| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| 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 two-part hypothesis testing question requiring standard procedures: a one-sample t-test and a chi-squared test for variance. Both are routine applications of learned techniques with clear hypotheses and given summary statistics, requiring only careful execution of standard formulas rather than problem-solving insight. Slightly above average difficulty due to being S4 (Further Maths) content and requiring two different tests. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \mu = 500\), \(H_1: \mu > 500\) | B1 | Both hypotheses correct |
| Test statistic: \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{502 - 500}{\sqrt{5.6/12}}\) | M1 | Correct formula used |
| \(t = \frac{2}{\sqrt{0.4\overline{6}}} = \frac{2}{0.6831...} = 2.928...\) | A1 | Correct value |
| Critical value: \(t_{11}\) at 1% one-tailed = \(2.718\) | B1 | Correct critical value |
| Since \(2.928 > 2.718\), reject \(H_0\). Evidence that mean amount is more than 500 ml | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \sigma = 3\), \(H_1: \sigma < 3\) | B1 | Both hypotheses correct |
| Test statistic: \(\chi^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{11 \times 5.6}{9}\) | M1 | Correct formula |
| \(\chi^2 = \frac{61.6}{9} = 6.844...\) | A1 | Correct value |
| Critical value: \(\chi^2_{11}\) at 1% lower tail = \(3.053\) | B1 | Correct critical value |
| Since \(6.844 > 3.053\), do not reject \(H_0\). No evidence that standard deviation is less than 3 ml | A1 | Correct conclusion in context |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 500$, $H_1: \mu > 500$ | B1 | Both hypotheses correct |
| Test statistic: $t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{502 - 500}{\sqrt{5.6/12}}$ | M1 | Correct formula used |
| $t = \frac{2}{\sqrt{0.4\overline{6}}} = \frac{2}{0.6831...} = 2.928...$ | A1 | Correct value |
| Critical value: $t_{11}$ at 1% one-tailed = $2.718$ | B1 | Correct critical value |
| Since $2.928 > 2.718$, reject $H_0$. Evidence that mean amount is more than 500 ml | A1 | Correct conclusion in context |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \sigma = 3$, $H_1: \sigma < 3$ | B1 | Both hypotheses correct |
| Test statistic: $\chi^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{11 \times 5.6}{9}$ | M1 | Correct formula |
| $\chi^2 = \frac{61.6}{9} = 6.844...$ | A1 | Correct value |
| Critical value: $\chi^2_{11}$ at 1% lower tail = $3.053$ | B1 | Correct critical value |
| Since $6.844 > 3.053$, do not reject $H_0$. No evidence that standard deviation is less than 3 ml | A1 | Correct conclusion in context |
---\begin{enumerate}
\item A machine fills bottles with water. The amount of water in each bottle is normally distributed. To check the machine is working properly, a random sample of 12 bottles is selected and the amount of water, in ml, in each bottle is recorded. Unbiased estimates for the mean and variance are
\end{enumerate}
$$\hat { \mu } = 502 \quad s ^ { 2 } = 5.6$$
Stating your hypotheses clearly, test at the 1\% level of significance\\
(a) whether or not the mean amount of water in a bottle is more than 500 ml ,\\
(b) whether or not the standard deviation of the amount of water in a bottle is less than 3 ml .\\
\hfill \mbox{\textit{Edexcel S4 2013 Q6 [10]}}