| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Standard +0.8 This S4 question requires constructing both mean and variance confidence intervals from summary statistics, then applying the normal distribution to estimate a proportion using confidence limits. The multi-step reasoning connecting confidence intervals to proportion estimation, plus the need to identify which limit gives the 'lowest estimate', requires solid understanding beyond routine application of formulas. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{x} = \frac{181}{9} = 20.1\overline{1}\) | B1 | |
| \(s^2 = \frac{1}{8}\left(3913 - \frac{181^2}{9}\right) = \frac{1}{8}(3913 - 3640.1\overline{1}) = 34.111\) | M1 A1 | Correct formula and value |
| \(s = 5.841\) | A1 | |
| CI: \(\bar{x} \pm t_8 \times \frac{s}{\sqrt{9}}\), with \(t_8 = 2.306\) | M1 B1 | Correct \(t\) value for 95%, 8 df |
| \(20.1\overline{1} \pm 2.306 \times \frac{5.841}{3}\) | M1 | |
| \((15.62, 24.60)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using \(\chi^2_8\): \(\chi^2_{0.025} = 17.535\), \(\chi^2_{0.975} = 2.180\) | B1 | |
| \(\left(\frac{8 \times 34.111}{17.535},\ \frac{8 \times 34.111}{2.180}\right)\) | M1 A1 | |
| \((15.56, 125.1)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use lower bound of mean \(= 15.62\) and upper bound of variance \(= 125.1\) | M1 | To maximise proportion needing extra feed |
| \(P(X < 16) = P\left(Z < \frac{16 - 15.62}{\sqrt{125.1}}\right)\) | M1 A1 | |
| \(= P(Z < 0.034) = 0.5135 \approx 0.514\) | A1 |
# Question 3:
**(a)(i) - CI for mean**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} = \frac{181}{9} = 20.1\overline{1}$ | B1 | |
| $s^2 = \frac{1}{8}\left(3913 - \frac{181^2}{9}\right) = \frac{1}{8}(3913 - 3640.1\overline{1}) = 34.111$ | M1 A1 | Correct formula and value |
| $s = 5.841$ | A1 | |
| CI: $\bar{x} \pm t_8 \times \frac{s}{\sqrt{9}}$, with $t_8 = 2.306$ | M1 B1 | Correct $t$ value for 95%, 8 df |
| $20.1\overline{1} \pm 2.306 \times \frac{5.841}{3}$ | M1 | |
| $(15.62, 24.60)$ | A1 | |
**(a)(ii) - CI for variance**
| Answer | Mark | Guidance |
|--------|------|----------|
| Using $\chi^2_8$: $\chi^2_{0.025} = 17.535$, $\chi^2_{0.975} = 2.180$ | B1 | |
| $\left(\frac{8 \times 34.111}{17.535},\ \frac{8 \times 34.111}{2.180}\right)$ | M1 A1 | |
| $(15.56, 125.1)$ | A1 | |
**(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Use lower bound of mean $= 15.62$ and upper bound of variance $= 125.1$ | M1 | To maximise proportion needing extra feed |
| $P(X < 16) = P\left(Z < \frac{16 - 15.62}{\sqrt{125.1}}\right)$ | M1 A1 | |
| $= P(Z < 0.034) = 0.5135 \approx 0.514$ | A1 | |
---3. A large number of chicks were fed a special diet for 10 days. A random sample of 9 of these chicks is taken and the weight gained, $x$ grams, by each chick is recorded. The results are summarised below.
$$\sum x = 181 \quad \sum x ^ { 2 } = 3913$$
You may assume that the weights gained by the chicks are normally distributed.\\
Calculate a 95\% confidence interval for
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item the mean of the weights gained by the chicks,
\item the variance of the weights gained by the chicks.
A chick which gains less than $16 g$ has to be given extra feed.
\end{enumerate}\item Using appropriate confidence limits from part (a), find the lowest estimate of the proportion of chicks that need extra feed.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2014 Q3 [14]}}