| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Challenging +1.2 This S4 question requires understanding of t-distribution confidence intervals and chi-squared intervals for variance, plus an application involving normal probability calculation. While it involves multiple statistical concepts and a non-routine part (c) requiring integration of confidence limits with normal distribution, the individual steps are methodical and follow standard procedures for Further Maths Statistics students. The conceptual demand is moderate rather than high. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) 95% CI for \(\mu\): \(560 \pm t_{14}(2.5\%)\sqrt{\frac{25.2}{15}} = 560 \pm 2.145\sqrt{\frac{25.2}{15}} = (557.2,\ 562.8)\) | B1, M1 A1 A1 | \(t\) value = 2.145 |
| (b) 95% CI for \(\sigma^2\): \(5.629 < \frac{14 \times 25.2}{\sigma^2} < 26.119\) | B1, M1, B1 | |
| \(\sigma^2 < 62.675\), \(\sigma^2 > 13.507\) | ||
| \(13.507 < \sigma^2 < 62.675\) | A1, A1 | awrt 13.5, 62.7 |
| (c) Require \(P(X > 565) = P\!\left(Z > \frac{565 - \mu}{\sigma}\right)\) to be as large as possible, so \(\frac{565-\mu}{\sigma}\) to be as small as possible; both imply highest \(\sigma\) and \(\mu\). | M1 | M1 for using their largest \(\sigma\) and \(\mu\) |
| \(\frac{565 - 562.8}{\sqrt{62.675}} = 0.28\) | M1 A1 | M1 for using \(\frac{x-\mu}{\sigma}\) |
| \(P(Z > 0.28) = 1 - 0.6103 = 0.3897\) | M1 A1 | awrt 0.39 – 0.40; M1 for \(1 -\) their prob |
# Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** 95% CI for $\mu$: $560 \pm t_{14}(2.5\%)\sqrt{\frac{25.2}{15}} = 560 \pm 2.145\sqrt{\frac{25.2}{15}} = (557.2,\ 562.8)$ | B1, M1 A1 A1 | $t$ value = 2.145 |
| **(b)** 95% CI for $\sigma^2$: $5.629 < \frac{14 \times 25.2}{\sigma^2} < 26.119$ | B1, M1, B1 | |
| $\sigma^2 < 62.675$, $\sigma^2 > 13.507$ | | |
| $13.507 < \sigma^2 < 62.675$ | A1, A1 | awrt 13.5, 62.7 |
| **(c)** Require $P(X > 565) = P\!\left(Z > \frac{565 - \mu}{\sigma}\right)$ to be as large as possible, so $\frac{565-\mu}{\sigma}$ to be as small as possible; both imply highest $\sigma$ and $\mu$. | M1 | M1 for using their largest $\sigma$ and $\mu$ |
| $\frac{565 - 562.8}{\sqrt{62.675}} = 0.28$ | M1 A1 | M1 for using $\frac{x-\mu}{\sigma}$ |
| $P(Z > 0.28) = 1 - 0.6103 = 0.3897$ | M1 A1 | awrt 0.39 – 0.40; M1 for $1 -$ their prob |\begin{enumerate}
\item A machine fills jars with jam. The weight of jam in each jar is normally distributed. To check the machine is working properly the contents of a random sample of 15 jars are weighed in grams. Unbiased estimates of the mean and variance are obtained as
\end{enumerate}
$$\hat { \mu } = 560 \quad s ^ { 2 } = 25.2$$
Calculate a 95\% confidence interval for,\\
(a) the mean weight of jam,\\
(b) the variance of the weight of jam.
A weight of more than 565 g is regarded as too high and suggests the machine is not working properly.\\
(c) Use appropriate confidence limits from parts (a) and (b) to find the highest estimate of the proportion of jars that weigh too much.\\
\hfill \mbox{\textit{Edexcel S4 2009 Q5 [14]}}