Question 18 - A-Level Maths

AQA Paper 3 2020 June — Question 18 14 marks

Exam Board	AQA
Module	Paper 3 (Paper 3)
Year	2020
Session	June
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hypothesis test of binomial distributions
Type	One-tailed test critical region
Difficulty	Standard +0.3 This is a standard A-level statistics question testing binomial distribution calculations and hypothesis testing. Part (a) involves routine binomial probability calculations using a calculator. Part (b) requires finding a critical region for a one-tailed test and applying it—both standard procedures covered extensively in A-level statistics. While it has multiple parts and requires careful setup, all techniques are textbook applications with no novel problem-solving required, making it slightly easier than average overall.
Spec	2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities 2.05b Hypothesis test for binomial proportion

Tiana is a quality controller in a clothes factory. She checks for four possible types of defects in shirts. Of the shirts with defects, the proportion of each type of defect is as shown in the table below.

Type of defect	Colour	Fabric	Sewing	Sizing
Probability	0.25	0.30	0.40	0.05

Shirts with defects are packed in boxes of 30 at random.

Find the probability that:
1. a box contains exactly 5 shirts with a colour defect [2 marks]
2. a box contains fewer than 15 shirts with a sewing defect [2 marks]
3. a box contains at least 20 shirts which do not have a fabric defect. [3 marks]
Tiana wants to investigate the proportion, \(p\), of defective shirts with a fabric defect. She wishes to test the hypotheses H\(_0\): \(p = 0.3\) H\(_1\): \(p < 0.3\) She takes a random sample of 60 shirts with a defect and finds that \(x\) of them have a fabric defect.
1. Using a 5\% level of significance, find the critical region for \(x\). [5 marks]
2. In her sample she finds 13 shirts with a fabric defect. Complete the test stating her conclusion in context. [2 marks]

Show mark scheme Show mark scheme source

Question 18:

Answer	Marks
18(a)(i)	States the binomial distribution

with n = 30, p = 0.25

or

states

or 𝑃𝑃(𝑋𝑋 = 5)= 𝑃𝑃(𝑋𝑋 ≤ 5)−

𝑃𝑃us(𝑋𝑋es≤ th4e) binomial formula with

P3I 0by corr5ect a2n5swer

� �0.25 0.75

Answer	Marks	Guidance
5	3.3	M1

∼

𝑃𝑃(𝑋𝑋 = 5) =0.1047

Obtains correct probability

AWRT 0.105

Answer	Marks	Guidance
ISW	1.1b	A1

Answer	Marks
18(a)(ii)	States the binomial distribution

with n = 30, p = 0.40

or

states

PI by 𝑃𝑃(𝑋𝑋 < 15)= =0.𝑃𝑃90(𝑋𝑋29≤ o1r 4b)y

correct answer

Answer	Marks	Guidance
𝑃𝑃(𝑋𝑋 ≤ 15)	3.3	M1

∼

0.8246

𝑃𝑃(𝑋𝑋 < 15)= 𝑃𝑃(𝑋𝑋 ≤ 14)

=

Obtains correct probability

AWRT 0.825

Answer	Marks	Guidance
ISW	1.1b	A1

Answer	Marks
18(a)(iii)	States the binomial distribution

with n = 30, p = 0.70

PI by = 0.2696 or 0.27

of = 0.5888 or by

correc𝑃𝑃t (a𝑋𝑋ns≤w1e9r )

or 𝑃𝑃(𝑋𝑋 ≥ 21)

states the binomial distribution

with n = 30, p = 0.30

PI by = 0.4112 or 0.41

or = 0.2696 or 0.27 or

𝑃𝑃(Y≥10)

by correct answer

Answer	Marks	Guidance
𝑃𝑃(Y≥11)	1.1b	B1

∼ = 1 – 0.2696

𝑃𝑃 ( 𝑋𝑋 ≥ 2 0 ) == 01.7−3𝑃𝑃04(𝑋𝑋 ≤ 19)

