| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2002 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.8 This S4 question requires constructing confidence intervals for both mean (using t-distribution) and variance (using chi-squared distribution), then applying these to estimate a proportion—a multi-step problem requiring knowledge of multiple distributions and careful interpretation. Part (c) particularly demands conceptual understanding of which confidence limits minimize the proportion estimate, going beyond routine calculation. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 95% confidence interval for \(\mu\) is | 2.064 | B1 |
| \(1.68 \pm t_{24}(2.5\%) \sqrt{\frac{1.79}{25}} = 1.68 \pm 2.064\sqrt{\frac{1.79}{25}} = (1.13, 2.23)\) | M1 A1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| 95% confidence interval for \(\sigma^2\) is | B1 M1 B1 | |
| \(12.401 < \frac{24 \times 1.79}{\sigma^2} < 39.364\) | A1 A1 | (5) |
| \(\sigma^2 > 1.09, \sigma^2 > 3.46\) |
| Answer | Marks | Guidance |
|---|---|---|
| Require \(P(X > 2.5) = P\left(Z > \frac{2.5 - \mu}{\sigma}\right)\) to be as small as possible OR | M1 M1 | |
| \(\frac{25 - \mu}{\sigma}\) to be as large as possible; both imply lowest \(\sigma\) and \(\mu\) | ||
| \(\frac{25 - 1.13}{\sqrt{1.09}} = 1.31\) | M1 | |
| \(P(Z > 1.31) = 1 - 0.9049 = 0.0951\) | A1 | (4) |
## Part (a)
95% confidence interval for $\mu$ is | 2.064 | B1 |
$1.68 \pm t_{24}(2.5\%) \sqrt{\frac{1.79}{25}} = 1.68 \pm 2.064\sqrt{\frac{1.79}{25}} = (1.13, 2.23)$ | M1 A1 A1 | (4)
## Part (b)
95% confidence interval for $\sigma^2$ is | B1 M1 B1 |
$12.401 < \frac{24 \times 1.79}{\sigma^2} < 39.364$ | A1 A1 | (5)
$\sigma^2 > 1.09, \sigma^2 > 3.46$ |
## Part (c)
Require $P(X > 2.5) = P\left(Z > \frac{2.5 - \mu}{\sigma}\right)$ to be as small as possible OR | M1 M1 |
$\frac{25 - \mu}{\sigma}$ to be as large as possible; both imply lowest $\sigma$ and $\mu$ | |
$\frac{25 - 1.13}{\sqrt{1.09}} = 1.31$ | M1 |
$P(Z > 1.31) = 1 - 0.9049 = 0.0951$ | A1 | (4)
**Total: (13 marks)**
---A nutritionist studied the levels of cholesterol, $X$ mg/cm³, of male students at a large college. She assumed that $X$ was distributed $\text{N}(\mu, \sigma^2)$ and examined a random sample of 25 male students. Using this sample she obtained unbiased estimates of $\mu$ and $\sigma^2$ as
$$\hat{\mu} = 1.68, \quad \hat{\sigma}^2 = 1.79.$$
\begin{enumerate}[label=(\alph*)]
\item Find a 95% confidence interval for $\mu$.
[4]
\item Obtain a 95% confidence interval for $\sigma^2$.
[5]
\end{enumerate}
A cholesterol reading of more than 2.5 mg/cm³ is regarded as high.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use appropriate confidence limits from parts $(a)$ and $(b)$ to find the lowest estimate of the proportion of male students in the college with high cholesterol.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2002 Q6 [13]}}