CAIE S2 2007 June — Question 6 9 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2007
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeConfidence intervals for population mean
DifficultyStandard +0.3 This is a straightforward confidence interval question requiring standard formulas for unbiased estimates, constructing a confidence interval using t-distribution, and solving for sample size. All steps are routine applications of S2 techniques with no novel problem-solving required, making it slightly easier than average.
Spec2.02g Calculate mean and standard deviation5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution

6 The daily takings, \(\\) x\(, for a shop were noted on 30 randomly chosen days. The takings are summarised by \)\Sigma x = 31500 , \Sigma x ^ { 2 } = 33141816$.
  1. Calculate unbiased estimates of the population mean and variance of the shop's daily takings.
  2. Calculate a \(98 \%\) confidence interval for the mean daily takings. The mean daily takings for a random sample of \(n\) days is found.
  3. Estimate the value of \(n\) for which it is approximately \(95 \%\) certain that the sample mean does not differ from the population mean by more than \(\\) 6$.

(i) \(\bar{x} = 1050\)
AnswerMarks Guidance
\(s^2 = \frac{1}{29}\left(33141816 - \frac{31500^2}{30}\right) = 2304\)B1, M1, A1 Correct mean. Correct formula with 29 in denom. Correct answer.
(ii) \(1050 \pm 2.326 \times \frac{48}{\sqrt{30}} = (1030, 1070)\)M1, B1, A1ℜ Correct shape with \(\sqrt{30}\) in denom. 2.326 seen. or equivalent, ft on their mean and variance.
(iii) \(1.96 \times \frac{48}{\sqrt{n}} = 6\)
AnswerMarks Guidance
\(n = 246\)M1*, M1dep, A1 Correct form of LHS of equation/inequality involving 1.96, 48, \(\sqrt{n}\). Equated to 6 and attempt to solve (accept factor of 2 errors). Correct answer. SR If M0 M0 scored but only error is in z value score M1.
**(i)** $\bar{x} = 1050$

$s^2 = \frac{1}{29}\left(33141816 - \frac{31500^2}{30}\right) = 2304$ | B1, M1, A1 | Correct mean. Correct formula with 29 in denom. Correct answer.

**(ii)** $1050 \pm 2.326 \times \frac{48}{\sqrt{30}} = (1030, 1070)$ | M1, B1, A1ℜ | Correct shape with $\sqrt{30}$ in denom. 2.326 seen. or equivalent, ft on their mean and variance.

**(iii)** $1.96 \times \frac{48}{\sqrt{n}} = 6$

$n = 246$ | M1*, M1dep, A1 | Correct form of LHS of equation/inequality involving 1.96, 48, $\sqrt{n}$. Equated to 6 and attempt to solve (accept factor of 2 errors). Correct answer. SR If M0 M0 scored but only error is in z value score M1.

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6 The daily takings, $\$ x$, for a shop were noted on 30 randomly chosen days. The takings are summarised by $\Sigma x = 31500 , \Sigma x ^ { 2 } = 33141816$.\\
(i) Calculate unbiased estimates of the population mean and variance of the shop's daily takings.\\
(ii) Calculate a $98 \%$ confidence interval for the mean daily takings.

The mean daily takings for a random sample of $n$ days is found.\\
(iii) Estimate the value of $n$ for which it is approximately $95 \%$ certain that the sample mean does not differ from the population mean by more than $\$ 6$.

\hfill \mbox{\textit{CAIE S2 2007 Q6 [9]}}