Uses B(30, 0.7) to calculate

= 0.2696 or 0.27

or

𝑃𝑃st(a𝑋𝑋te≤s 19)

or

𝑃𝑃(𝑋𝑋 ≥20)=1−𝑃𝑃(𝑋𝑋 ≤19)

uses B(30, 0.3) to calculate

= 0.4111 or 0.41 or

= 0.2696 or 0.27

Answer	Marks	Guidance
𝑃𝑃(Y≥10)	3.4	M1

𝑃𝑃(Y≥11)

Obtains correct probability

Answer	Marks	Guidance
AWFW [0.73, 0.7304]	1.1b	A1

Finds of P(X ≤ x) where

x = [0, 15] correct to at least 3dp

Do not accept P(X = x)

P(X ≤ x) = [0, 0.244], x = [0, 15]

Answer	Marks	Guidance
OE	1.1a	M1

Obtains P(X ≤ 12) = 0.0568

AWFW [0.05677, 0.0568]

Condone wrong inequality sign if

Answer	Marks	Guidance
[0.05677, 0.0568] seen	1.1b	A1

Obtains P(X ≤ 11) = 0.0295

AWFW [0.02947, 0.0295]

Condone wrong inequality sign if

Answer	Marks	Guidance
[0.02947, 0.0295] seen	1.1b	A1

Obtains correct critical region x

Accept critical region of 0≤ x ≤11

Answer	Marks	Guidance
or x <12 or	3.2a	R1

Answer	Marks
18(b)(ii)	C{0o,1m,2p,a3r,4e,s5 ,163,7 ,w8i,t9h,1 t0h,e1i1r }critical

region from b(i) and makes their

correct inference

or

compares = 0.09996

(accept AWRT 0.1) with 0.05

and infers 𝑃𝑃(𝑋𝑋 is≤ no1t3 r)ejected

Allow reference to

𝐻𝐻0

Answer	Marks	Guidance
𝐻𝐻1	2.2b	M1

There is insufficien𝐻𝐻t 0 evidence to

suggest that the proportion of shirts

with a fabric defect has decreased.

Concludes correctly in context

that there is insufficient

evidence to suggest that the

proportion of shirts with a

fabric defect has decreased.

