Edexcel S2 2015 June — Question 5 12 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2015
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of a Poisson distribution
TypeTwo-tailed test setup or execution
DifficultyStandard +0.3 This is a standard S2 hypothesis testing question with routine Poisson calculations. Parts (a)-(c) involve textbook critical region finding and hypothesis testing procedures requiring only table lookups. Part (d) requires more steps (calculating quarterly probabilities then combining for a year) but uses straightforward Poisson distribution work without novel insight. Slightly easier than average due to being methodical rather than conceptually challenging.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion
\emph{Liftsforall} claims that the lift they maintain in a block of flats breaks down at random at a mean rate of 4 times per month. To test this, the number of times the lift breaks down in a month is recorded.
  1. Using a 5\% level of significance, find the critical region for a two-tailed test of the null hypothesis that 'the mean rate at which the lift breaks down is 4 times per month'. The probability of rejection in each of the tails should be as close to 2.5\% as possible. [3]
Over a randomly selected 1 month period the lift broke down 3 times.
  1. Test, at the 5\% level of significance, whether \emph{Liftsforall}'s claim is correct. State your hypotheses clearly. [2]
  2. State the actual significance level of this test. [1]
The residents in the block of flats have a maintenance contract with \emph{Liftsforall}. The residents pay \emph{Liftsforall} £500 for every quarter (3 months) in which there are at most 3 breakdowns. If there are 4 or more breakdowns in a quarter then the residents do not pay for that quarter. \emph{Liftsforall} installs a new lift in the block of flats. Given that the new lift breaks down at a mean rate of 2 times per month,
  1. find the probability that the residents do not pay more than £500 to \emph{Liftsforall} in the next year. [6]
Question 5:
AnswerMarks
5(a)X~Po(4)
P(X = 0) = 0.0183 P(X  8) = 0.0511
AnswerMarks
P(X  1) = 0.0916 P(X  9) = 0.0214M1: using Po(4), need to see a
probability from Po(4), need not be
one of the 4 given here. May be
AnswerMarks
implied by a single correct CRM1
A1
A1
CR X = 0
AnswerMarks
X 9A1: X = 0 or X  0 or X < 1
A1: X 9or X > 8
Any letter(s) may be used instead of X
eg CR or Y or in words
SC candidates who write P(X = 0) and
PX 9award M1A1 A0
NB Candidates who write8 x 0 oe
get M1A0A0
AnswerMarks
(b)H :  = 4 H :   4
0 1B1: both hypotheses correct, labelled
H or NH or H and H or AH or H
0 n 1 a
may use or . These must be seen
AnswerMarks
in part (b)B1
B1ft
There is evidence that Liftsforall’s claim is
true
or There is insufficient evidence to doubt
AnswerMarks
Liftforall’s claimB1: ft their CR only, Do not ft
hypotheses.Needs to include the word
Liftsforall. If no Critical region stated
in part (a) award B0
or PX 3awrt 0.434 and a
correct conclusion.
AnswerMarks Guidance
(c)0.0183 + 0.0214 = 0.0397 B1: Awrt 0.0397
(d)P(B  3 B ~ Po(6) ) = 0.1512
P(B  3) oe. A1: awrt 0.151M1 A1
X ~ B(4, 0.1512)B1ft: dep on M1 being awarded.
Using or writing B(4,“their 0.151”)
for use they need (1 – p)4 or p (1 – p)3
AnswerMarks
or p2(1 – p)2dB1ft
Alternative method for first 3 marks
AnswerMarks Guidance
P(B  4B ~ Po(6) ) = 0.8488 M1: using Po(6) and writing or using
P(B  4) oe A1: awrt 0.849M1 A1
Y ~ B(4, 0.849)B1ft: dep on M1 being awarded.
Using or writing B(4,“their 0.849”)
for use they need (p)4 or p3(1 – p) or
AnswerMarks
p2(1 – p)2dB1ft
If 0 < p < 0.5
AnswerMarks Guidance
P(X  1) = P(X = 0) + P(X = 1)M1: using or writing
P(X = 0) + P(X = 1) oeM1
(10.1512)4 410.15123 0.1512M1: 1 p4 41 p3  p oe dM1
= 0.889A1: awrt 0.889 A1
If 0.5 < p < 1
AnswerMarks Guidance
P(Y  3) = P(Y = 3) + P(Y = 4)M1: using or writing
P(X = 3) + P(X = 4) oeM1
