CAIE S2 2014 June — Question 6 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2014
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeInterpret test conclusion or error
DifficultyStandard +0.3 This is a straightforward hypothesis testing question requiring understanding of Type I error definition (rejecting H₀ when true) and recognizing which error type is possible given observed data. Part (i) is a direct binomial probability calculation P(X < 17 | p = 0.9), and part (ii) tests conceptual understanding that Type II errors occur when we fail to reject H₀. Both parts are standard textbook applications with no novel problem-solving required.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail

6 Stephan is an athlete who competes in the high jump. In the past, Stephan has succeeded in \(90 \%\) of jumps at a certain height. He suspects that his standard has recently fallen and he decides to carry out a hypothesis test to find out whether he is right. If he succeeds in fewer than 17 of his next 20 jumps at this height, he will conclude that his standard has fallen.
  1. Find the probability of a Type I error.
  2. In fact Stephan succeeds in 18 of his next 20 jumps. Which of the errors, Type I or Type II, is possible? Explain your answer.

Question 6:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(H_0:\) Rate \(= 0.9\); \(H_1:\) Rate \(< 0.9\)B1 \(p = 0.9\); \(p < 0.9\); Use of \(B(20, 0.1)\)
\(1 - P(17, 18, 19, 20)\)M1 Allow \(1-P(18,19,20)\) or \(1-P(16,17,18,19,20)\)
\(1-(^{20}C_{17} \times 0.1^3 \times 0.9^{17} + ^{20}C_{18} \times 0.1^2 \times 0.9^{18} + 20 \times 0.1 \times 0.9^{19} + 0.9^{20})\)M1
\(= 0.133\) (3 sf)A1 [4]
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Type IIB1 or Stephan will conclude standard not fallen
\(H_0\) will not be rejectedB1 [2] No contradictions
## Question 6:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0:$ Rate $= 0.9$; $H_1:$ Rate $< 0.9$ | B1 | $p = 0.9$; $p < 0.9$; Use of $B(20, 0.1)$ |
| $1 - P(17, 18, 19, 20)$ | M1 | Allow $1-P(18,19,20)$ or $1-P(16,17,18,19,20)$ |
| $1-(^{20}C_{17} \times 0.1^3 \times 0.9^{17} + ^{20}C_{18} \times 0.1^2 \times 0.9^{18} + 20 \times 0.1 \times 0.9^{19} + 0.9^{20})$ | M1 | |
| $= 0.133$ (3 sf) | A1 [4] | |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Type II | B1 | or Stephan will conclude standard not fallen |
| $H_0$ will not be rejected | B1 [2] | No contradictions |

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6 Stephan is an athlete who competes in the high jump. In the past, Stephan has succeeded in $90 \%$ of jumps at a certain height. He suspects that his standard has recently fallen and he decides to carry out a hypothesis test to find out whether he is right. If he succeeds in fewer than 17 of his next 20 jumps at this height, he will conclude that his standard has fallen.\\
(i) Find the probability of a Type I error.\\
(ii) In fact Stephan succeeds in 18 of his next 20 jumps. Which of the errors, Type I or Type II, is possible? Explain your answer.

\hfill \mbox{\textit{CAIE S2 2014 Q6 [6]}}