| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| 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 +0.8 This S4 question requires chi-squared test for variance, then both z-test and t-test for means, followed by interpretation considering the variance test result. It involves multiple hypothesis tests with correct distribution selection, calculation from summary statistics, and critical thinking about test validity - substantially harder than typical A-level questions but standard for Further Maths S4. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s_x^2 = \frac{214856 - 20 \times \left(\frac{2072}{20}\right)^2}{19} = 10.357...\) awrt 10.4 | B1 | |
| \(H_0: \sigma = 2.8\) (or \(\sigma^2 = ...\)) \(\quad H_1: \sigma \neq 2.8\) (or \(\sigma^2 \neq ...\)) | B1 | |
| \(\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{19}\), test statistic \(= 25.102...\) awrt 25.1 | M1A1 | M1 for use of correct test statistic |
| \(\chi^2_{19}(0.025) = 32.852\), \(\quad \chi^2_{19}(0.975) = 8.907\) | B1B1 | |
| Not significant so no evidence of a change in standard deviation | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu = 102.3 \quad H_1: \mu \neq 102.3\) | B1 | |
| \(z = \frac{\frac{2072}{20} - 102.3}{\frac{2.8}{\sqrt{20}}} = 2.0763...\) aw rt 2.08 | M1A1 | For use of correct test statistic |
| Critical value is \(z = 1.96\) or awrt \(0.019 < 0.025\) | B1 | |
| So a significant result, there is evidence of a change in mean length | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = \frac{\frac{2072}{20} - 102.3}{\sqrt{\frac{10.357...}{20}}} = 1.8064...\) awrt 1.81 | M1A1 | For use of correct test statistic |
| Critical value of \(t_{19} = 2.093\) | B1 | |
| Not significant, there is insufficient evidence of a change in mean length | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (a) suggests that \(\sigma\) is unchanged so can use \(\sigma = 2.8\) so normal test can be used | B1ft | 1st B1 for reason for selecting (i) or (ii) based on conclusion from test in (a) |
| So using (i) conclude that there is evidence of an increase in mean length | B1ft | 2nd B1 for final conclusion about mean lengths based on (a) and (b); NB if both conclusions are the same it needs to be clear they have chosen (i) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s_x^2 = \frac{214856 - 20 \times \left(\frac{2072}{20}\right)^2}{19} = 10.357...$ awrt **10.4** | B1 | |
| $H_0: \sigma = 2.8$ (or $\sigma^2 = ...$) $\quad H_1: \sigma \neq 2.8$ (or $\sigma^2 \neq ...$) | B1 | |
| $\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{19}$, test statistic $= 25.102...$ awrt **25.1** | M1A1 | M1 for use of correct test statistic |
| $\chi^2_{19}(0.025) = 32.852$, $\quad \chi^2_{19}(0.975) = 8.907$ | B1B1 | |
| Not significant so no evidence of a change in standard deviation | A1 | |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 102.3 \quad H_1: \mu \neq 102.3$ | B1 | |
| $z = \frac{\frac{2072}{20} - 102.3}{\frac{2.8}{\sqrt{20}}} = 2.0763...$ aw rt **2.08** | M1A1 | For use of correct test statistic |
| Critical value is $z = 1.96$ or awrt $0.019 < 0.025$ | B1 | |
| So a significant result, there is evidence of a change in mean length | A1ft | |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = \frac{\frac{2072}{20} - 102.3}{\sqrt{\frac{10.357...}{20}}} = 1.8064...$ awrt **1.81** | M1A1 | For use of correct test statistic |
| Critical value of $t_{19} = 2.093$ | B1 | |
| Not significant, there is insufficient evidence of a change in mean length | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (a) suggests that $\sigma$ is unchanged so can use $\sigma = 2.8$ so normal test can be used | B1ft | 1st B1 for reason for selecting (i) or (ii) based on conclusion from test in (a) |
| So using (i) conclude that there is evidence of an increase in mean length | B1ft | 2nd B1 for final conclusion about mean lengths based on (a) and (b); NB if both conclusions are the same it needs to be clear they have chosen (i) |
The image you've shared appears to be the **back cover/final page** of an Edexcel publication (Order Code UA028843, June 2011). It contains only publisher contact information, logos (Ofqual, Welsh Assembly Government, CEA), and no mark scheme content.
There are **no questions, answers, mark allocations, or marking guidance** on this page to extract.
If you have additional pages from this mark scheme, please share them and I'll extract the content as requested.\begin{enumerate}
\item A machine produces components whose lengths are normally distributed with mean 102.3 mm and standard deviation 2.8 mm . After the machine had been serviced, a random sample of 20 components were tested to see if the mean and standard deviation had changed. The lengths, $x \mathrm {~mm}$, of each of these 20 components are summarised as
\end{enumerate}
$$\sum x = 2072 \quad \sum x ^ { 2 } = 214856$$
(a) Stating your hypotheses clearly, test, at the $5 \%$ level of significance, whether or not there is evidence of a change in standard deviation.\\
(b) Stating your hypotheses clearly, test, at the $5 \%$ level of significance, whether or not the mean length of the components has changed from the original value of 102.3 mm using\\
(i) a normal distribution,\\
(ii) a $t$ distribution.\\
(c) Comment on the mean length of components produced after the service in the light of the tests from part (a) and part (b). Give a reason for your answer.\\
\hfill \mbox{\textit{Edexcel S4 2011 Q7 [18]}}