| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Standard CI with summary statistics |
| Difficulty | Standard +0.3 This is a standard two-sample t-test confidence interval question with summary statistics provided. Part (a) is routine variance calculation, part (b) follows a textbook procedure (pooled variance, standard error, t-critical value), and part (c) requires basic interpretation. While it has multiple steps and 12 marks, it requires no novel insight—just careful application of memorized formulas from S4 syllabus. Slightly easier than average due to its procedural nature. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| \(\Sigma x\) | \(\Sigma x ^ { 2 }\) | |||
| Stored outside | 20 | 2340 | 274050 | ||
| Stored inside | 37 | 4884 | 645282 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(S_I = \frac{1}{19}\left(274050 - \frac{(2340)^2}{20}\right) = 14.2\) | M1 | |
| \(S_O = \frac{1}{36}\left(645282 - \frac{(4884)^2}{37}\right) = 165\) | A1 | (2) awrt, y=tr |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(s_p = \sqrt{\frac{19\times14.2 + 36\times165}{55}} = \sqrt{15.705} = 3.963\ldots\) | M1, A1 | |
| Mean outside \(= \frac{2340}{20} = 117\), Mean inside \(= 132\) | B1, B1 | |
| Confidence limits \(= (132-117) \pm 2.0\overline{04} \times 3.93\ldots\sqrt{\frac{1}{20}+\frac{1}{37}}\) | M1, A1 | |
| \(= (12.8, 17.2)\) | A1, A1 | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(0\) lies outside confidence interval. The means are different. | B1, B1 | (2) |
| TOTAL | 12 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $S_I = \frac{1}{19}\left(274050 - \frac{(2340)^2}{20}\right) = 14.2$ | M1 | |
| $S_O = \frac{1}{36}\left(645282 - \frac{(4884)^2}{37}\right) = 165$ | A1 | (2) awrt, y=tr |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $s_p = \sqrt{\frac{19\times14.2 + 36\times165}{55}} = \sqrt{15.705} = 3.963\ldots$ | M1, A1 | |
| Mean outside $= \frac{2340}{20} = 117$, Mean inside $= 132$ | B1, B1 | |
| Confidence limits $= (132-117) \pm 2.0\overline{04} \times 3.93\ldots\sqrt{\frac{1}{20}+\frac{1}{37}}$ | M1, A1 | |
| $= (12.8, 17.2)$ | A1, A1 | (8) |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $0$ lies outside confidence interval. The means are different. | B1, B1 | (2) |
| **TOTAL** | **12** | |
---6. A tree is cut down and sawn into pieces. Half of the pieces are stored outside and half of the pieces are stored inside. After a year, a random sample of pieces is taken from each location and the hardness is measured. The hardness $x$ units are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& \begin{tabular}{ c }
Number of \\
pieces sampled \\
\end{tabular} & $\Sigma x$ & $\Sigma x ^ { 2 }$ \\
\hline
Stored outside & 20 & 2340 & 274050 \\
\hline
Stored inside & 37 & 4884 & 645282 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that unbiased estimates for the variance of the values of hardness for wood stored outside and for the wood stored inside are 14.2 and 16.5 , to 1 decimal place, respectively.\\
(2)
The hardness of wood stored outside and the hardness of wood stored inside can be assumed to be normally distributed with equal variances.
\item Calculate $95 \%$ confidence limits for the difference in mean hardness between the wood that was stored outside and the wood that was stored inside.\\
(8)
\item Using your answer to part (b), comment on the means of the hardness of wood stored outside and inside. Give a reason for your answer.\\
(2)\\
(Total 12 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2006 Q6 [12]}}