| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Easy -1.2 This is a straightforward application of a standard confidence interval formula with known sample statistics (n=50, mean=234, s=25.1). Part (a) requires only substituting values into the formula x̄ ± 1.96(s/√n), which is routine recall for S1. Part (b) is a simple conceptual question about sampling bias. No problem-solving or novel insight required—purely procedural. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(95\% \Rightarrow z = 1.96\) OR \(95\% \Rightarrow t = 2.0\) to \(2.01\) | B1 (B1) | CAO; AWFW (2.009). Knowledge of \(t\)-distribution not required in this unit |
| CI for \(\mu\) is \(\bar{x} \pm (z \text{ or } t) \times \frac{(s_{n-1} \text{ or } s_n)}{\sqrt{n}}\) | M1 | Used; must have \(\sqrt{n}\) with \(n > 1\) |
| Note: \(25.1 \times \sqrt{\frac{50}{49}} = 25.35483\) | \(25.1 \times \frac{50}{49} = 25.61224\); Max of B1 M1 \(\text{A0}\sqrt{}\) A1 | |
| \(234 \pm (1.96 \text{ or } 2.009) \times \frac{(25.1 \text{ or } 25.3 \text{ to } 25.4)}{(\sqrt{50} \text{ or } \sqrt{49})}\) | \(\text{A1}\sqrt{}\) | \(\sqrt{}\) on \(z\) or \(t\) only |
| Hence \(234 \pm (6.95 \text{ to } 7.30)\) i.e. \(234 \pm 7\) or \((227, 241)\) | A1 | 4; AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Customers are likely to choose large / similar sized potatoes | B1 | 1; OE; accept any sensible alternative |
## Question 3:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $95\% \Rightarrow z = 1.96$ OR $95\% \Rightarrow t = 2.0$ to $2.01$ | B1 (B1) | CAO; AWFW (2.009). Knowledge of $t$-distribution not required in this unit |
| CI for $\mu$ is $\bar{x} \pm (z \text{ or } t) \times \frac{(s_{n-1} \text{ or } s_n)}{\sqrt{n}}$ | M1 | Used; must have $\sqrt{n}$ with $n > 1$ |
| Note: $25.1 \times \sqrt{\frac{50}{49}} = 25.35483$ | | $25.1 \times \frac{50}{49} = 25.61224$; Max of B1 M1 $\text{A0}\sqrt{}$ A1 |
| $234 \pm (1.96 \text{ or } 2.009) \times \frac{(25.1 \text{ or } 25.3 \text{ to } 25.4)}{(\sqrt{50} \text{ or } \sqrt{49})}$ | $\text{A1}\sqrt{}$ | $\sqrt{}$ on $z$ or $t$ only |
| Hence $234 \pm (6.95 \text{ to } 7.30)$ i.e. $234 \pm 7$ or $(227, 241)$ | A1 | 4; AWRT |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Customers are likely to choose large / similar sized potatoes | B1 | 1; OE; accept any sensible alternative |3
\begin{enumerate}[label=(\alph*)]
\item A sample of 50 washed baking potatoes was selected at random from a large batch.\\
The weights of the 50 potatoes were found to have a mean of 234 grams and a standard deviation of 25.1 grams.
Construct a $95 \%$ confidence interval for the mean weight of potatoes in the batch.\\
(4 marks)
\item The batch of potatoes is purchased by a market stallholder. He sells them to his customers by allowing them to choose any 5 potatoes for $\pounds 1$.
Give a reason why such chosen potatoes are unlikely to represent a random sample from the batch.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2007 Q3 [5]}}