and X = 473
40
P(X < 473) = 0.01640 or z = –2.13(45)
or CV = 470.6
p > 0.01 or z > –2.326 or 473 > 470.6
Do not reject H . Insufficient evidence that
0
greatest weight that new design can support is
less than the greatest weight that the traditional
design can support. | B1
B1
M1
A1
A1
M1ft
A1ft
[7] | 1.1
2.5
3.3
3.4
1.1
1.1
2.2b | One error, e.g. H :   500, or  not
1
defined, or all in words: B1
40 needed but allow  errors, e.g.
variance 80/40 etc. If CV found, not
centred on 473.
p or z correct to 3 sf.
Compare p with 0.01 or z with –
2.326, or 2.326 used in CV
Correct first conclusion, needs
correct method and like-with-like, ft
on test statistic if method correct
Contextualised, not too definite | x or x : 0 unless defined as
population mean (then B1)
Can be implied by 0.0164,
0.9836, 0.433, 0.198, 0.000
but not 0.3679 or 0.00127
NOT 0.9836
Must be like-with-like,
Not e.g. 0.9836 > 0.01
or p < 2.326
But BOD if no explicit
comparison of p with 0.01
Not “the new design does
not have a smaller greatest
weight …”
(b) | Standard deviation/variance remains unchanged,
or sample must be random | B1
[1] | 1.2 | No extras. Not “same distribution”. | Not “assume normal”; this is
not needed
(c) | Either: Yes as we do not know that the
distribution of weights for the new
design is normal
Or: No as the population distribution known
to be normal | B1
[1] | 2.1 | Allow “population distribution
assumed to be normal”. No extras,
e.g. “and sample size is large”. | Allow “yes as we do not
know that the distribution
for the new design is
normal” only if clearly refers
to the new design only
4 The greatest weight $W N$ that can be supported by a shelving bracket of traditional design is a normally distributed random variable with mean 500 and standard deviation 80 .

A sample of 40 shelving brackets of a new design are tested and it is found that the mean of the greatest weights that the brackets in the sample can support is 473.0 N .
\begin{enumerate}[label=(\alph*)]
\item Test at the $1 \%$ significance level whether the mean of the greatest weight that a bracket of the new design can support is less than the mean of the greatest weight that a bracket of the traditional design can support.
\item State an assumption needed in carrying out the test in part (a).
\item Explain whether it is necessary to use the central limit theorem in carrying out the test.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2019 Q4 [9]}}
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