CAIE S2 2016 June — Question 2 4 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2016
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeFind or state significance level
DifficultyModerate -0.3 This is a straightforward hypothesis testing question requiring understanding of Type I error definition and a single binomial probability calculation P(X < 12 | p = 0.9). Both parts are standard textbook exercises with no problem-solving insight needed, making it slightly easier than average.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion

2 Jacques is a chef. He claims that \(90 \%\) of his customers are satisfied with his cooking. Marie suspects that the true percentage is lower than \(90 \%\). She asks a random sample of 15 of Jacques' customers whether they are satisfied. She then performs a hypothesis test of the null hypothesis \(p = 0.9\) against the alternative hypothesis \(p < 0.9\), where \(p\) is the population proportion of customers who are satisfied. She decides to reject the null hypothesis if fewer than 12 customers are satisfied.
  1. In the context of the question, explain what is meant by a Type I error.
  2. Find the probability of a Type I error in Marie's test.

Question 2:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Conclude less than 90% satisfied when this is not true oeB11 In context
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(1-(0.9^{15} + 15 \times 0.9^{14} \times 0.1 + {}^{15}C_2 \times 0.9^{13} \times 0.1^2 + {}^{15}C_3 \times 0.9^{12} \times 0.1^3)\)M1 Attempt \((1-)P(X=15,14,13,12)\) allow 1 end error
M1Attempt fully correct expression
\(= 0.0556\) (3 sf) or \(0.0555\)A1 [3]
## Question 2:

### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Conclude less than 90% satisfied when this is not true oe | B11 | In context |

### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1-(0.9^{15} + 15 \times 0.9^{14} \times 0.1 + {}^{15}C_2 \times 0.9^{13} \times 0.1^2 + {}^{15}C_3 \times 0.9^{12} \times 0.1^3)$ | M1 | Attempt $(1-)P(X=15,14,13,12)$ allow 1 end error |
| | M1 | Attempt fully correct expression |
| $= 0.0556$ (3 sf) or $0.0555$ | A1 [3] | |
2 Jacques is a chef. He claims that $90 \%$ of his customers are satisfied with his cooking. Marie suspects that the true percentage is lower than $90 \%$. She asks a random sample of 15 of Jacques' customers whether they are satisfied. She then performs a hypothesis test of the null hypothesis $p = 0.9$ against the alternative hypothesis $p < 0.9$, where $p$ is the population proportion of customers who are satisfied. She decides to reject the null hypothesis if fewer than 12 customers are satisfied.\\
(i) In the context of the question, explain what is meant by a Type I error.\\
(ii) Find the probability of a Type I error in Marie's test.

\hfill \mbox{\textit{CAIE S2 2016 Q2 [4]}}