| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a straightforward application of standard hypothesis testing procedures from S4: a chi-squared test for variance and a t-test for the mean. Both are routine textbook exercises requiring recall of test statistics and critical values, with no conceptual challenges or novel problem-solving. The multi-part structure and 13 marks reflect thoroughness rather than difficulty. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks |
|---|---|
| \(H_0: \sigma^2 = 36\); \(H_1: \sigma^2 > 36\) | B1 |
| \(\nu = 24\), \(X_{24}^2\) (0.05) = 36.415 | B1 |
| \(\frac{(n-1)s^2}{\sigma^2} = \frac{24 \times 55}{36} = 36.67\) | M1 A1 |
| Since 36.67 > 36.415 there is sufficient evidence to reject \(H_0\). | A1 ft |
| There is evidence to suggest that the variance is greater than 36. | A1 ft |
| (6 marks total) |
| Answer | Marks |
|---|---|
| \(H_0: \mu = 450\) \(H_1: \mu > 450\) | B1 |
| \(t_{24} = 1.711\) | B1 |
| \(t = \pm \frac{455 - 450}{\sqrt{\frac{55}{25}}} = \pm 3.37...\) | M1 A1 |
| Significant; The mean weight of chocolates is greater than 450. Or \(\mu\) is more than 450 | A1ft; A1ft |
| (6 marks total) |
| Answer | Marks |
|---|---|
| The weights are normally distributed | B1 |
| (1 mark total) |
**Part (a)**
| $H_0: \sigma^2 = 36$; $H_1: \sigma^2 > 36$ | B1 |
| $\nu = 24$, $X_{24}^2$ (0.05) = 36.415 | B1 |
| $\frac{(n-1)s^2}{\sigma^2} = \frac{24 \times 55}{36} = 36.67$ | M1 A1 |
| Since 36.67 > 36.415 there is sufficient evidence to reject $H_0$. | A1 ft |
| There is evidence to suggest that the variance is greater than 36. | A1 ft |
| | **(6 marks total)** |
**Part (b)**
| $H_0: \mu = 450$ $H_1: \mu > 450$ | B1 |
| $t_{24} = 1.711$ | B1 |
| $t = \pm \frac{455 - 450}{\sqrt{\frac{55}{25}}} = \pm 3.37...$ | M1 A1 |
| Significant; The **mean weight** of chocolates is **greater than 450.** Or $\mu$ **is more than 450** | A1ft; A1ft |
| | **(6 marks total)** |
**Part (c)**
| The **weights** are normally distributed | B1 |
| | **(1 mark total)** |
**Guidance notes for 5(a):**
- B1 both correct. Also allow $H_0: \sigma = 6$; $H_1: \sigma > 6$
- B1 36.415
- M1 use of $\frac{(n-1)s^2}{\sigma^2}$
- A1 awrt 36.7
- A1ft any statement – no conflicting
- A1ft contextual statement must include "weight of chocolate" and is "greater than 50"
**Guidance notes for 5(b):**
- M1 $\pm \frac{455-450}{\sqrt{\frac{55}{25}}}$
- A1 awrt 3.4
**Total: 13 marks**
---Boxes of chocolates manufactured by Philippe have a mean weight of $\mu$ grams and a standard deviation of $\sigma$ grams. A random sample of 25 of these boxes are weighed. Using this sample, the unbiased estimate of $\mu$ is 455 and the unbiased estimate of $\sigma^2$ is 55.
\begin{enumerate}[label=(\alph*)]
\item Test, at the 5\% level of significance, whether or not $\sigma$ is greater than 6. State your hypotheses clearly. [6]
\item Test, at the 5\% level of significance, whether or not $\mu$ is more than 450. [6]
\item State an assumption you have made in order to carry out the above tests. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2012 Q5 [13]}}