| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Moderate -0.3 This is a straightforward application of confidence intervals with known sample statistics (n=78, large sample so CLT applies). Part (a) is routine calculation using z-critical values. Part (b) requires basic understanding of sampling assumptions but the points are fairly obvious (day-of-week effects, play-specific factors). Slightly easier than average due to large sample size eliminating t-distribution complications and predictable commentary points. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(90\% \Rightarrow z = 1.64\) to \(1.65\) | B1 | AWFW (1.6449) |
| or \(90\% \Rightarrow t = 1.66\) to \(1.67\) | (B1) | AWFW (1.6649); knowledge of \(t\)-distribution not required |
| 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 \(n > 1\) |
| \(184 \pm (1.6449 \text{ or } 1.6649) \times \frac{(32 \text{ or } 32.2)}{(\sqrt{78} \text{ or } \sqrt{77})}\) | A1\(\sqrt{}\) | \(\sqrt{}\) on \(z\) or \(t\) only |
| Hence \(184 \pm (5.94 \text{ to } 6.13)\) or £\(184 \pm\) £\(6\) or (£178, £190) | A1 | AWRT; ignore units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Likely to be valid | B1 | Accept 'valid' or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Different plays have different: programme prices, sales, marketing, etc; theatre or audience sizes, etc; popularity, artists, etc | B1 | |
| so Unlikely to be valid | \(\uparrow\)Dep\(\uparrow\) B1 | Accept 'not valid' or equivalent |
# Question 4:
## Part 4(a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $90\% \Rightarrow z = 1.64$ to $1.65$ | B1 | AWFW (1.6449) |
| **or** $90\% \Rightarrow t = 1.66$ to $1.67$ | (B1) | AWFW (1.6649); knowledge of $t$-distribution **not** required |
| 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 $n > 1$ |
| $184 \pm (1.6449 \text{ or } 1.6649) \times \frac{(32 \text{ or } 32.2)}{(\sqrt{78} \text{ or } \sqrt{77})}$ | A1$\sqrt{}$ | $\sqrt{}$ on $z$ or $t$ only |
| Hence $184 \pm (5.94 \text{ to } 6.13)$ **or** £$184 \pm$ £$6$ **or** (£178, £190) | A1 | AWRT; ignore units |
## Part 4(b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Likely** to be **valid** | B1 | Accept 'valid' or equivalent |
## Part 4(b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Different plays have different: programme prices, sales, marketing, etc; theatre or audience sizes, etc; popularity, artists, etc | B1 | |
| so **Unlikely** to be **valid** | $\uparrow$Dep$\uparrow$ B1 | Accept 'not valid' or equivalent |4 A very popular play has been performed at a London theatre on each of 6 evenings per week for about a year. Over the past 13 weeks ( 78 performances), records have been kept of the proceeds from the sales of programmes at each performance. An analysis of these records has found that the mean was $\pounds 184$ and the standard deviation was $\pounds 32$.
\begin{enumerate}[label=(\alph*)]
\item Assuming that the 78 performances may be considered to be a random sample, construct a $90 \%$ confidence interval for the mean proceeds from the sales of programmes at an evening performance of this play.
\item Comment on the likely validity of the assumption in part (a) when constructing a confidence interval for the mean proceeds from the sales of programmes at an evening performance of:
\begin{enumerate}[label=(\roman*)]
\item this particular play;
\item any play.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2007 Q4 [7]}}