| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Standard +0.8 This S4 question requires multiple confidence interval techniques including t-distribution CI, margin of error manipulation, chi-squared distribution for variance estimation, and sample size calculation. While each component uses standard formulas, the multi-part structure requiring integration of concepts (particularly part c's chi-squared percentile and part d's synthesis) and the unfamiliar context of deliberately choosing an upper bound for variance elevate this above routine S4 exercises. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks |
|---|---|
| \(\bar{x} = \dfrac{847.89}{10} = 84.79\); \(\quad s_x^2 = \dfrac{103712.6151 - (847.89)^2/10}{9}\) | B1 |
| \(s_x^2 = 3535.6522\ldots\) or \(s_x = 59.461\ldots\) | B1 |
| \(t\) critical value \(= 2.262\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| accept \((42.3, 127.3)\) | M1, A1, A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Chief accountant requires \(1.96\dfrac{\sigma}{\sqrt{n}} < 10\) | M1 A1 | |
| i.e. \(\dfrac{\sigma^2}{n} < \left(\dfrac{10}{1.96}\right)^2 = 26.0308\ldots = 26.03\) (2 d.p.) | A1 cso | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\chi^2_9 = 2.088\); i.e. \(\dfrac{9s^2}{\sigma^2} > 2.088\), \(\Rightarrow \sigma^2 < 15239.88\ldots\) awrt 15240 | B1; M1, A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute into part (b), \(n > \dfrac{15240}{26.03} \Rightarrow n = 586\) | M1, A1 | (2 marks) (14 marks total) |
# Question 4:
## Part (a)
$\bar{x} = \dfrac{847.89}{10} = 84.79$; $\quad s_x^2 = \dfrac{103712.6151 - (847.89)^2/10}{9}$ | B1 |
$s_x^2 = 3535.6522\ldots$ or $s_x = 59.461\ldots$ | B1 |
$t$ critical value $= 2.262$ | B1 |
95% confidence interval for $\mu = 84.79 \pm 2.262 \times \dfrac{59.461}{\sqrt{10}} = (42.25, 127.33)$
accept $(42.3, 127.3)$ | M1, A1, A1 | (6 marks)
## Part (b)
95% confidence interval $\bar{x} \pm 1.96\dfrac{\sigma}{\sqrt{n}}$
Chief accountant requires $1.96\dfrac{\sigma}{\sqrt{n}} < 10$ | M1 A1 |
i.e. $\dfrac{\sigma^2}{n} < \left(\dfrac{10}{1.96}\right)^2 = 26.0308\ldots = 26.03$ (2 d.p.) | A1 cso | (3 marks)
## Part (c)
Require the upper confidence limit of 98% confidence interval for $\sigma^2$
$\chi^2_9 = 2.088$; i.e. $\dfrac{9s^2}{\sigma^2} > 2.088$, $\Rightarrow \sigma^2 < 15239.88\ldots$ awrt 15240 | B1; M1, A1 | (3 marks)
## Part (d)
Substitute into part (b), $n > \dfrac{15240}{26.03} \Rightarrow n = 586$ | M1, A1 | (2 marks) **(14 marks total)**
---4. Gill, a member of the accounts department in a large company, is studying the expenses claims of company employees. She assumes that the claims, in $\pounds$, follow a normal distribution with mean $\mu$ and variance $\sigma ^ { 2 }$. As a first stage in her investigation she took the following random sample of 10 claims.
$$30.85,99.75,142.73,223.16,75.43,28.57,53.90,81.43,68.62,43.45 .$$
\begin{enumerate}[label=(\alph*)]
\item Find a 95\% confidence interval for $\mu$.
The chief accountant would like a $95 \%$ confidence interval where the difference between the upper confidence limit and the lower confidence limit is less than 20 .
\item Show that $\frac { \sigma ^ { 2 } } { n } < 26.03$ (to 2 decimal places), where $n$ is the size of the sample required to achieve this.
Gill decides to use her original sample of 10 to obtain a value for $\sigma ^ { 2 }$ so that the chance of her value being an underestimate is 0.01 .
\item Find such a value for $\sigma ^ { 2 }$.
\item Use this value for $\sigma ^ { 2 }$ to estimate the size of sample the chief accountant requires.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 Q4 [14]}}