| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2002 |
| Session | June |
| Marks | 13 |
| 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 two-sample hypothesis testing question requiring an F-test for variances followed by a t-test for means. While it involves multiple steps and calculation of sample variances from summary statistics, the procedures are routine S4 content with no novel problem-solving required. The question clearly signposts both tests needed, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| No. of competitors | Sample Mean \(\overline{x}\) | \(\Sigma x^2\) | |
| Girls | 8 | 83.10 | 55746 |
| Boys | 7 | 88.90 | 56130 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \sigma_B^2 = \sigma_B^2\), \(H_1: \sigma_B^2 \neq \sigma_B^2\) | B1 | |
| \(s_B^2 = \frac{1}{6}(56130 - 7 \times 88.9^2) = \frac{807.53}{6} = 134.6\) | M1 A1 | |
| \(s_G^2 = \frac{1}{7}(55746 - 8 \times 83.1^2) = \frac{501.12}{7} = 71.58\) | A1 | |
| \(\frac{s_B^2}{s_G^2} = 1.880\ldots\) | M1 | |
| Critical value \(F_{6,7} = 3.87\) | B1 | |
| Not significant, variances are the same | A1 ft | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu_B = \mu_G\), \(H_1: \mu_B > \mu_G\) | B1 | |
| Pooled estimate of variance \(s^2 = \frac{6 \times 134.6 + 7 \times 71.58}{13} = 100.6653\ldots\) | M1 | |
| Test statistic \(t = \frac{88.9 - 83.1}{s\sqrt{\frac{1}{7} + \frac{1}{8}}}\) | M1 A1 | |
| Critical value \(t_{13}(5\%) = 1.771\) | B1 | |
| Insufficient evidence to support parent's claim | A1 ft | (6) |
## Part (a)
$H_0: \sigma_B^2 = \sigma_B^2$, $H_1: \sigma_B^2 \neq \sigma_B^2$ | B1 |
$s_B^2 = \frac{1}{6}(56130 - 7 \times 88.9^2) = \frac{807.53}{6} = 134.6$ | M1 A1 |
$s_G^2 = \frac{1}{7}(55746 - 8 \times 83.1^2) = \frac{501.12}{7} = 71.58$ | A1 |
$\frac{s_B^2}{s_G^2} = 1.880\ldots$ | M1 |
Critical value $F_{6,7} = 3.87$ | B1 |
Not significant, variances are the same | A1 ft | (7)
## Part (b)
$H_0: \mu_B = \mu_G$, $H_1: \mu_B > \mu_G$ | B1 |
Pooled estimate of variance $s^2 = \frac{6 \times 134.6 + 7 \times 71.58}{13} = 100.6653\ldots$ | M1 |
Test statistic $t = \frac{88.9 - 83.1}{s\sqrt{\frac{1}{7} + \frac{1}{8}}}$ | M1 A1 |
Critical value $t_{13}(5\%) = 1.771$ | B1 |
Insufficient evidence to support parent's claim | A1 ft | (6)
**Total: (13 marks)**
---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.
\begin{tabular}{|l|c|c|c|}
\hline
& No. of competitors & Sample Mean $\overline{x}$ & $\Sigma x^2$ \\
\hline
Girls & 8 & 83.10 & 55746 \\
\hline
Boys & 7 & 88.90 & 56130 \\
\hline
\end{tabular}
Following the gala a proud parent claims that girls are faster swimmers than boys. Assuming that the times taken by the competitors are two independent random samples from normal distributions,
\begin{enumerate}[label=(\alph*)]
\item test, at the 10% level of significance, whether or not the variances of the two distributions are the same. State your hypotheses clearly.
[7]
\item Stating your hypotheses clearly, test the parent's claim. Use a 5% level of significance.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2002 Q5 [13]}}