40.84883 0.1512(0.8488)4p4 4p3 1 p
M1: oedM1
= 0.889A1: awrt 0.889 A1
NB: a correct answer implies full marks, lose the final A mark if got awrt 0.889 and go on to do more work
NB: All powers of 1 must be simplified for the Accuracy(A) marks
notes
Question 5:
--- 5(a) ---
5(a) | X~Po(4)
P(X = 0) = 0.0183 P(X  8) = 0.0511
P(X  1) = 0.0916 P(X  9) = 0.0214 | M1: using Po(4), need to see a
probability from Po(4), need not be
one of the 4 given here. May be
implied by a single correct CR | M1
A1
A1
CR X = 0
X 9 | A1: X = 0 or X  0 or X < 1
A1: X 9or X > 8
Any letter(s) may be used instead of X
eg CR or Y or in words
SC candidates who write P(X = 0) and
PX 9award M1A1 A0
NB Candidates who write8 x 0 oe
get M1A0A0
(b) | H :  = 4 H :   4
0 1 | B1: both hypotheses correct, labelled
H or NH or H and H or AH or H
0 n 1 a
may use or . These must be seen
in part (b) | B1
B1ft
There is evidence that Liftsforall’s claim is
true
or There is insufficient evidence to doubt
Liftforall’s claim | B1: ft their CR only, Do not ft
hypotheses.Needs to include the word
Liftsforall. If no Critical region stated
in part (a) award B0
or PX 3awrt 0.434 and a
correct conclusion.
(c) | 0.0183 + 0.0214 = 0.0397 | B1: Awrt 0.0397 | B1
(d) | P(B  3| B ~ Po(6) ) = 0.1512 | M1: using Po(6) and writing or using
P(B  3) oe. A1: awrt 0.151 | M1 A1
X ~ B(4, 0.1512) | B1ft: dep on M1 being awarded.
Using or writing B(4,“their 0.151”)
for use they need (1 – p)4 or p (1 – p)3
or p2(1 – p)2 | dB1ft
Alternative method for first 3 marks
P(B  4| B ~ Po(6) ) = 0.8488 | M1: using Po(6) and writing or using
P(B  4) oe A1: awrt 0.849 | M1 A1
Y ~ B(4, 0.849) | B1ft: dep on M1 being awarded.
Using or writing B(4,“their 0.849”)
for use they need (p)4 or p3(1 – p) or
p2(1 – p)2 | dB1ft
If 0 < p < 0.5
P(X  1) = P(X = 0) + P(X = 1) | M1: using or writing
P(X = 0) + P(X = 1) oe | M1
(10.1512)4 410.15123 0.1512 | M1: 1 p4 41 p3  p oe | dM1
= 0.889 | A1: awrt 0.889 | A1
If 0.5 < p < 1
P(Y  3) = P(Y = 3) + P(Y = 4) | M1: using or writing
P(X = 3) + P(X = 4) oe | M1
40.84883 0.1512(0.8488)4 | p4 4p3 1 p
M1: oe | dM1
= 0.889 | A1: awrt 0.889 | A1
NB: a correct answer implies full marks, lose the final A mark if got awrt 0.889 and go on to do more work
NB: All powers of 1 must be simplified for the Accuracy(A) marks
notes
\emph{Liftsforall} claims that the lift they maintain in a block of flats breaks down at random at a mean rate of 4 times per month. To test this, the number of times the lift breaks down in a month is recorded.

\begin{enumerate}[label=(\alph*)]
\item Using a 5\% level of significance, find the critical region for a two-tailed test of the null hypothesis that 'the mean rate at which the lift breaks down is 4 times per month'. The probability of rejection in each of the tails should be as close to 2.5\% as possible. [3]
\end{enumerate}

Over a randomly selected 1 month period the lift broke down 3 times.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Test, at the 5\% level of significance, whether \emph{Liftsforall}'s claim is correct. State your hypotheses clearly. [2]

\item State the actual significance level of this test. [1]
\end{enumerate}

The residents in the block of flats have a maintenance contract with \emph{Liftsforall}. The residents pay \emph{Liftsforall} £500 for every quarter (3 months) in which there are at most 3 breakdowns. If there are 4 or more breakdowns in a quarter then the residents do not pay for that quarter.

\emph{Liftsforall} installs a new lift in the block of flats.

Given that the new lift breaks down at a mean rate of 2 times per month,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item find the probability that the residents do not pay more than £500 to \emph{Liftsforall} in the next year. [6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2015 Q5 [12]}}
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