| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | CI from raw data list |
| Difficulty | Standard +0.3 This is a standard S4 confidence interval question requiring calculation of CI for mean (using t-distribution) and variance (using chi-squared distribution), followed by straightforward interpretation. The calculations are routine applications of formulas with no conceptual challenges, though the chi-squared CI for variance is less commonly seen than the mean CI, making it slightly above average difficulty for A-level but still a textbook exercise. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks |
|---|---|
| \(\bar{x} = 4.9\) | B1 |
| \(s = \sqrt{0.191..}\) (0.437...) | B1 |
| (NB: \(\Sigma x = 49\); \(\Sigma x^2 = 241.82\)) |
| Answer | Marks |
|---|---|
| 95% confidence interval is given by \(4.9 \pm 2.262 \times \sqrt{\frac{0.191..}{10}}\) | M1A1ft B1 |
| i.e: (4.587..., 5.212 ...) | A1 A1 |
| (13 marks total) |
| Answer | Marks |
|---|---|
| 95% confidence interval is given by \(\frac{9 \times 0.437...^2}{19.023} < \sigma^2 < \frac{9 \times 0.437...^2}{2.7}\) use of \(\frac{(n-1)s^2}{\chi_{n-1}^2}\) | M1B1B1A1 |
| i.e; (0.0904, 0.63704) | A1 A1 |
| (13 marks total) |
| Answer | Marks |
|---|---|
| 5 lies inside the confidence interval | B1ft |
| 0.49(0.7)² lies inside the confidence interval | B1ft |
| Yes it does meet the time requirement | B1 ft |
| (3 marks total) |
**Part (a)**
| $\bar{x} = 4.9$ | B1 |
| $s = \sqrt{0.191..}$ (0.437...) | B1 |
| (NB: $\Sigma x = 49$; $\Sigma x^2 = 241.82$) | |
**Part (a)(i)**
| 95% confidence interval is given by $4.9 \pm 2.262 \times \sqrt{\frac{0.191..}{10}}$ | M1A1ft B1 |
| i.e: (4.587..., 5.212 ...) | A1 A1 |
| | **(13 marks total)** |
**Part (a)(ii)**
| 95% confidence interval is given by $\frac{9 \times 0.437...^2}{19.023} < \sigma^2 < \frac{9 \times 0.437...^2}{2.7}$ use of $\frac{(n-1)s^2}{\chi_{n-1}^2}$ | M1B1B1A1 |
| i.e; (0.0904, 0.63704) | A1 A1 |
| | **(13 marks total)** |
**Part (b)**
| 5 lies inside the confidence interval | B1ft |
| 0.49(0.7)² lies inside the confidence interval | B1ft |
| Yes it does meet the time requirement | B1 ft |
| | **(3 marks total)** |
**Guidance notes for 4(a):**
- B1 B1 may be implied by correct a correct answer to (i) or (ii)
- M1 "their 4.9" $\pm$ t value $\times \sqrt{\frac{\text{their }0.191..}{10}}$
- A1ft "their 4.9" $\pm 2.262 \times \sqrt{\frac{\text{their }0.191..}{10}}$
- B1 2.262
- A1 either correct to 3sf or better or both correct to 2sf or better
- A1 both correct to 3sf or better
**Guidance notes for 4(a)(ii):**
- M1 writing and attempting to use $\frac{(n-1)s^2}{\chi_{n-1}^2}$ or may be implied by correct formula
- B1 19.023
- B1 2.7
- A1ft follow through their 0.437 and two chi squared values
- A1 either correct to 2sf or better
- A1 awrt (0.09, 0.637)
**Guidance notes for 4(b):**
- For the second B1. If both 0.7 and 0.49 lie in interval they must state variance = 0.49 or the interval for standard deviation.
- For the third B1 their must not be two conflicting conclusions unless they give just one overall as well.
**Total: 16 marks**
---A newspaper runs a daily Sudoku. A random sample of 10 people took the following times, in minutes, to complete the Sudoku.
5.0 \quad 4.5 \quad 4.7 \quad 5.3 \quad 5.2 \quad 4.1 \quad 5.3 \quad 4.8 \quad 5.5 \quad 4.6
Given that the times to complete the Sudoku follow a normal distribution,
\begin{enumerate}[label=(\alph*)]
\item calculate a 95\% confidence interval for
\begin{enumerate}[label=(\roman*)]
\item the mean,
\item the variance,
\end{enumerate}
of the times taken by people to complete the Sudoku. [13]
The newspaper requires the average time needed to complete the Sudoku to be 5 minutes with a standard deviation of 0.7 minutes.
\item Comment on whether or not the Sudoku meets this requirement. Give a reason for your answer. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2012 Q4 [16]}}