| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| 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 hypothesis test for variance using the chi-squared distribution. Students need to calculate sample variance from given summaries, set up H₀: σ=2.4 vs H₁: σ≠2.4, compute the test statistic (n-1)s²/σ₀², and compare to critical values. While it requires multiple steps, it's a standard S4 procedure with no conceptual surprises—slightly easier than average due to clear structure and routine application of a learned technique. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \mu = \mu_0\) (some stated value), \(H_1: \sigma^2 \neq \sigma_0^2\) — actually: \(H_0: \sigma^2 = 5.76\), \(H_1: \sigma^2 \neq 5.76\) | B1 | Both hypotheses correct with \(\sigma^2 = 5.76\) (or \(\sigma = 2.4\)) |
| \(S^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{9}\left(1414.08 - \frac{113.4^2}{10}\right)\) | M1 | Correct formula used |
| \(= \frac{1}{9}(1414.08 - 1285.956) = \frac{128.124}{9} = 14.236\) | A1 | Correct value of \(s^2\) |
| Test statistic: \(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{9 \times 14.236}{5.76}\) | M1 | Correct formula |
| \(= \frac{128.124}{5.76} = 22.24\) | A1 | Correct value |
| Critical values: \(\chi^2_9\) at 5% two-tailed: \(\chi^2_{0.025} = 19.023\) and \(\chi^2_{0.975} = 2.700\) | B1 | Correct critical values stated |
| \(22.24 > 19.023\), so reject \(H_0\). There is evidence at 5% significance level that the standard deviation is not 2.4 thousand miles. | A1 | Correct conclusion in context |
# Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = \mu_0$ (some stated value), $H_1: \sigma^2 \neq \sigma_0^2$ — actually: $H_0: \sigma^2 = 5.76$, $H_1: \sigma^2 \neq 5.76$ | B1 | Both hypotheses correct with $\sigma^2 = 5.76$ (or $\sigma = 2.4$) |
| $S^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{9}\left(1414.08 - \frac{113.4^2}{10}\right)$ | M1 | Correct formula used |
| $= \frac{1}{9}(1414.08 - 1285.956) = \frac{128.124}{9} = 14.236$ | A1 | Correct value of $s^2$ |
| Test statistic: $\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{9 \times 14.236}{5.76}$ | M1 | Correct formula |
| $= \frac{128.124}{5.76} = 22.24$ | A1 | Correct value |
| Critical values: $\chi^2_9$ at 5% two-tailed: $\chi^2_{0.025} = 19.023$ and $\chi^2_{0.975} = 2.700$ | B1 | Correct critical values stated |
| $22.24 > 19.023$, so reject $H_0$. There is evidence at 5% significance level that the standard deviation is not 2.4 thousand miles. | A1 | Correct conclusion in context |
---\begin{enumerate}
\item George owns a garage and he records the mileage of cars, $x$ thousands of miles, between services. The results from a random sample of 10 cars are summarised below.
\end{enumerate}
$$\sum x = 113.4 \quad \sum x ^ { 2 } = 1414.08$$
The mileage of cars between services is normally distributed and George believes that the standard deviation is 2.4 thousand miles.
Stating your hypotheses clearly, test, at the $5 \%$ level of significance, whether or not these data support George's belief.\\
\hfill \mbox{\textit{Edexcel S4 2013 Q1 [7]}}