| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring recall of t-distribution critical values and understanding of CI interpretation. Part (a) is direct lookup, part (b) tests conceptual understanding (common misconception), and part (c) involves simple algebraic manipulation of the CI width formula. All parts are standard textbook exercises with no novel problem-solving required. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1.83\) | B1 | States 1.83 AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| David's statement may not be true | R1 | Comments that David's statement is not or may not be true |
| Interval is centred on sample mean, not population mean; or interval is for population mean length, not lizard lengths | R1 | Mentions that interval is based on mean from sample; or mentions that interval is for population mean length, not for population of lizard lengths |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{b \times s}{\sqrt{n}} = 4.27\) or \(2 \times \frac{b \times s}{\sqrt{n}} = 8.54\) | M1 | Uses correct expression in \(b\), \(s\) and \(n\) in equation relating to width value |
| \(\frac{b \times 4.15(45691)}{\sqrt{10}} = 4.27\), \(b = 3.25\) | A1F | Solves for \(b\); FT incorrect \(s\) |
| From \(t\) tables, \(\beta = 99\) | R1 | Deduces that \(\beta = 99\) (condone 99%) |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1.83$ | B1 | States 1.83 AWRT |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| David's statement may not be true | R1 | Comments that David's statement is not or may not be true |
| Interval is centred on sample mean, not population mean; or interval is for population mean length, not lizard lengths | R1 | Mentions that interval is based on mean from sample; or mentions that interval is for population mean length, not for population of lizard lengths |
## Question 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{b \times s}{\sqrt{n}} = 4.27$ or $2 \times \frac{b \times s}{\sqrt{n}} = 8.54$ | M1 | Uses correct expression in $b$, $s$ and $n$ in equation relating to width value |
| $\frac{b \times 4.15(45691)}{\sqrt{10}} = 4.27$, $b = 3.25$ | A1F | Solves for $b$; FT incorrect $s$ |
| From $t$ tables, $\beta = 99$ | R1 | Deduces that $\beta = 99$ (condone 99%) |
**Total: 6 marks**4 David, a zoologist, is investigating a particular species of monitor lizard. He measures the lengths, in centimetres, of a random sample of this particular species of lizard. His measured lengths are
$$\begin{array} { l l l l l l l l l l }
53.2 & 57.8 & 55.3 & 58.9 & 59.0 & 60.2 & 61.8 & 62.3 & 65.4 & 66.5
\end{array}$$
The lengths may be assumed to be normally distributed.\\
David correctly constructed a 90\% confidence interval for the mean length of lizard using the measured lengths given and the formula $\bar { x } \pm \left( b \times \frac { s } { \sqrt { n } } \right)$
This interval had limits of 57.63 and 62.45, correct to two decimal places.\\
4
\begin{enumerate}[label=(\alph*)]
\item State the value for $b$ used in David's formula.
4
\item David interprets his interval and states,\\
"My confidence interval indicates that exactly 90\% of the population of lizard lengths for this particular species lies between 57.63 cm and $62.45 \mathrm {~cm} ^ { \prime \prime }$.
Do you think David's statement is true? Explain your reasoning.
4
\item David's assistant, Amina, correctly constructs a $\beta \%$ confidence interval from David's random sample of measured lengths.
Amina informs David that the width of her confidence interval is 8.54 .\\
Find the value of $\beta$.\\[0pt]
[3 marks]\\
Turn over for the next question
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics Q4 [6]}}