| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | Confidence interval supports assertion |
| Difficulty | Standard +0.3 This is a straightforward application of the chi-squared confidence interval formula for variance with given sample statistics. Students need to recall the formula (n-1)s²/χ² and look up critical values, but no problem-solving or conceptual insight is required beyond standard S4 procedure. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1}\), so \(\frac{29 \times 0.36}{\sigma^2} \sim \chi^2_{29}\) | M1 | Correct pivot |
| \(\chi^2_{29}\) critical values at 2.5% and 97.5%: \(16.047\) and \(45.722\) | B1 | Both correct values stated |
| \(\frac{29 \times 0.36}{45.722} < \sigma^2 < \frac{29 \times 0.36}{16.047}\) | M1 | Correct structure of interval |
| \(0.228 < \sigma^2 < 0.650\) | A1A1 | A1 each correct endpoint (awrt) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Since \(0.495\) lies within the interval \((0.228, 0.650)\) | B1 | Correct comparison |
| The confidence interval supports the assertion that \(\sigma^2 = 0.495\) | B1 | Correct conclusion with reason |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1}$, so $\frac{29 \times 0.36}{\sigma^2} \sim \chi^2_{29}$ | M1 | Correct pivot |
| $\chi^2_{29}$ critical values at 2.5% and 97.5%: $16.047$ and $45.722$ | B1 | Both correct values stated |
| $\frac{29 \times 0.36}{45.722} < \sigma^2 < \frac{29 \times 0.36}{16.047}$ | M1 | Correct structure of interval |
| $0.228 < \sigma^2 < 0.650$ | A1A1 | A1 each correct endpoint (awrt) |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Since $0.495$ lies within the interval $(0.228, 0.650)$ | B1 | Correct comparison |
| The confidence interval supports the assertion that $\sigma^2 = 0.495$ | B1 | Correct conclusion with reason |
---2. The time, $t$ hours, that a typist can sit before incurring back pain is modelled by $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$. A random sample of 30 typists gave unbiased estimates for $\mu$ and $\sigma ^ { 2 }$ as shown below.
$$\hat { \mu } = 2.5 \quad s ^ { 2 } = 0.36$$
\begin{enumerate}[label=(\alph*)]
\item Find a 95\% confidence interval for $\sigma ^ { 2 }$
\item State with a reason whether or not the confidence interval supports the assertion that $\sigma ^ { 2 } = 0.495$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2013 Q2 [7]}}