| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Two-sample t-test |
| Difficulty | Standard +0.3 This is a standard two-part hypothesis testing question requiring an F-test for equal variances followed by a two-sample t-test. While it involves multiple steps and calculation of sample variances from summary statistics, the procedures are routine for Further Statistics 2 students with no novel problem-solving required. The question clearly signposts both tests needed and provides all necessary data in organized form. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | No. of competitors | Sample mean \(\overline { \boldsymbol { x } }\) | \(\sum \boldsymbol { x } ^ { \mathbf { 2 } }\) |
| Girls | 8 | 83.1 | 55746 |
| Boys | 7 | 88.9 | 56130 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \sigma_B^2 = \sigma_C^2, H_1: \sigma_B^2 \neq \sigma_C^2\) | B1 | Both hypotheses correct using the notation \(\sigma^2\). Allow \(s\) rather than \(\sigma^2\) |
| Using a correct Method for either \(s_B^2\) or \(s_C^2\) May be implied by a correct value | M1 | |
| \(s_B^2 = \frac{1}{6}(56130 - 7 \times 88.9^2) = \frac{807.53}{6} = 134.6\) | A1 | awrt 135 |
| \(s_C^2 = \frac{1}{7}(55746 - 8 \times 83.1^2) = \frac{501.12}{7} = 71.58\) | A1 | awrt 71.6 |
| \(\frac{s_B^2}{s_C^2} = 1.880...\) | M1 | Using the F-distribution as the model e.g. \(\frac{s_B^2}{s_C^2}\) |
| Critical value \(F_{6,7} = 3.87\) | B1 | awrt 3.87 |
| Not significant, variances can be treated as the same | A1ft | Drawing a correct inference following through their CV and value for \(\frac{s_B^2}{s_C^2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu_B = \mu_C, H_1: \mu_B > \mu_C\) | B1 | Both hypotheses correct using the notation \(\mu\) |
| Pooled estimate of variance \(s^2 = \frac{6 \times 1346 + 7 \times 71.58}{13} = 100.6653...\) | M1 | For realising the need to find the pooled estimate for the test require from a correct interpretation of the question |
| Test statistic \(t = \frac{88.9 - 83.1}{s\sqrt{\frac{1}{7}+\frac{1}{8}}} = \) awrt 1.12 | M1 | Correct method for test statistic \(t = \frac{88.9-83.1}{\text{"their }s"\sqrt{\frac{1}{7}+\frac{1}{8}}}\). May be implied by a correct value |
| Critical value \(t_{13}(5\%) = 1.771\) | B1 | awrt 1.77 |
| Insufficient evidence to support mother's claim | A1ft | Drawing a correct inference following through their CV and test statistic |
**(a)**
| $H_0: \sigma_B^2 = \sigma_C^2, H_1: \sigma_B^2 \neq \sigma_C^2$ | B1 | Both hypotheses correct using the notation $\sigma^2$. Allow $s$ rather than $\sigma^2$ |
| Using a correct Method for either $s_B^2$ or $s_C^2$ May be implied by a correct value | M1 | |
| $s_B^2 = \frac{1}{6}(56130 - 7 \times 88.9^2) = \frac{807.53}{6} = 134.6$ | A1 | awrt 135 |
| $s_C^2 = \frac{1}{7}(55746 - 8 \times 83.1^2) = \frac{501.12}{7} = 71.58$ | A1 | awrt 71.6 |
| $\frac{s_B^2}{s_C^2} = 1.880...$ | M1 | Using the F-distribution as the model e.g. $\frac{s_B^2}{s_C^2}$ |
| Critical value $F_{6,7} = 3.87$ | B1 | awrt 3.87 |
| Not significant, variances can be treated as the same | A1ft | Drawing a correct inference following through their CV and value for $\frac{s_B^2}{s_C^2}$ |
**(b)**
| $H_0: \mu_B = \mu_C, H_1: \mu_B > \mu_C$ | B1 | Both hypotheses correct using the notation $\mu$ |
| Pooled estimate of variance $s^2 = \frac{6 \times 1346 + 7 \times 71.58}{13} = 100.6653...$ | M1 | For realising the need to find the pooled estimate for the test require from a correct interpretation of the question |
| Test statistic $t = \frac{88.9 - 83.1}{s\sqrt{\frac{1}{7}+\frac{1}{8}}} = $ awrt 1.12 | M1 | Correct method for test statistic $t = \frac{88.9-83.1}{\text{"their }s"\sqrt{\frac{1}{7}+\frac{1}{8}}}$. May be implied by a correct value |
| Critical value $t_{13}(5\%) = 1.771$ | B1 | awrt 1.77 |
| Insufficient evidence to support mother's claim | A1ft | Drawing a correct inference following through their CV and test statistic |
---\begin{enumerate}
\item The times, $x$ seconds, taken by the competitors in the 100 m freestyle events at a school swimming gala are recorded. The following statistics are obtained from the data.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & No. of competitors & Sample mean $\overline { \boldsymbol { x } }$ & $\sum \boldsymbol { x } ^ { \mathbf { 2 } }$ \\
\hline
Girls & 8 & 83.1 & 55746 \\
\hline
Boys & 7 & 88.9 & 56130 \\
\hline
\end{tabular}
\end{center}
Following the gala, a mother claims that girls are faster swimmers than boys. Assuming that the times taken by the competitors are two independent random samples from normal distributions,\\
(a) test, at the $10 \%$ level of significance, whether or not the variances of the two distributions are the same. State your hypotheses clearly.\\
(b) Stating your hypotheses clearly, test the mother's claim. Use a $5 \%$ level of significance.
\hfill \mbox{\textit{Edexcel FS2 Q4 [13]}}