CAIE S2 2013 November — Question 1 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2013
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeConfidence intervals for population mean
DifficultyModerate -0.8 This is a straightforward application of standard formulas for unbiased estimates and confidence intervals. Students need only recall the formulas for sample mean, unbiased variance (dividing by n-1), and the 95% CI using the normal distribution. All calculations are routine with no problem-solving or conceptual challenges beyond basic recall and substitution.
Spec5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution
1 A random sample of 80 values of a variable \(X\) is taken and these values are summarised below. $$n = 80 \quad \Sigma x = 150.2 \quad \Sigma x ^ { 2 } = 820.24$$ Calculate unbiased estimates of the population mean and variance of \(X\) and hence find a \(95 \%\) confidence interval for the population mean of \(X\).
Question 1:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\text{Est}(\mu) = 1.8775\) or \(1.88\) (3 sf)B1 Accept \(751/400\) (not \(150.2/80\))
\(\text{Est}(\sigma^2) = \frac{80}{79}\left(\frac{820.24}{80} - 1.8775^2\right)\)M1 Correct substitution in correct formula \(\frac{1}{79}(820.24 - 150.2^2/80)\)
\(= 6.81316\) or \(6.81\) (3 sf)A1
\(z = 1.96\)B1 Seen
\(1.8775 \pm z \times \sqrt{\frac{6.81316}{80}}\)M1
\(= 1.31\) to \(2.45\) (3 sf)A1 [6] Must be an interval. NB use of biased var can still score A1. 
## Question 1:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{Est}(\mu) = 1.8775$ or $1.88$ (3 sf) | B1 | Accept $751/400$ (not $150.2/80$) |
| $\text{Est}(\sigma^2) = \frac{80}{79}\left(\frac{820.24}{80} - 1.8775^2\right)$ | M1 | Correct substitution in correct formula $\frac{1}{79}(820.24 - 150.2^2/80)$ |
| $= 6.81316$ or $6.81$ (3 sf) | A1 | |
| $z = 1.96$ | B1 | Seen |
| $1.8775 \pm z \times \sqrt{\frac{6.81316}{80}}$ | M1 | |
| $= 1.31$ to $2.45$ (3 sf) | A1 | **[6]** Must be an interval. NB use of biased var can still score A1. |

---
1 A random sample of 80 values of a variable $X$ is taken and these values are summarised below.

$$n = 80 \quad \Sigma x = 150.2 \quad \Sigma x ^ { 2 } = 820.24$$

Calculate unbiased estimates of the population mean and variance of $X$ and hence find a $95 \%$ confidence interval for the population mean of $X$.

\hfill \mbox{\textit{CAIE S2 2013 Q1 [6]}}
This paper (6 questions)
View full paper
Q1 6 Q2 8 Q3 8 Q4 9 Q5 9 Q6 10