CAIE S2 2013 June — Question 3 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2013
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeCI from raw data list
DifficultyModerate -0.8 This is a straightforward confidence interval question requiring only standard formula application (calculating sample mean, then using z-value for 92% CI with known σ). Parts (ii) and (iii) test basic statistical definitions rather than problem-solving. The calculation is routine with no conceptual challenges, making it easier than average A-level material.
Spec2.01a Population and sample: terminology5.05d Confidence intervals: using normal distribution

3 Each of a random sample of 15 students was asked how long they spent revising for an exam. The results, in minutes, were as follows. $$\begin{array} { l l l l l l l l l l l l l l l } 50 & 70 & 80 & 60 & 65 & 110 & 10 & 70 & 75 & 60 & 65 & 45 & 50 & 70 & 50 \end{array}$$ Assume that the times for all students are normally distributed with mean \(\mu\) minutes and standard deviation 12 minutes.
  1. Calculate a \(92 \%\) confidence interval for \(\mu\).
  2. Explain what is meant by a \(92 \%\) confidence interval for \(\mu\).
  3. Explain what is meant by saying that a sample is 'random'.

Question 3:
AnswerMarks Guidance
(i)\(\bar{x} = 930/15 = 62\); \(z = 1.751\); \(62 \pm z \times \frac{12}{\sqrt{15}}\); \(= 56.6\) to \(67.4\) (3 sf) B1, B1, M1, A1 [4]
(ii)92% of such intervals will contain \(\mu\) B1 [1]
(iii)Each possible sample of this size is equally likely B1 [1]
## Question 3:

**(i)** | $\bar{x} = 930/15 = 62$; $z = 1.751$; $62 \pm z \times \frac{12}{\sqrt{15}}$; $= 56.6$ to $67.4$ (3 sf) | B1, B1, M1, A1 [4] | Any $z$; must be an interval |

**(ii)** | 92% of such intervals will contain $\mu$ | B1 [1] | Accept $P(\text{this interval contains}\ \mu) = 0.92$ |

**(iii)** | Each possible sample of this size is equally likely | B1 [1] | Each member of pop equally likely to be chosen |

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3 Each of a random sample of 15 students was asked how long they spent revising for an exam. The results, in minutes, were as follows.

$$\begin{array} { l l l l l l l l l l l l l l l } 
50 & 70 & 80 & 60 & 65 & 110 & 10 & 70 & 75 & 60 & 65 & 45 & 50 & 70 & 50
\end{array}$$

Assume that the times for all students are normally distributed with mean $\mu$ minutes and standard deviation 12 minutes.\\
(i) Calculate a $92 \%$ confidence interval for $\mu$.\\
(ii) Explain what is meant by a $92 \%$ confidence interval for $\mu$.\\
(iii) Explain what is meant by saying that a sample is 'random'.

\hfill \mbox{\textit{CAIE S2 2013 Q3 [6]}}