| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample t-test (unknown variances) |
| Difficulty | Challenging +1.2 This is a standard two-stage hypothesis testing question requiring an F-test for equal variances followed by a pooled variance confidence interval calculation. While it involves multiple steps and careful calculation of sample variances, the procedures are routine for S4 students with no novel problem-solving required. The computational demands and need to apply two distinct statistical tests elevate it slightly above average difficulty. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \sigma_A^2 = \sigma_B^2\), \(H_1: \sigma_A^2 \neq \sigma_B^2\) | B1 | Both hypotheses correct |
| \(\bar{x}_A = \frac{1.5+0.9+1.3+1.2}{4} = 1.225\) | M1 | Attempt to find sample variance for A |
| \(S_A^2 = \frac{1}{3}\left[(1.5^2+0.9^2+1.3^2+1.2^2) - 4(1.225)^2\right] = \frac{0.0875}{3} \approx 0.0492\) | A1 | Correct \(s_A^2\) (accept 0.049) |
| \(\bar{x}_B = \frac{0.4+0.6+0.8+0.3+0.5+0.4}{6} = 0.5\) | M1 | Attempt to find sample variance for B |
| \(S_B^2 = \frac{1}{5}\left[(0.4^2+0.6^2+0.8^2+0.3^2+0.5^2+0.4^2) - 6(0.5)^2\right] = \frac{0.22}{5} = 0.044\) | A1 | Correct \(s_B^2\) |
| \(F = \frac{s_A^2}{s_B^2} = \frac{0.0492}{0.044} \approx 1.118\) (or reciprocal \(\approx 0.895\)) | M1 | Form \(F\) ratio, larger variance on top |
| Critical value \(F_{3,5}\) at 10% two-tailed = \(F_{3,5}(5\%) \approx 5.41\) | A1 | Correct critical value |
| Since \(1.118 < 5.41\), do not reject \(H_0\) | A1 | Correct conclusion: common variance can be assumed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Pooled estimate: \(s^2 = \frac{(n_A-1)s_A^2 + (n_B-1)s_B^2}{n_A+n_B-2} = \frac{3(0.0492)+5(0.044)}{8} = \frac{0.3675}{8} \approx 0.04594\) | M1 A1 | Correct pooled variance using both samples, \(\nu = 8\) degrees of freedom |
| 99% CI for \(\sigma^2\): \(\left(\frac{\nu s^2}{\chi^2_{\nu,0.005}}, \frac{\nu s^2}{\chi^2_{\nu,0.995}}\right)\) | M1 | Correct form of CI using \(\chi^2\) with 8 d.f. |
| \(\chi^2_{8, 0.005} = 21.955\), \(\chi^2_{8, 0.995} = 1.344\) | B1 | Both critical values stated |
| Lower bound: \(\frac{8 \times 0.04594}{21.955} \approx 0.01675\) | A1 | Correct lower bound |
| Upper bound: \(\frac{8 \times 0.04594}{1.344} \approx 0.2735\) | A1 | Correct upper bound |
| 99% CI for \(\sigma^2\) is approximately \((0.0168, 0.274)\) | Accept awrt these values |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \sigma_A^2 = \sigma_B^2$, $H_1: \sigma_A^2 \neq \sigma_B^2$ | B1 | Both hypotheses correct |
| $\bar{x}_A = \frac{1.5+0.9+1.3+1.2}{4} = 1.225$ | M1 | Attempt to find sample variance for A |
| $S_A^2 = \frac{1}{3}\left[(1.5^2+0.9^2+1.3^2+1.2^2) - 4(1.225)^2\right] = \frac{0.0875}{3} \approx 0.0492$ | A1 | Correct $s_A^2$ (accept 0.049) |
| $\bar{x}_B = \frac{0.4+0.6+0.8+0.3+0.5+0.4}{6} = 0.5$ | M1 | Attempt to find sample variance for B |
| $S_B^2 = \frac{1}{5}\left[(0.4^2+0.6^2+0.8^2+0.3^2+0.5^2+0.4^2) - 6(0.5)^2\right] = \frac{0.22}{5} = 0.044$ | A1 | Correct $s_B^2$ |
| $F = \frac{s_A^2}{s_B^2} = \frac{0.0492}{0.044} \approx 1.118$ (or reciprocal $\approx 0.895$) | M1 | Form $F$ ratio, larger variance on top |
| Critical value $F_{3,5}$ at 10% two-tailed = $F_{3,5}(5\%) \approx 5.41$ | A1 | Correct critical value |
| Since $1.118 < 5.41$, do not reject $H_0$ | A1 | Correct conclusion: common variance can be assumed |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Pooled estimate: $s^2 = \frac{(n_A-1)s_A^2 + (n_B-1)s_B^2}{n_A+n_B-2} = \frac{3(0.0492)+5(0.044)}{8} = \frac{0.3675}{8} \approx 0.04594$ | M1 A1 | Correct pooled variance using both samples, $\nu = 8$ degrees of freedom |
| 99% CI for $\sigma^2$: $\left(\frac{\nu s^2}{\chi^2_{\nu,0.005}}, \frac{\nu s^2}{\chi^2_{\nu,0.995}}\right)$ | M1 | Correct form of CI using $\chi^2$ with 8 d.f. |
| $\chi^2_{8, 0.005} = 21.955$, $\chi^2_{8, 0.995} = 1.344$ | B1 | Both critical values stated |
| Lower bound: $\frac{8 \times 0.04594}{21.955} \approx 0.01675$ | A1 | Correct lower bound |
| Upper bound: $\frac{8 \times 0.04594}{1.344} \approx 0.2735$ | A1 | Correct upper bound |
| 99% CI for $\sigma^2$ is approximately $(0.0168, 0.274)$ | | Accept awrt these values |6. The carbon content, measured in suitable units, of steel is normally distributed. Two independent random samples of steel were taken from a refining plant at different times and their carbon content recorded. The results are given below.
Sample A: $\quad 1.5 \quad 0.9 \quad 1.3 \quad 1.2$\\
$\begin{array} { l l l l l l l } \text { Sample } B : & 0.4 & 0.6 & 0.8 & 0.3 & 0.5 & 0.4 \end{array}$
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, carry out a suitable test, at the $10 \%$ level of significance, to show that both samples can be assumed to have come from populations with a common variance $\sigma ^ { 2 }$.
\item Showing your working clearly, find the $99 \%$ confidence interval for $\sigma ^ { 2 }$ based on both samples.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2013 Q6 [13]}}