| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample t-test (unknown variances) |
| Difficulty | Standard +0.3 This is a standard two-sample t-test procedure with preliminary F-test for equal variances. All steps are routine S4 material: F-test for variances, pooled variance calculation, confidence interval construction, and interpretation. The calculations are straightforward with small sample sizes and clearly presented data, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Sample size | Mean \(\bar { x }\) |
| |||
| Machine \(A\) | 9 | 4.83 | 0.721 | ||
| Machine \(B\) | 10 | 4.85 | 0.572 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \sigma_A^2 = \sigma_B^2\), \(H_1: \sigma_A^2 \neq \sigma_B^2\) | ||
| \(\frac{S_B^2}{S_g^2} = \frac{0.721^2}{0.572^2} = 1.588...\) | M1A1 | Awrt 1.59 |
| \(F_{8,9}\) (5%) c.v. \([= 10\% \text{ 2-tail}] = 3.23\) | B1 | |
| Not significant, can assume variances are equal (accept \(\sigma_A^2 = \sigma_B^2\)) | B1 cao | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_p^2 = \frac{8\times0.721^2 + 9\times0.572^2}{8+9} = 0.41784...\) | M1A1 | \(0\mathord{\cdot}417...\) or Awrt \(0.418\) |
| \(t_{17}\) (2.5%) c.v. \(= 2.110\) | B1 | |
| \(95\%\) CI \(= \bar{x}_B - \bar{x}_A \pm 2.110 \times S_p \times \sqrt{\frac{1}{9}+\frac{1}{10}}\) | M1 | |
| \(= 0.02 \pm 2.110 \times \sqrt{0.417...} \times \sqrt{\frac{1}{9}+\frac{1}{10}}\) | A1ft | |
| \(= (-0.6066...\,,\ 0.6466...)\) | A1, A1 | Awrt \((-0.607, 0.647)\) (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm 0.7\) is outside interval | B1ft | |
| \(\therefore\) manager need not be concerned | B1ft (dep) | Allow if \(0.7\) inside (2) |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \sigma_A^2 = \sigma_B^2$, $H_1: \sigma_A^2 \neq \sigma_B^2$ | | |
| $\frac{S_B^2}{S_g^2} = \frac{0.721^2}{0.572^2} = 1.588...$ | M1A1 | Awrt 1.59 |
| $F_{8,9}$ (5%) c.v. $[= 10\% \text{ 2-tail}] = 3.23$ | B1 | |
| Not significant, can assume variances are equal (accept $\sigma_A^2 = \sigma_B^2$) | B1 cao | (4) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_p^2 = \frac{8\times0.721^2 + 9\times0.572^2}{8+9} = 0.41784...$ | M1A1 | $0\mathord{\cdot}417...$ or Awrt $0.418$ |
| $t_{17}$ (2.5%) c.v. $= 2.110$ | B1 | |
| $95\%$ CI $= \bar{x}_B - \bar{x}_A \pm 2.110 \times S_p \times \sqrt{\frac{1}{9}+\frac{1}{10}}$ | M1 | |
| $= 0.02 \pm 2.110 \times \sqrt{0.417...} \times \sqrt{\frac{1}{9}+\frac{1}{10}}$ | A1ft | |
| $= (-0.6066...\,,\ 0.6466...)$ | A1, A1 | Awrt $(-0.607, 0.647)$ (7) |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm 0.7$ is outside interval | B1ft | |
| $\therefore$ manager need **not** be concerned | B1ft (dep) | Allow if $0.7$ inside (2) |
---4. Two machines $A$ and $B$ produce the same type of component in a factory. The factory manager wishes to know whether the lengths, $x \mathrm {~cm}$, of the components produced by the two machines have the same mean. The manager took a random sample of components from each machine and the results are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& Sample size & Mean $\bar { x }$ & \begin{tabular}{ c }
Standard \\
deviation $s$ \\
\end{tabular} \\
\hline
Machine $A$ & 9 & 4.83 & 0.721 \\
\hline
Machine $B$ & 10 & 4.85 & 0.572 \\
\hline
\end{tabular}
\end{center}
The lengths of components produced by the machines can be assumed to follow normal distributions.
\begin{enumerate}[label=(\alph*)]
\item Use a two tail test to show, at the $10 \%$ significance level, that the variances of the lengths of components produced by each machine can be assumed to be equal.\\
(4)
\item Showing your working clearly, find a $95 \%$ confidence interval for $\mu _ { B } - \mu _ { A }$, where $\mu _ { A }$ and $\mu _ { B }$ are the mean lengths of the populations of components produced by machine $A$ and machine $B$ respectively.
There are serious consequences for the production at the factory if the difference in mean lengths of the components produced by the two machines is more than 0.7 cm .
\item State, giving your reason, whether or not the factory manager should be concerned.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2006 Q4 [13]}}