CAIE S2 2022 November — Question 1 7 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2022
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeFind alpha from CI width
DifficultyStandard +0.3 This is a straightforward confidence interval question requiring standard formulas. Part (a) involves direct calculation of sample mean and unbiased variance from summary statistics (routine bookwork). Part (b) requires working backwards from a confidence interval boundary to find the confidence level, which adds a small problem-solving element but uses standard normal tables and algebraic manipulation. The question is slightly easier than average as it's purely procedural with no conceptual challenges or novel insights required.
Spec5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution

Each of a random sample of 80 adults gave an estimate, \(h\) metres, of the height of a particular building. The results were summarised as follows. $$n = 80 \quad \sum h = 2048 \quad \sum h^2 = 52760$$
  1. Calculate unbiased estimates of the population mean and variance. [3]
  2. Using this sample, the upper boundary of an \(\alpha\%\) confidence interval for the population mean is 26.0. Find the value of \(\alpha\). [4]

Question 1:

AnswerMarks
1(a)2048 128
Est μ = 25.6 or or
AnswerMarks
80 5B1
80  52760 2048 2 1  20482 
Est σ2 =  −   or 52760− 
 
AnswerMarks Guidance
79  80  80   79 80 M1 Substitution into a correct formula.
Biased 4.14 scores M0.
1656
= 4.19 (3 sf) or
AnswerMarks
395A1
3

AnswerMarks
1(b)'4.19'
‘25.6’ + z = 26.0
AnswerMarks Guidance
80M1 Use of correct equation with their values.
z = 1.748 or 1.747A1 Accept 3sf.
FT Biased z = 1.758.
AnswerMarks Guidance
(ɸ(‘1.748’) = 0.960) ‘0.960’ – (1 – ‘0.960’)M1 Correct area using their values.
α = 92.0 or 91.9A1 Allow 92 .
FT Biased 92.1.
A final answer of 0.92 or 0.919 scores A0.
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
--- 1(a) ---
1(a) | 2048 128
Est μ = 25.6 or or
80 5 | B1
80  52760 2048 2 1  20482 
Est σ2 =  −   or 52760− 
 
79  80  80   79 80  | M1 | Substitution into a correct formula.
Biased 4.14 scores M0.
1656
= 4.19 (3 sf) or
395 | A1
3
--- 1(b) ---
1(b) | '4.19'
‘25.6’ + z = 26.0
80 | M1 | Use of correct equation with their values.
z = 1.748 or 1.747 | A1 | Accept 3sf.
FT Biased z = 1.758.
(ɸ(‘1.748’) = 0.960) ‘0.960’ – (1 – ‘0.960’) | M1 | Correct area using their values.
α = 92.0 or 91.9 | A1 | Allow 92 .
FT Biased 92.1.
A final answer of 0.92 or 0.919 scores A0.
4
Question | Answer | Marks | Guidance
Each of a random sample of 80 adults gave an estimate, $h$ metres, of the height of a particular building. The results were summarised as follows.

$$n = 80 \quad \sum h = 2048 \quad \sum h^2 = 52760$$

\begin{enumerate}[label=(\alph*)]
\item Calculate unbiased estimates of the population mean and variance. [3]

\item Using this sample, the upper boundary of an $\alpha\%$ confidence interval for the population mean is 26.0.

Find the value of $\alpha$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2022 Q1 [7]}}