| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Challenging +1.3 This S4 question requires calculating confidence intervals for mean (using t-distribution) and variance (using chi-squared), then applying normal distribution to estimate a proportion. Part (c) requires insight to maximize the proportion by choosing appropriate confidence limits, which elevates it above routine calculation. However, the individual techniques are standard for Further Maths Statistics, making it moderately challenging rather than highly difficult. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = 4.01\) | B1 | |
| \(s = 0.7992...\) | M1 A1 | |
| \(4.01 \pm t_9(2.5\%) \frac{0.7992...}{\sqrt{10}} = 4.01 \pm 2.262 \frac{0.7992...}{\sqrt{10}}\) | B1 | 2.262 |
| M1 A1 √ | their \(\bar{x}\) and \(s\) and \(\sqrt{10}\) | |
| \(= 4.5816...\) and \(3.4383...\) | awrt 4.58 and 3.44 | A1 |
Total: 7 marks
| Answer | Marks | Guidance |
|---|---|---|
| \(2.700 < \frac{9 \times 0.7992...^2}{s^2} < 19.023\) | B1 B1 | 2.7,19.023 |
| \(9 \times s^2/\sigma^2\) | ||
| \(\sigma^2 < 2.13, \quad \sigma^2 > 0.302\) | A1 | both awrt 2.13, 0.302 |
Total: 4 marks
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 7) = P\left(Z > \frac{7 - \mu}{\sigma}\right)\) needs to be as high as possible | M1 | |
| Therefore \(\mu\) and \(\sigma\) must be as big as possible | M1 | |
| \(= P\left(Z > \frac{7 - 4.581}{\sqrt{2.13}}\right)\) | A1 √ | |
| \(= 1 - 0.9515\) | ||
| \(= 0.0485\) | ||
| \(= 4.85\%\) | A1 | 4.8 to 4.9 |
Total: 4 marks
**Part a:**
$\bar{x} = 4.01$ | B1 |
$s = 0.7992...$ | M1 A1 |
$4.01 \pm t_9(2.5\%) \frac{0.7992...}{\sqrt{10}} = 4.01 \pm 2.262 \frac{0.7992...}{\sqrt{10}}$ | B1 | 2.262 |
| M1 A1 √ | their $\bar{x}$ and $s$ and $\sqrt{10}$ |
$= 4.5816...$ and $3.4383...$ | awrt 4.58 and 3.44 | A1 |
Total: 7 marks
**Part b:**
$2.700 < \frac{9 \times 0.7992...^2}{s^2} < 19.023$ | B1 B1 | 2.7,19.023 |
| | | $9 \times s^2/\sigma^2$ |
$\sigma^2 < 2.13, \quad \sigma^2 > 0.302$ | A1 | both awrt 2.13, 0.302 |
Total: 4 marks
**Part c:**
$P(X > 7) = P\left(Z > \frac{7 - \mu}{\sigma}\right)$ needs to be as high as possible | M1 |
Therefore $\mu$ and $\sigma$ must be as big as possible | M1 |
$= P\left(Z > \frac{7 - 4.581}{\sqrt{2.13}}\right)$ | A1 √ |
$= 1 - 0.9515$ | |
$= 0.0485$ | |
$= 4.85\%$ | A1 | 4.8 to 4.9 |
Notes:
- (a) $s^2 = 0.63877...$
- (c) M1 may be implied by them using their highest $\mu$ and $\sigma$.
Total: 4 marks7. A doctor wishes to study the level of blood glucose in males. The level of blood glucose is normally distributed. The doctor measured the blood glucose of 10 randomly selected male students from a school. The results, in mmol/litre, are given below.
$$\begin{array} { l l l l l l l l l l }
4.7 & 3.6 & 3.8 & 4.7 & 4.1 & 2.2 & 3.6 & 4.0 & 4.4 & 5.0
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Calculate a $95 \%$ confidence interval for the mean.
\item Calculate a 95\% confidence interval for the variance.
A blood glucose reading of more than 7 mmol/litre is counted as high.
\item Use appropriate confidence limits from parts (a) and (b) to find the highest estimate of the proportion of male students in the school with a high blood glucose level.
\section*{END}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2007 Q7 [15]}}