| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2003 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a standard S4 question testing understanding of two-sample vs paired t-tests with all summary statistics provided. Part (a) is routine calculation with given values, part (b) requires computing differences and performing a paired test (standard procedure), and part (c) tests conceptual understanding of experimental design. The question is slightly easier than average because the calculations are straightforward and the experimental design clearly suggests pairing, making it a textbook application rather than requiring novel insight. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Orange | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Method A | 29 | 30 | 26 | 25 | 26 | 22 | 23 | 28 |
| Method B | 27 | 25 | 28 | 24 | 23 | 26 | 22 | 25 |
| Answer | Marks | Guidance |
|---|---|---|
| \(s_p^2 = \frac{7 \times 7.84 + 7 \times 4}{7 + 7} = 5.92\) | M1 | |
| \(s_p = 2.433105\) | awrt 2.43 | A1 |
| \(H_0 : \mu_A = \mu_B\), \(H_1 : \mu_A \neq \mu_B\) | both | B1 |
| \(t = \frac{26.125 - 25}{2.43\sqrt{\frac{1}{8} + \frac{1}{8}}} = 0.92474\) | awrt 0.925 | M1A1 |
| \(t_{14}(2.5\%) = 2.145\) | B1 | |
| Insufficient evidence to reject \(H_0\) that there is no difference in the means. | A1\(\mathcal{f}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(d = M1 - M2\): \(2.5, -2, 1.3, -4, 1.3\) | M1 | |
| \(\bar{d} = \frac{9}{8} = 1.125\) | B1 | |
| \(s_d^2 = \frac{69 - 8 \times 1.125^2}{7} = 8.410714\) | awrt 8.41 | M1A1 |
| \(H_0 : \delta = 0\), \(H_1 : \delta \neq 0\) | both | B1 |
| \(t = \frac{1.125}{\sqrt{\frac{8.41}{8}}} = 1.0972\) | awrt 1.10 | M1A1 |
| \(t_7(2.5\%) = 2.365\) | B1 | |
| There is no significant evidence of a difference between method A and method B. | A1\(\mathcal{f}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Paired sample as they are two measurements on the same orange | B1 |
## Part (a)
$s_p^2 = \frac{7 \times 7.84 + 7 \times 4}{7 + 7} = 5.92$ | | M1 |
$s_p = 2.433105$ | awrt 2.43 | A1 |
$H_0 : \mu_A = \mu_B$, $H_1 : \mu_A \neq \mu_B$ | both | B1 |
$t = \frac{26.125 - 25}{2.43\sqrt{\frac{1}{8} + \frac{1}{8}}} = 0.92474$ | awrt 0.925 | M1A1 |
$t_{14}(2.5\%) = 2.145$ | | B1 |
Insufficient evidence to reject $H_0$ that there is no difference in the means. | | A1$\mathcal{f}$ | (7)
## Part (b)
$d = M1 - M2$: $2.5, -2, 1.3, -4, 1.3$ | | M1 |
$\bar{d} = \frac{9}{8} = 1.125$ | | B1 |
$s_d^2 = \frac{69 - 8 \times 1.125^2}{7} = 8.410714$ | awrt 8.41 | M1A1 |
$H_0 : \delta = 0$, $H_1 : \delta \neq 0$ | both | B1 |
$t = \frac{1.125}{\sqrt{\frac{8.41}{8}}} = 1.0972$ | awrt 1.10 | M1A1 |
$t_7(2.5\%) = 2.365$ | | B1 |
There is no significant evidence of a difference between method A and method B. | | A1$\mathcal{f}$ | (9)
## Part (c)
Paired sample as they are two measurements on the same orange | | B1 | (1)Two methods of extracting juice from an orange are to be compared. Eight oranges are halved. One half of each orange is chosen at random and allocated to Method $A$ and the other half is allocated to Method $B$. The amounts of juice extracted, in ml, are given in the table.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
Orange & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
Method A & 29 & 30 & 26 & 25 & 26 & 22 & 23 & 28 \\
\hline
Method B & 27 & 25 & 28 & 24 & 23 & 26 & 22 & 25 \\
\hline
\end{tabular}
\end{center}
One statistician suggests performing a two-sample $t$-test to investigate whether or not there is a difference between the mean amounts of juice extracted by the two methods.
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly and using a 5\% significance level, carry out this test.
(You may assume $\bar{x}_A = 26.125$, $s_A^2 = 7.84$, $\bar{x}_B = 25$, $s_B^2 = 4$ and $\sigma_A^2 = \sigma_B^2$) [7]
\end{enumerate}
Another statistician suggests analysing these data using a paired $t$-test.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Using a 5\% significance level, carry out this test. [9]
\item State which of these two tests you consider to be more appropriate. Give a reason for your choice. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2003 Q7 [17]}}