| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample t-test |
| Difficulty | Standard +0.3 This is a standard paired t-test with clearly paired data, requiring calculation of differences, their mean and standard deviation, then applying the t-test formula. While it's S4 (Further Maths), the procedure is routine and methodical with no conceptual surprises—slightly above average difficulty only due to the computational steps and being a Further Maths topic. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank5.07d Paired vs two-sample: selection |
| Student | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| Score on paper I | 57 | 63 | 68 | 81 | 43 | 65 | 52 | 31 |
| Score on paper II | 53 | 62 | 61 | 78 | 44 | 64 | 43 | 29 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differences \(d_i\) (Paper I \(-\) Paper II): 4, 1, 7, 3, −1, 1, 9, 2 | B1 | Correct differences |
| \(\sum d = 26\), \(\bar{d} = 3.25\) | B1 | Correct mean of differences |
| \(\sum d^2 = 16+1+49+9+1+1+81+4 = 162\) | M1 | Correct method for \(S_{dd}\) |
| \(S_{dd} = 162 - \frac{26^2}{8} = 162 - 84.5 = 77.5\) | A1 | Correct \(S_{dd}\) |
| \(s_d^2 = \frac{77.5}{7} = 11.071...\), \(s_d = 3.327...\) | A1 | Correct \(s_d\) |
| \(H_0: \mu_d = 1\); \(H_1: \mu_d > 1\) | B1 | Both hypotheses correct |
| \(t = \frac{\bar{d} - 1}{s_d/\sqrt{n}} = \frac{3.25 - 1}{3.327.../\sqrt{8}}\) | M1 | Correct method |
| \(t = \frac{2.25}{1.1760...} \approx 1.91\) | A1 | Correct test statistic |
| Critical value: \(t_7(0.05) = 1.895\); \(1.91 > 1.895\), reject \(H_0\) | B1 | Correct CV and conclusion — evidence to support teacher's belief |
# Question 5:
## Paired t-test
| Answer | Mark | Guidance |
|--------|------|----------|
| Differences $d_i$ (Paper I $-$ Paper II): 4, 1, 7, 3, −1, 1, 9, 2 | B1 | Correct differences |
| $\sum d = 26$, $\bar{d} = 3.25$ | B1 | Correct mean of differences |
| $\sum d^2 = 16+1+49+9+1+1+81+4 = 162$ | M1 | Correct method for $S_{dd}$ |
| $S_{dd} = 162 - \frac{26^2}{8} = 162 - 84.5 = 77.5$ | A1 | Correct $S_{dd}$ |
| $s_d^2 = \frac{77.5}{7} = 11.071...$, $s_d = 3.327...$ | A1 | Correct $s_d$ |
| $H_0: \mu_d = 1$; $H_1: \mu_d > 1$ | B1 | Both hypotheses correct |
| $t = \frac{\bar{d} - 1}{s_d/\sqrt{n}} = \frac{3.25 - 1}{3.327.../\sqrt{8}}$ | M1 | Correct method |
| $t = \frac{2.25}{1.1760...} \approx 1.91$ | A1 | Correct test statistic |
| Critical value: $t_7(0.05) = 1.895$; $1.91 > 1.895$, reject $H_0$ | B1 | Correct CV and conclusion — evidence to support teacher's belief |\begin{enumerate}
\item Students studying for their Mathematics GCSE are assessed by two examination papers. A teacher believes that on average the score on paper I is more than 1 mark higher than the score on paper II. To test this belief the scores of 8 randomly selected students are recorded. The results are given in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Student & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Score on paper I & 57 & 63 & 68 & 81 & 43 & 65 & 52 & 31 \\
\hline
Score on paper II & 53 & 62 & 61 & 78 & 44 & 64 & 43 & 29 \\
\hline
\end{tabular}
\end{center}
Assuming that the scores are normally distributed and stating your hypotheses clearly, test at the $5 \%$ level of significance whether or not there is evidence to support the teacher's belief.
\hfill \mbox{\textit{Edexcel S4 2013 Q5 [8]}}