| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | F-test then t-test sequential |
| Difficulty | Standard +0.3 This is a standard two-sample hypothesis testing question requiring F-test for variances followed by a t-test for means. While it involves multiple parts and requires understanding of when to use pooled variance, the procedures are routine S4 material with straightforward calculations and no novel problem-solving required. Slightly easier than average due to clear structure and standard methodology. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank5.07c Single-sample tests5.07d Paired vs two-sample: selection |
| \(A\) | 11 | 12 | 12 | 13 | 14 | 15 | 16 | 17 | 17 |
| \(B\) | 8 | 10 | 11 | 13 | 13 | 14 | 14 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(S_A^2 = 5.11\), \(S_B^2 = 5.14\) | B1, B1 | |
| \(H_0: \sigma_A^2 = \sigma_B^2\), \(H_1: \sigma_A^2 \neq \sigma_B^2\) | B1 | |
| Critical value \(F_{6,8} = 3.58\) | B1 | |
| \(\frac{S_B^2}{S_A^2} = 1.00621\ldots\) | M1, A1 | awrt 1.01 |
| No evidence to reject \(H_0\). The variances are equal. | A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(s_p^2 = \frac{8\times5.14 + 6\times5.11}{9+7-2} = 5.1247\) | M1, A1 | awrt 5.12 |
| \(\mu_A = 14.11\ldots\), \(\mu_B = 11.857\ldots\) | B1 | |
| \(H_0: \mu_A = \mu_B\), \(H_1: \mu_A > \mu_B\) | B1 | |
| Critical value \(t_{14}(5\%) = 1.761\) | ||
| \(t = \frac{14.11\ldots - 11.857\ldots}{\sqrt{5.1247\ldots\left(\frac{1}{9}+\frac{1}{7}\right)}} = 1.9757\ldots\) | M1, A1 | awrt 1.98 |
| There is evidence to reject \(H_0\). Mean time taken from school A is greater than school B. | A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Equal variances are a condition for the test in part (b) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Groups not equal ability | B1 | (1) |
| TOTAL | 16 |
## Question 7:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $S_A^2 = 5.11$, $S_B^2 = 5.14$ | B1, B1 | |
| $H_0: \sigma_A^2 = \sigma_B^2$, $H_1: \sigma_A^2 \neq \sigma_B^2$ | B1 | |
| Critical value $F_{6,8} = 3.58$ | B1 | |
| $\frac{S_B^2}{S_A^2} = 1.00621\ldots$ | M1, A1 | awrt 1.01 |
| No evidence to reject $H_0$. The variances are equal. | A1 | (7) |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $s_p^2 = \frac{8\times5.14 + 6\times5.11}{9+7-2} = 5.1247$ | M1, A1 | awrt 5.12 |
| $\mu_A = 14.11\ldots$, $\mu_B = 11.857\ldots$ | B1 | |
| $H_0: \mu_A = \mu_B$, $H_1: \mu_A > \mu_B$ | B1 | |
| Critical value $t_{14}(5\%) = 1.761$ | | |
| $t = \frac{14.11\ldots - 11.857\ldots}{\sqrt{5.1247\ldots\left(\frac{1}{9}+\frac{1}{7}\right)}} = 1.9757\ldots$ | M1, A1 | awrt 1.98 |
| There is evidence to reject $H_0$. Mean time taken from school A is greater than school B. | A1 | (7) |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| Equal variances are a condition for the test in part (b) | B1 | (1) |
### Part (d):
| Answer/Working | Marks | Notes |
|---|---|---|
| Groups not equal ability | B1 | (1) |
| **TOTAL** | **16** | |7. A psychologist gives a test to students from two different schools, $A$ and $B$.
A group of 9 students is randomly selected from school $A$ and given instructions on how to do the test.\\
A group of 7 students is randomly selected from school $B$ and given the test without the instructions.
The table shows the time taken, to the nearest second, to complete the test by the two groups.
\begin{center}
\begin{tabular}{ | c | c c c c c c c c c | }
\hline
$A$ & 11 & 12 & 12 & 13 & 14 & 15 & 16 & 17 & 17 \\
\hline
$B$ & 8 & 10 & 11 & 13 & 13 & 14 & 14 & & \\
\hline
\end{tabular}
\end{center}
Stating your hypotheses clearly,
\begin{enumerate}[label=(\alph*)]
\item test at the $10 \%$ significance level, whether or not the variance of the times taken to complete the test by students from school $A$ is the same as the variance of the times taken to complete the test by students from school $B$. (You may assume that times taken for each school are normally distributed.)
\item test at the $5 \%$ significance level, whether or not the mean time taken to complete the test by students from school $A$ is greater than the mean time taken to complete the test by students from school $B$.
\item Why does the result to part (a) enable you to carry out the test in part (b)?
\item Give one factor that has not been taken into account in your analysis.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2006 Q7 [16]}}