| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | Chi-squared test then t-test sequential |
| Difficulty | Standard +0.3 This is a straightforward two-part hypothesis test using standard chi-squared and t-test procedures with clearly given data. Part (a) requires a two-tailed chi-squared test for variance (routine S4 content), and part (b) requires a one-tailed t-test for the mean. Both are standard textbook applications with no conceptual challenges—students simply need to state hypotheses, calculate test statistics using given formulas, and compare to critical values. Slightly above average difficulty only because it combines two tests and requires careful handling of the chi-squared distribution, which is less familiar than the normal distribution. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks |
|---|---|
| \(H_0 : \sigma^2 = 0.9\) | B1 |
| \(H_1 : \sigma^2 \neq 0.9\) | |
| \(\nu = 19\) | |
| CR (Lower tail 10.117) | B1 |
| CR (Upper tail 30.144) | B1 |
| Test statistic \(= \frac{19 \times 1.5}{0.9} = 31.6666,\) significant | M1 A1 |
| There is sufficient evidence that the variance of the length of spring is different to 0.9 | A1 |
Total: 6 marks
| Answer | Marks |
|---|---|
| \(H_0 : \mu = 100\) | B1 |
| \(H_1 : \mu > 100\) | |
| \(t_{19} = 1.328\) | B1 |
| \(t = \frac{100.6 - 100}{\sqrt{\frac{1.5}{20}}} = 2.19\) | M1 A1 A1 |
| Significant. The mean length of spring is greater than 100 | B1 |
Total: 6 marks
**Part a:**
$H_0 : \sigma^2 = 0.9$ | B1 |
$H_1 : \sigma^2 \neq 0.9$ | |
$\nu = 19$ | |
CR (Lower tail 10.117) | B1 |
CR (Upper tail 30.144) | B1 |
Test statistic $= \frac{19 \times 1.5}{0.9} = 31.6666,$ significant | M1 A1 |
There is sufficient evidence that the variance of the length of spring is different to 0.9 | A1 |
Total: 6 marks
**Part b:**
$H_0 : \mu = 100$ | B1 |
$H_1 : \mu > 100$ | |
$t_{19} = 1.328$ | B1 |
$t = \frac{100.6 - 100}{\sqrt{\frac{1.5}{20}}} = 2.19$ | M1 A1 A1 |
Significant. The mean length of spring is greater than 100 | B1 |
Notes:
- (a) only need to see 30.144
- need variance in conclusion
- (b) conclusion must be in context. Length of spring needed
Total: 6 marks
---4. The length $X \mathrm {~mm}$ of a spring made by a machine is normally distributed $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$. A random sample of 20 springs is selected and their lengths measured in mm . Using this sample the unbiased estimates of $\mu$ and $\sigma ^ { 2 }$ are
$$\bar { x } = 100.6 , \quad s ^ { 2 } = 1.5 .$$
Stating your hypotheses clearly test, at the $10 \%$ level of significance,
\begin{enumerate}[label=(\alph*)]
\item whether or not the variance of the lengths of springs is different from 0.9 ,
\item whether or not the mean length of the springs is greater than 100 mm .
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2007 Q4 [12]}}