| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Challenging +1.2 This is a multi-part S4 question requiring t-distribution confidence intervals for mean and variance, plus an application involving normal probability. While it tests multiple techniques (calculating sample statistics, using t-tables, chi-squared for variance CI, and normal distribution), each step follows standard procedures with no novel insight required. The final part (b) is slightly non-routine but still algorithmic once you recognize to use the upper confidence limit for the mean. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = \frac{291}{15} = 19.4\) | M1 | \(\frac{291}{15}\) |
| \(s = \sqrt{\frac{5968 - 15\bar{x}^2}{14}} = 4.800\) | M1 | \(\sqrt{\frac{5968 - 15\bar{x}^2}{14}}\) |
| \(t_{14} = 2.145\) | B1 | |
| \(95\%\ \text{CI} = 19.4 \pm 2.145 \times \frac{4.800}{\sqrt{15}}\) | M1 A1ft | A1ft for \(19.4 \pm 2.145 \times \frac{\text{"their s"}}{\sqrt{15}}\) |
| \(= (16.7,\ 22.1)\) | A1 A1 | A1 awrt 16.7, A1 awrt 22.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{14 \times 4.800^2}{26.119} < \sigma^2 < \frac{14 \times 4.800^2}{5.629}\) | M1 B1 B1 | M1 for \(\frac{14 \times s^2}{\chi^2}\); B1 for 26.119; B1 for 5.629 |
| \((12.4,\ 57.3)\) accept 12.3 | A1 A1 | A1 awrt 12.4/12.3, A1 awrt 57.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Require \(P(X>23) = P\!\left(Z > \frac{23-\mu}{\sigma}\right)\) to be as large as possible, so \(\frac{23-\mu}{\sigma}\) to be as small as possible; both imply highest \(\sigma\) and \(\mu\) | M1 M1 | M1 use of highest mean and sigma; M1 standardising using values from intervals |
| \(\frac{23 - 22.1}{\sqrt{57.3}} = 0.124\) | ||
| \(P(Z > 0.124) = 1 - 0.5478\) | M1 | M1 finding \(1 - P(z > \text{their value})\) |
| \(= 0.4522\) | A1 | A1 awrt 0.45 |
# Question 4:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = \frac{291}{15} = 19.4$ | M1 | $\frac{291}{15}$ |
| $s = \sqrt{\frac{5968 - 15\bar{x}^2}{14}} = 4.800$ | M1 | $\sqrt{\frac{5968 - 15\bar{x}^2}{14}}$ |
| $t_{14} = 2.145$ | B1 | |
| $95\%\ \text{CI} = 19.4 \pm 2.145 \times \frac{4.800}{\sqrt{15}}$ | M1 A1ft | A1ft for $19.4 \pm 2.145 \times \frac{\text{"their s"}}{\sqrt{15}}$ |
| $= (16.7,\ 22.1)$ | A1 A1 | A1 awrt 16.7, A1 awrt 22.1 |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{14 \times 4.800^2}{26.119} < \sigma^2 < \frac{14 \times 4.800^2}{5.629}$ | M1 B1 B1 | M1 for $\frac{14 \times s^2}{\chi^2}$; B1 for 26.119; B1 for 5.629 |
| $(12.4,\ 57.3)$ accept 12.3 | A1 A1 | A1 awrt 12.4/12.3, A1 awrt 57.3 |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Require $P(X>23) = P\!\left(Z > \frac{23-\mu}{\sigma}\right)$ to be as large as possible, so $\frac{23-\mu}{\sigma}$ to be as small as possible; both imply highest $\sigma$ and $\mu$ | M1 M1 | M1 use of highest mean and sigma; M1 standardising using values from intervals |
| $\frac{23 - 22.1}{\sqrt{57.3}} = 0.124$ | | |
| $P(Z > 0.124) = 1 - 0.5478$ | M1 | M1 finding $1 - P(z > \text{their value})$ |
| $= 0.4522$ | A1 | A1 awrt 0.45 |
---4. A random sample of 15 strawberries is taken from a large field and the weight $x$ grams of each strawberry is recorded. The results are summarised below.
$$\sum x = 291 \quad \sum x ^ { 2 } = 5968$$
Assume that the weights of strawberries are normally distributed. Calculate a 95\% confidence interval for
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item the mean of the weights of the strawberries in the field,
\item the variance of the weights of the strawberries in the field.
Strawberries weighing more than 23 g are considered to be less tasty.
\end{enumerate}\item Use appropriate confidence limits from part (a) to find the highest estimate of the proportion of strawberries that are considered to be less tasty.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2010 Q4 [16]}}