Pre-U Pre-U 9795/2 Specimen — Question 8 5 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
SessionSpecimen
Marks5
TopicConfidence intervals
TypeComment on claim using CI
DifficultyStandard +0.3 This is a standard two-sample t-test confidence interval problem with pooled variance. While it requires multiple computational steps (calculating means, pooled variance, standard error, and CI), the procedure is routine for Further Maths statistics. The interpretation in part (ii) is straightforward. Slightly above average difficulty due to the pooled variance calculation and being a Further Maths topic, but still a textbook exercise.
Spec5.05d Confidence intervals: using normal distribution

8 Specimens of rain were collected at random from the north and south sides of an island and analysed for sulphur content. The results (in suitable units) are given below.
North side0.120.610.790.160.08
South side1.120.270.060.120.240.78
Assume that the sulphur contents have normal distributions with population means \(\mu _ { N }\) and \(\mu _ { S }\) and a common, but unknown, variance.
  1. Calculate a symmetric \(95 \%\) confidence interval for the difference in population mean sulphur contents of the rain on the north and south sides of the island, \(\mu _ { S } - \mu _ { N }\).
  2. Comment on a claim that the mean sulphur content is the same on both sides of the island.

Question 8
(i) North: \(\bar{x}_N = 0.352\), \(s_N = 0.290\,888\,29\), \(n=5\) B1
South: \(\bar{x}_S = 0.431\,666\,66\), \(s_S = 0.385\,850\,43\), \(n=6\) B1
\[\widehat{\sigma^2} = \frac{5 \times 0.290\,888\,29^2 + 6 \times 0.385\,850\,43^2}{5+6-2} = 0.3824^2\] M1A1
\(\nu = 9\), \(\ t_9(0.975) = 2.262\) B1
\(\bar{x}_S - \bar{x}_N = 0.079\,666\,66\) B1
95% CI: \(\ 0.079\,666\,66 \pm 2.262 \times 0.3824\sqrt{\frac{1}{5}+\frac{1}{6}} = (-0.444,\ 0.603)\) M1A1 [8 marks]
(ii) Zero falls within the confidence interval, so the claim that sulphur content is the same on both sides of the island has some credence. B1 [1 mark]
**Question 8**

**(i)** North: $\bar{x}_N = 0.352$, $s_N = 0.290\,888\,29$, $n=5$ B1

South: $\bar{x}_S = 0.431\,666\,66$, $s_S = 0.385\,850\,43$, $n=6$ B1

$$\widehat{\sigma^2} = \frac{5 \times 0.290\,888\,29^2 + 6 \times 0.385\,850\,43^2}{5+6-2} = 0.3824^2$$ M1A1

$\nu = 9$, $\ t_9(0.975) = 2.262$ B1

$\bar{x}_S - \bar{x}_N = 0.079\,666\,66$ B1

95% CI: $\ 0.079\,666\,66 \pm 2.262 \times 0.3824\sqrt{\frac{1}{5}+\frac{1}{6}} = (-0.444,\ 0.603)$ M1A1 **[8 marks]**

**(ii)** Zero falls within the confidence interval, so the claim that sulphur content is the same on both sides of the island has some credence. B1 **[1 mark]**
8 Specimens of rain were collected at random from the north and south sides of an island and analysed for sulphur content. The results (in suitable units) are given below.

\begin{center}
\begin{tabular}{ l l l l l l l }
North side & 0.12 & 0.61 & 0.79 & 0.16 & 0.08 &  \\
South side & 1.12 & 0.27 & 0.06 & 0.12 & 0.24 & 0.78 \\
\end{tabular}
\end{center}

Assume that the sulphur contents have normal distributions with population means $\mu _ { N }$ and $\mu _ { S }$ and a common, but unknown, variance.\\
(i) Calculate a symmetric $95 \%$ confidence interval for the difference in population mean sulphur contents of the rain on the north and south sides of the island, $\mu _ { S } - \mu _ { N }$.\\
(ii) Comment on a claim that the mean sulphur content is the same on both sides of the island.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2  Q8 [5]}}