CAIE S2 2015 November — Question 7 11 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2015
SessionNovember
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. Part (i) requires basic understanding of sampling bias (temporal clustering), parts (ii)-(v) are routine applications of binomial hypothesis testing procedures with no computational complexity or novel reasoning required. Slightly above average difficulty only because it's a complete hypothesis test with multiple standard components.
Spec2.04c Calculate binomial probabilities2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

7 At a certain hospital it was found that the probability that a patient did not arrive for an appointment was 0.2 . The hospital carries out some publicity in the hope that this probability will be reduced. They wish to test whether the publicity has worked.
  1. It is suggested that the first 30 appointments on a Monday should be used for the test. Give a reason why this is not an appropriate sample. A suitable sample of 30 appointments is selected and the number of patients that do not arrive is noted. This figure is used to carry out a test at the 5\% significance level.
  2. Explain why the test is one-tail and state suitable null and alternative hypotheses.
  3. State what is meant by a Type I error in this context.
  4. Use the binomial distribution to find the critical region, and find the probability of a Type I error.
  5. In fact 3 patients out of the 30 do not arrive. State the conclusion of the test, explaining your answer.

Question 7:
Part (i):
AnswerMarks Guidance
AnswerMark Guidance
Prob could be different later in day or on a different dayB1 [1] or any explanation why not random or "Not random" or "Not representative"
Part (ii):
AnswerMarks Guidance
AnswerMark Guidance
Looking for decrease (or improvement)B1 oe, Allow "\(p = 0.2\)"
\(H_0: P(\text{not arrive}) = 0.2\)
\(H_1: P(\text{not arrive}) < 0.2\)B1 [2]
Part (iii):
AnswerMarks Guidance
AnswerMark Guidance
Concluding that prob has decreased (or publicity has worked) when it hasn'tB1 [1] In context
Part (iv):
AnswerMarks Guidance
AnswerMark Guidance
\(P(X=0)\) and \(P(X=1)\) attemptedM1 \(B(30, 0.2)\) Not nec'y added. May be implied by calc \(P(X \leqslant 2)\) or \(P(X \leqslant 3)\)
\(P(X \leqslant 2) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2\) \((= 0.0442)\)M1 Attempt \(P(X \leqslant 2)\)
\(P(X \leqslant 3) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2 + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123\)B1 Or \(`0.0442' + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123\)
cr is \(X \leqslant 2\)A1
\(P(\text{Type I}) = 0.0442\) (3 sf)A1 [5]
Part (v):
AnswerMarks Guidance
AnswerMark Guidance
3 is outside crM1 Comparison of 3 with their cr or \(P(X \leqslant 3) = 0.123\) which is \(> 0.05\)
No evidence that \(p\) has decreased (or that publicity has worked)A1\(\checkmark\) [2] Correct conclusion. No contradictions
# Question 7:

## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Prob could be different later in day or on a different day | B1 [1] | or any explanation why not random or "Not random" or "Not representative" |

## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Looking for decrease (or improvement) | B1 | oe, Allow "$p = 0.2$" |
| $H_0: P(\text{not arrive}) = 0.2$ | | |
| $H_1: P(\text{not arrive}) < 0.2$ | B1 [2] | |

## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Concluding that prob has decreased (or publicity has worked) when it hasn't | B1 [1] | In context |

## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=0)$ and $P(X=1)$ attempted | M1 | $B(30, 0.2)$ Not nec'y added. May be implied by calc $P(X \leqslant 2)$ or $P(X \leqslant 3)$ |
| $P(X \leqslant 2) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2$ $(= 0.0442)$ | M1 | Attempt $P(X \leqslant 2)$ |
| $P(X \leqslant 3) = 0.8^{30} + 30 \times 0.8^{29} \times 0.2 + {}^{30}C_2 \times 0.8^{28} \times 0.2^2 + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123$ | B1 | Or $`0.0442' + {}^{30}C_3 \times 0.8^{27} \times 0.2^3 = 0.123$ |
| cr is $X \leqslant 2$ | A1 | |
| $P(\text{Type I}) = 0.0442$ (3 sf) | A1 [5] | |

## Part (v):
| Answer | Mark | Guidance |
|--------|------|----------|
| 3 is outside cr | M1 | Comparison of 3 with their cr or $P(X \leqslant 3) = 0.123$ which is $> 0.05$ |
| No evidence that $p$ has decreased (or that publicity has worked) | A1$\checkmark$ [2] | Correct conclusion. No contradictions |
7 At a certain hospital it was found that the probability that a patient did not arrive for an appointment was 0.2 . The hospital carries out some publicity in the hope that this probability will be reduced. They wish to test whether the publicity has worked.\\
(i) It is suggested that the first 30 appointments on a Monday should be used for the test. Give a reason why this is not an appropriate sample.

A suitable sample of 30 appointments is selected and the number of patients that do not arrive is noted. This figure is used to carry out a test at the 5\% significance level.\\
(ii) Explain why the test is one-tail and state suitable null and alternative hypotheses.\\
(iii) State what is meant by a Type I error in this context.\\
(iv) Use the binomial distribution to find the critical region, and find the probability of a Type I error.\\
(v) In fact 3 patients out of the 30 do not arrive. State the conclusion of the test, explaining your answer.

\hfill \mbox{\textit{CAIE S2 2015 Q7 [11]}}