Edexcel S2 2016 October — Question 1 10 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2016
SessionOctober
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeOne-tailed hypothesis test (upper tail, H₁: p > p₀)
DifficultyModerate -0.3 This is a standard S2 hypothesis testing question with routine procedures: part (a) requires a straightforward one-tailed binomial test with clear hypotheses and p-value calculation from tables, while part (b) involves a standard normal approximation with continuity correction. Both parts follow textbook methods with no novel problem-solving required, making it slightly easier than average for A-level.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

  1. A mobile phone company claims that each year \(5 \%\) of its customers have their mobile phone stolen. An insurance company claims this percentage is higher. A random sample of 30 of the mobile phone company's customers is taken and 4 of them have had their mobile phone stolen during the last year.
    1. Test the insurance company's claim at the \(10 \%\) level of significance. State your hypotheses clearly.
    A new random sample of 90 customers is taken. A test is carried out using these 90 customers, to see if the percentage of customers who have had a mobile phone stolen in the last year is more than 5\%
  2. Using a suitable approximation and a \(10 \%\) level of significance, find the critical region for this test.

Question 1:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(H_0: p = 0.05\), \(H_1: p > 0.05\)B1 Both hypotheses correct (must use \(p\) or \(\pi\))
\(X \sim B(30, 0.05)\)B1 May be implied
\(P(X \geq 4) = 1 - P(X \leq 3)\) or \(P(X \leq 3) = 0.9392\)M1 For writing or using \(1 - P(X \leq 3)\) or giving \(P(X \leq 3) = 0.9392\) for CR method
\(= 1 - 0.9392 = 0.0608\) or \(CR: X \geq 4\)A1 For \(0.0608\) or \(CR: X \geq 4\)
Reject \(H_0\) or Significant or 4 lies in the Critical RegiondM1 Dependent on 1st M1; correct statement. Do not allow contradictory statements e.g. 'significant, accept \(H_0\)'
The claim is supported/true or there is evidence that the percentage/\%/number/rate of customers who have had a mobile phone stolen is more than 5%A1cso Correct contextual conclusion and no errors seen. Two tailed test could score B0B1M1A1dM1 (if comparing p-value with 0.05 and accepting \(H_0\)) A0
(6 marks)
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(Y \sim B(90, 0.05) \rightarrow Po(4.5)\)M1 A1 M1 for writing or using a Poisson approximation; A1 for 4.5
\(P(Y \leq 6) = 0.8311\), \(P(Y \geq 7) = 0.1689 > 0.10\)dM1 Dependent on 1st M1 for using \(Po(4.5)\) to find a relevant probability for determining the CR: \(P(Y \leq 6) = 0.8311\), \(P(Y \geq 7) = 0.1689\), \(P(Y \leq 7) = 0.9134\), \(P(Y \geq 8) = 0.0866\)
\(P(Y \leq 7) = 0.9134\), \(P(Y \geq 8) = 0.0866 < 0.10\)
\(CR: Y \geq 8\)A1 Allow any letter for \(Y\)
(4 marks)
[10 marks total]
# Question 1:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.05$, $H_1: p > 0.05$ | **B1** | Both hypotheses correct (must use $p$ or $\pi$) |
| $X \sim B(30, 0.05)$ | **B1** | May be implied |
| $P(X \geq 4) = 1 - P(X \leq 3)$ or $P(X \leq 3) = 0.9392$ | **M1** | For writing or using $1 - P(X \leq 3)$ or giving $P(X \leq 3) = 0.9392$ for CR method |
| $= 1 - 0.9392 = 0.0608$ or $CR: X \geq 4$ | **A1** | For $0.0608$ or $CR: X \geq 4$ |
| Reject $H_0$ or Significant or 4 lies in the Critical Region | **dM1** | Dependent on 1st M1; correct statement. Do not allow contradictory statements e.g. 'significant, accept $H_0$' |
| The claim is supported/true **or** there is evidence that the percentage/\%/number/rate of customers who have had a mobile phone stolen is more than 5% | **A1cso** | Correct contextual conclusion and no errors seen. Two tailed test could score B0B1M1A1dM1 (if comparing p-value with 0.05 and accepting $H_0$) A0 |

**(6 marks)**

---

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y \sim B(90, 0.05) \rightarrow Po(4.5)$ | **M1 A1** | M1 for writing or using a Poisson approximation; A1 for 4.5 |
| $P(Y \leq 6) = 0.8311$, $P(Y \geq 7) = 0.1689 > 0.10$ | **dM1** | Dependent on 1st M1 for using $Po(4.5)$ to find a relevant probability for determining the CR: $P(Y \leq 6) = 0.8311$, $P(Y \geq 7) = 0.1689$, $P(Y \leq 7) = 0.9134$, $P(Y \geq 8) = 0.0866$ |
| $P(Y \leq 7) = 0.9134$, $P(Y \geq 8) = 0.0866 < 0.10$ | | |
| $CR: Y \geq 8$ | **A1** | Allow any letter for $Y$ |

**(4 marks)**

**[10 marks total]**
\begin{enumerate}
  \item A mobile phone company claims that each year $5 \%$ of its customers have their mobile phone stolen. An insurance company claims this percentage is higher. A random sample of 30 of the mobile phone company's customers is taken and 4 of them have had their mobile phone stolen during the last year.\\
(a) Test the insurance company's claim at the $10 \%$ level of significance. State your hypotheses clearly.
\end{enumerate}

A new random sample of 90 customers is taken. A test is carried out using these 90 customers, to see if the percentage of customers who have had a mobile phone stolen in the last year is more than 5\%\\
(b) Using a suitable approximation and a $10 \%$ level of significance, find the critical region for this test.\\

\hfill \mbox{\textit{Edexcel S2 2016 Q1 [10]}}