| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | CI from raw data list |
| Difficulty | Moderate -0.3 This is a standard S2 confidence interval question requiring routine application of formulas for known and unknown variance cases, plus basic interpretation. The calculations are straightforward (finding sample mean, applying z and t critical values), and part (b) requires only simple comparison of the intervals to the stated weight. While it involves multiple parts and some interpretation, it demands no novel insight beyond textbook procedures. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = \frac{910.4+908.7+\cdots+907.9}{8} = 909.175\) | B1 | Correct sample mean |
| \(CI: \bar{x} \pm 1.96 \times \frac{2.2}{\sqrt{8}}\) | M1 A1 | M1 for correct \(z\) interval structure, A1 for \(z = 1.96\) and \(\sigma = 2.2\) |
| \((907.6, 910.7)\) | A1 | Correct interval to 4 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s^2\) calculated from sample | M1 | Must use \(t\)-distribution |
| \(CI: \bar{x} \pm t_7 \times \frac{s}{\sqrt{8}}\) using \(t_7 = 2.365\) | A1 A1 | A1 for \(t = 2.365\), A1 for correct \(s\) used |
| Correct interval | A1 | Correct final interval to 4 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Both confidence intervals lie entirely above 907g | B1 | First comment about CI and 907 |
| All 8 bags weigh more than 907g / sample mean well above 907g | B1 | Second comment supporting no concern about underweight bags |
# Question 3:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = \frac{910.4+908.7+\cdots+907.9}{8} = 909.175$ | B1 | Correct sample mean |
| $CI: \bar{x} \pm 1.96 \times \frac{2.2}{\sqrt{8}}$ | M1 A1 | M1 for correct $z$ interval structure, A1 for $z = 1.96$ and $\sigma = 2.2$ |
| $(907.6, 910.7)$ | A1 | Correct interval to 4 s.f. |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s^2$ calculated from sample | M1 | Must use $t$-distribution |
| $CI: \bar{x} \pm t_7 \times \frac{s}{\sqrt{8}}$ using $t_7 = 2.365$ | A1 A1 | A1 for $t = 2.365$, A1 for correct $s$ used |
| Correct interval | A1 | Correct final interval to 4 s.f. |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Both confidence intervals lie entirely above 907g | B1 | First comment about CI and 907 |
| All 8 bags weigh more than 907g / sample mean well above 907g | B1 | Second comment supporting no concern about underweight bags |3 A machine fills bags with frozen peas. Measurements taken over several weeks have shown that the standard deviation of the weights of the filled bags of peas has been 2.2 grams.
Following maintenance on the machine, a quality control inspector selected 8 bags of peas. The weights, in grams, of the bags were
$$\begin{array} { l l l l l l l l }
910.4 & 908.7 & 907.2 & 913.2 & 905.6 & 911.1 & 909.5 & 907.9
\end{array}$$
It may be assumed that the bags constitute a random sample from a normal distribution.
\begin{enumerate}[label=(\alph*)]
\item Giving the limits to four significant figures, calculate a 95\% confidence interval for the mean weight of a bag of frozen peas filled by the machine following the maintenance:
\begin{enumerate}[label=(\roman*)]
\item assuming that the standard deviation of the weights of the bags of peas is known to be 2.2 grams;
\item assuming that the standard deviation of the weights of the bags of peas may no longer be 2.2 grams.
\end{enumerate}\item The weight printed on the bags of peas is 907 grams. One of the inspector's concerns is that bags should not be underweight.
Make two comments about this concern with regard to the data and your calculated confidence intervals.\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2015 Q3 [10]}}