Follow though correct

conclusion for their critical

Answer	Marks	Guidance
region	3.2a	R1F
Total	14

Question 18:
--- 18(a)(i) ---
18(a)(i) | States the binomial distribution
with n = 30, p = 0.25
or
states
or 𝑃𝑃(𝑋𝑋 = 5)= 𝑃𝑃(𝑋𝑋 ≤ 5)−
𝑃𝑃us(𝑋𝑋es≤ th4e) binomial formula with
P3I 0by corr5ect a2n5swer
� �0.25 0.75
5 | 3.3 | M1 | X B(30, 0.25)
∼
𝑃𝑃(𝑋𝑋 = 5) =0.1047
Obtains correct probability
AWRT 0.105
ISW | 1.1b | A1
--- 18(a)(ii) ---
18(a)(ii) | States the binomial distribution
with n = 30, p = 0.40
or
states
PI by 𝑃𝑃(𝑋𝑋 < 15)= =0.𝑃𝑃90(𝑋𝑋29≤ o1r 4b)y
correct answer
𝑃𝑃(𝑋𝑋 ≤ 15) | 3.3 | M1 | X B(30, 0.40)
∼
0.8246
𝑃𝑃(𝑋𝑋 < 15)= 𝑃𝑃(𝑋𝑋 ≤ 14)
=
Obtains correct probability
AWRT 0.825
ISW | 1.1b | A1
--- 18(a)(iii) ---
18(a)(iii) | States the binomial distribution
with n = 30, p = 0.70
PI by = 0.2696 or 0.27
of = 0.5888 or by
correc𝑃𝑃t (a𝑋𝑋ns≤w1e9r )
or 𝑃𝑃(𝑋𝑋 ≥ 21)
states the binomial distribution
with n = 30, p = 0.30
PI by = 0.4112 or 0.41
or = 0.2696 or 0.27 or
𝑃𝑃(Y≥10)
by correct answer
𝑃𝑃(Y≥11) | 1.1b | B1 | X B(30, 0.70)
∼ = 1 – 0.2696
𝑃𝑃 ( 𝑋𝑋 ≥ 2 0 ) == 01.7−3𝑃𝑃04(𝑋𝑋 ≤ 19)
Uses B(30, 0.7) to calculate
= 0.2696 or 0.27
or
𝑃𝑃st(a𝑋𝑋te≤s 19)
or
𝑃𝑃(𝑋𝑋 ≥20)=1−𝑃𝑃(𝑋𝑋 ≤19)
uses B(30, 0.3) to calculate
= 0.4111 or 0.41 or
= 0.2696 or 0.27
𝑃𝑃(Y≥10) | 3.4 | M1
𝑃𝑃(Y≥11)
Obtains correct probability
AWFW [0.73, 0.7304] | 1.1b | A1
Finds of P(X ≤ x) where
x = [0, 15] correct to at least 3dp
Do not accept P(X = x)
P(X ≤ x) = [0, 0.244], x = [0, 15]
OE | 1.1a | M1
Obtains P(X ≤ 12) = 0.0568
AWFW [0.05677, 0.0568]
Condone wrong inequality sign if
[0.05677, 0.0568] seen | 1.1b | A1
Obtains P(X ≤ 11) = 0.0295
AWFW [0.02947, 0.0295]
Condone wrong inequality sign if
[0.02947, 0.0295] seen | 1.1b | A1
Obtains correct critical region x
Accept critical region of 0≤ x ≤11
or x <12 or | 3.2a | R1
--- 18(b)(ii) ---
18(b)(ii) | C{0o,1m,2p,a3r,4e,s5 ,163,7 ,w8i,t9h,1 t0h,e1i1r }critical
region from b(i) and makes their
correct inference
or
compares = 0.09996
(accept AWRT 0.1) with 0.05
and infers 𝑃𝑃(𝑋𝑋 is≤ no1t3 r)ejected
Allow reference to
𝐻𝐻0
𝐻𝐻1 | 2.2b | M1 | 13 > 11 so accept
There is insufficien𝐻𝐻t 0 evidence to
suggest that the proportion of shirts
with a fabric defect has decreased.
Concludes correctly in context
that there is insufficient
evidence to suggest that the
proportion of shirts with a
fabric defect has decreased.
Follow though correct
conclusion for their critical
region | 3.2a | R1F
Total | 14

Show LaTeX source

Tiana is a quality controller in a clothes factory. She checks for four possible types of defects in shirts.

Of the shirts with defects, the proportion of each type of defect is as shown in the table below.

\begin{tabular}{|l|c|c|c|c|}
\hline
Type of defect & Colour & Fabric & Sewing & Sizing \\
\hline
Probability & 0.25 & 0.30 & 0.40 & 0.05 \\
\hline
\end{tabular}

Shirts with defects are packed in boxes of 30 at random.

\begin{enumerate}[label=(\alph*)]
\item Find the probability that:

\begin{enumerate}[label=(\roman*)]
\item a box contains exactly 5 shirts with a colour defect
[2 marks]

\item a box contains fewer than 15 shirts with a sewing defect
[2 marks]

\item a box contains at least 20 shirts which do not have a fabric defect.
[3 marks]
\end{enumerate}

\item Tiana wants to investigate the proportion, $p$, of defective shirts with a fabric defect.

She wishes to test the hypotheses

H$_0$: $p = 0.3$

H$_1$: $p < 0.3$

She takes a random sample of 60 shirts with a defect and finds that $x$ of them have a fabric defect.

\begin{enumerate}[label=(\roman*)]
\item Using a 5\% level of significance, find the critical region for $x$.
[5 marks]

\item In her sample she finds 13 shirts with a fabric defect.

Complete the test stating her conclusion in context.
[2 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2020 Q18 [14]}}

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