CAIE S2 2021 June — Question 8 11 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2021
SessionJune
Marks11
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Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeCritique inappropriate sampling methods
DifficultyStandard +0.3 This is a straightforward hypothesis testing question covering standard S2 content: identifying appropriate sampling, stating hypotheses, calculating Type I error probability using binomial tables, drawing conclusions, and distinguishing error types. All parts follow textbook procedures with no novel problem-solving required, though part (c) requires careful justification of the critical region which elevates it slightly above pure recall.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

At a certain large school it was found that the proportion of students not wearing correct uniform was 0.15. The school sent a letter to parents asking them to ensure that their children wear the correct uniform. The school now wishes to test whether the proportion not wearing correct uniform has been reduced.
  1. It is suggested that a random sample of the students in Grade 12 should be used for the test. Give a reason why this would not be an appropriate sample. [1]
  2. State suitable null and alternative hypotheses. [1]
  3. Use a binomial distribution to find the probability of a Type I error. You must justify your answer fully. [5]
  4. In fact 4 students out of the 50 are not wearing correct uniform. State the conclusion of the test, explaining your answer. [2]
  5. State, with a reason, which of the errors, Type I or Type II, may have been made. [2]
A suitable sample of 50 students is selected and the number not wearing correct uniform is noted. This figure is used to carry out a test at the 5% significance level.

Question 8:

AnswerMarks Guidance
8(a)Not representative (of all students in the school) B1
grades in the school have different
characteristics/proportions....
Don’t accept ‘not random’ or ‘biased’ without
further explanation.
1

AnswerMarks
8(b)H : P(not correct uniform) = 0.15
0
H : P(not correct uniform) < 0.15
AnswerMarks Guidance
1B1 Allow "p"
1

AnswerMarks Guidance
8(c)Any two probs attempted using B(50,0.15) M1
P(X ⩽ 3) = 0.8550 + 50 × 0.8549 × 0.15 + 50C × 0.8548 × 0.152 + 50C × 0.8547 ×
2 3
AnswerMarks Guidance
0.153M1 Attempt the tail probability P(0,1,2,3) with
B(50,0.15) must be added.
P(X ⩽ 4) = 0.04605 + 50C ×0.8546×0.154
AnswerMarks Guidance
4M1 OE. Their P(X ⩽ 3) + P(X = 4) or P(0,1,2,3,4) with
B(50,0.15) must be added.
P(X ⩽ 3) = 0.0460 or 0.0461 [<0.05]
AnswerMarks Guidance
P(X ⩽ 4) = 0.112 or [>0.05]A1 Both correct.
OR if P(X ⩽ 4) not seen; P(4)=0.06606 and
0.06606>0.05 and P(X ⩽ 3)=0.0460 scores M1 A1
AnswerMarks Guidance
P(Type I) = 0.0460 or 0.0461 (3 sf)A1 Dependent on second M1.
SC If M1M1M1A0 scored allow A1FT for
incorrect P(X ⩽ 3) as long as <0.05
5
AnswerMarks Guidance
QuestionAnswer Marks

AnswerMarks Guidance
8(d)4 is outside critical region (⩽3) OE or P(X ⩽ 4) = 0.112 which is > 0.05 M1
No evidence that proportion not wearing the correct uniform has decreased
AnswerMarks Guidance
(Accept Ho)A1 In context not definite, e.g. not ‘Proportion has not
decreased’. No contradiction.
2

AnswerMarks Guidance
8(e)Not rejected H
0*B1 FT FT If Reject H in (d)
0
AnswerMarks Guidance
Type IIDB1 FT FT Type I
2
Question 8:
--- 8(a) ---
8(a) | Not representative (of all students in the school) | B1 | OE idea of ‘not being representative’ e.g. different
grades in the school have different
characteristics/proportions....
Don’t accept ‘not random’ or ‘biased’ without
further explanation.
1
--- 8(b) ---
8(b) | H : P(not correct uniform) = 0.15
0
H : P(not correct uniform) < 0.15
1 | B1 | Allow "p"
1
--- 8(c) ---
8(c) | Any two probs attempted using B(50,0.15) | M1
P(X ⩽ 3) = 0.8550 + 50 × 0.8549 × 0.15 + 50C × 0.8548 × 0.152 + 50C × 0.8547 ×
2 3
0.153 | M1 | Attempt the tail probability P(0,1,2,3) with
B(50,0.15) must be added.
P(X ⩽ 4) = 0.04605 + 50C ×0.8546×0.154
4 | M1 | OE. Their P(X ⩽ 3) + P(X = 4) or P(0,1,2,3,4) with
B(50,0.15) must be added.
P(X ⩽ 3) = 0.0460 or 0.0461 [<0.05]
P(X ⩽ 4) = 0.112 or [>0.05] | A1 | Both correct.
OR if P(X ⩽ 4) not seen; P(4)=0.06606 and
0.06606>0.05 and P(X ⩽ 3)=0.0460 scores M1 A1
P(Type I) = 0.0460 or 0.0461 (3 sf) | A1 | Dependent on second M1.
SC If M1M1M1A0 scored allow A1FT for
incorrect P(X ⩽ 3) as long as <0.05
5
Question | Answer | Marks | Guidance
--- 8(d) ---
8(d) | 4 is outside critical region (⩽3) OE or P(X ⩽ 4) = 0.112 which is > 0.05 | M1 | FT working from (c).
No evidence that proportion not wearing the correct uniform has decreased
(Accept Ho) | A1 | In context not definite, e.g. not ‘Proportion has not
decreased’. No contradiction.
2
--- 8(e) ---
8(e) | Not rejected H
0 | *B1 FT | FT If Reject H in (d)
0
Type II | DB1 FT | FT Type I
2
At a certain large school it was found that the proportion of students not wearing correct uniform was 0.15. The school sent a letter to parents asking them to ensure that their children wear the correct uniform. The school now wishes to test whether the proportion not wearing correct uniform has been reduced.

\begin{enumerate}[label=(\alph*)]
\item It is suggested that a random sample of the students in Grade 12 should be used for the test.

Give a reason why this would not be an appropriate sample. [1]

\item State suitable null and alternative hypotheses. [1]

\item Use a binomial distribution to find the probability of a Type I error. You must justify your answer fully. [5]

\item In fact 4 students out of the 50 are not wearing correct uniform.

State the conclusion of the test, explaining your answer. [2]

\item State, with a reason, which of the errors, Type I or Type II, may have been made. [2]
\end{enumerate}

A suitable sample of 50 students is selected and the number not wearing correct uniform is noted. This figure is used to carry out a test at the 5% significance level.

\hfill \mbox{\textit{CAIE S2 2021 Q8 [11]}}