| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Standard CI with summary statistics |
| Difficulty | Challenging +1.2 This is a two-sample t-test confidence interval problem requiring students to extract information from a given CI, calculate pooled variance, and construct a new CI for the difference of means. While it involves multiple steps and careful extraction of summary statistics from the first CI, the procedure is standard S4 material with no novel insight required—just systematic application of the pooled two-sample method. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \dfrac{492 + 507}{2} = 499.5\) | M1 | |
| A1cao | ||
| \(2.093 \cdot \dfrac{s}{\sqrt{20}} = 7.5\) | M1, B1 | B1 for 2.093 |
| \(s = 16.02533\ldots\) (\(s^2 = 256.81\ldots6\)) | A1 | awrt 16.0 for \(s\) or 257 for \(s^2\) |
| \(s_p^2 = \dfrac{19 \times 16.025^2 + 9 \times 280}{28} = 264.26\ldots\) | M1A1ft | M1: \(\dfrac{(n_1-1)(s \text{ or } s^2)+(n_2-1)(s \text{ or } s^2)}{n_1+n_2-2}\); A1 ft their \(s^2\) |
| \(t_{28(0.05)} = 1.701\) | B1 | awrt 1.701 |
| \(90\%\ CI = (499.5 - 480) \pm 1.701 \times \sqrt{264.26} \times \sqrt{\dfrac{1}{20}+\dfrac{1}{10}}\) | M1A1ft | M1: \((\bar{x}-480) \pm t\text{-value} \times \sqrt{s_p^2} \times \sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}\); A1 ft their \(s_p^2\) and \(\bar{x}\) |
| \(= (8.8,\ 30.2)\) | A1cao | awrt 8.8 and awrt 30.2 |
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \dfrac{492 + 507}{2} = 499.5$ | M1 | |
| | A1cao | |
| $2.093 \cdot \dfrac{s}{\sqrt{20}} = 7.5$ | M1, B1 | B1 for 2.093 |
| $s = 16.02533\ldots$ ($s^2 = 256.81\ldots6$) | A1 | awrt 16.0 for $s$ or 257 for $s^2$ |
| $s_p^2 = \dfrac{19 \times 16.025^2 + 9 \times 280}{28} = 264.26\ldots$ | M1A1ft | M1: $\dfrac{(n_1-1)(s \text{ or } s^2)+(n_2-1)(s \text{ or } s^2)}{n_1+n_2-2}$; A1 ft their $s^2$ |
| $t_{28(0.05)} = 1.701$ | B1 | awrt 1.701 |
| $90\%\ CI = (499.5 - 480) \pm 1.701 \times \sqrt{264.26} \times \sqrt{\dfrac{1}{20}+\dfrac{1}{10}}$ | M1A1ft | M1: $(\bar{x}-480) \pm t\text{-value} \times \sqrt{s_p^2} \times \sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}$; A1 ft their $s_p^2$ and $\bar{x}$ |
| $= (8.8,\ 30.2)$ | A1cao | awrt 8.8 and awrt 30.2 |
---\begin{enumerate}
\item Jamland and Goodjam are two suppliers of jars of jam. The weights of the jars of jam produced by each supplier can be assumed to be normally distributed with unknown, but equal, variances. A random sample of 20 jars of jam is taken from those supplied by Jamland.
\end{enumerate}
Based on this sample, the 95\% confidence interval for the mean weight of a jar of Jamland jam, in grams, is\\[0pt]
[ 492, 507 ]
A random sample of 10 jars of jam is selected from those supplied by Goodjam. The weight of each jar of Goodjam jam, $y$ grams, is recorded. The results are summarised as follows
$$\bar { y } = 480 \quad s _ { y } ^ { 2 } = 280$$
Find a 90\% confidence interval for the value by which the mean weight of a jar of jam supplied by Jamland exceeds the mean weight of a jar of jam supplied by Goodjam.\\
\hfill \mbox{\textit{Edexcel S4 2017 Q5 [11]}}