Pre-U Pre-U 9795/2 2019 Specimen — Question 3 4 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2019
SessionSpecimen
Marks4
TopicConfidence intervals
TypeCI from raw data list
DifficultyStandard +0.3 This is a straightforward confidence interval question requiring standard procedures: calculating sample mean/variance, applying z-interval (known variance) and t-interval (unknown variance) formulas with given confidence level, then interpreting results. All steps are routine textbook applications with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec5.05d Confidence intervals: using normal distribution

3 Small amounts of a potentially hazardous chemical are discharged into a river from a nearby industrial site. A random sample of size 6 was taken from the river and the concentration of the chemical present in each item was measured in grams per litre. The results are shown below. $$\begin{array} { l l l l l l } 1.64 & 1.53 & 1.78 & 1.60 & 1.73 & 1.77 \end{array}$$
  1. Assuming that the sample was taken from a normal distribution with known variance 0.01 , construct a \(99 \%\) confidence interval for the mean concentration of the chemical present in the river.
  2. If instead the sample was taken from a normal distribution, but with unknown variance, construct a revised \(99 \%\) confidence interval for the mean concentration of the chemical present in the river.
  3. If the mean concentration of the chemical in the river exceeds 1.8 grams per litre, then remedial action needs to be taken. Comment briefly on the need for remedial action in the light of the results in parts (a) and (b).

Question 3(a)
\(\bar{x} = 1.675\) [B1]
99% confidence limits are \(1.675 \pm 2.576 \times \dfrac{0.1}{\sqrt{6}}\) (ft on wrong mean) [M1A1, M1A1ft]
99% confidence interval is \((1.57, 1.78)\) art [A1]
Total: 4 marks
Question 3(b)
\(s_n = 0.09215\) OR \(s_{n-1} = 0.1009\) [B1]
\(\nu = 5 \Rightarrow t_5(0.99) = 4.032\) [B1]
99% confidence limits are \(1.675 \pm 4.032 \times \dfrac{0.1009}{\sqrt{6}}\)
OR \(1.675 \pm 4.032 \times \dfrac{0.09215}{\sqrt{5}}\) [M1A1]
99% confidence interval is \((1.51, 1.84)\) art [A1]
Total: 5 marks
Question 3(c)
Sensible comment referring to the fact that 1.8 is outside the 1st interval but inside 2nd. (ft on their confidence intervals) [B1, B1ft]
Total: 1 mark
**Question 3(a)**

$\bar{x} = 1.675$ [B1]

99% confidence limits are $1.675 \pm 2.576 \times \dfrac{0.1}{\sqrt{6}}$ (**ft** on wrong mean) [M1A1, M1A1ft]

99% confidence interval is $(1.57, 1.78)$ art [A1]

**Total: 4 marks**

**Question 3(b)**

$s_n = 0.09215$ **OR** $s_{n-1} = 0.1009$ [B1]

$\nu = 5 \Rightarrow t_5(0.99) = 4.032$ [B1]

99% confidence limits are $1.675 \pm 4.032 \times \dfrac{0.1009}{\sqrt{6}}$

**OR** $1.675 \pm 4.032 \times \dfrac{0.09215}{\sqrt{5}}$ [M1A1]

99% confidence interval is $(1.51, 1.84)$ art [A1]

**Total: 5 marks**

**Question 3(c)**

Sensible comment referring to the fact that 1.8 is outside the 1st interval but inside 2nd. (**ft** on their confidence intervals) [B1, B1ft]

**Total: 1 mark**
3 Small amounts of a potentially hazardous chemical are discharged into a river from a nearby industrial site. A random sample of size 6 was taken from the river and the concentration of the chemical present in each item was measured in grams per litre. The results are shown below.

$$\begin{array} { l l l l l l } 
1.64 & 1.53 & 1.78 & 1.60 & 1.73 & 1.77
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Assuming that the sample was taken from a normal distribution with known variance 0.01 , construct a $99 \%$ confidence interval for the mean concentration of the chemical present in the river.
\item If instead the sample was taken from a normal distribution, but with unknown variance, construct a revised $99 \%$ confidence interval for the mean concentration of the chemical present in the river.
\item If the mean concentration of the chemical in the river exceeds 1.8 grams per litre, then remedial action needs to be taken. Comment briefly on the need for remedial action in the light of the results in parts (a) and (b).
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2019 Q3 [4]}}