| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2020 |
| 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 +0.8 This question requires understanding of chi-squared tests for variance (a less common topic even in Further Maths), converting from a t-distribution confidence interval to extract sample statistics, then applying both hypothesis testing and normal distribution confidence intervals. The multi-step nature and the need to work backwards from given CI to find sample variance makes this moderately challenging, though the individual calculations are standard once set up correctly. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| From CI: \(\bar{x} = \frac{1.193+1.367}{2} = 1.28\), or width \(= 1.367-1.193 = 0.174\) | B1 | For finding mean from CI or calculation of width of CI |
| \(1.367 = \text{"1.28"} \pm 2.064 \times \frac{s}{\sqrt{25}}\) or \(\text{"0.174"} = 2\times 2.064\times\frac{s}{\sqrt{25}}\) | M1; A1 | M1 for using the given \(t\) model; A1 for correct use of \(t_{24}=2.064\) |
| \(\Rightarrow s = 0.210755\ldots\) | A1 | For \(s=0.21\) or better |
| \(H_0: \sigma = 0.175\), \(H_1: \sigma \neq 0.175\) | B1 | For correct hypotheses in terms of \(\sigma\) |
| \(\chi_{24}^2 = \frac{24s^2}{\sigma^2} = 34.8092\ldots\) awrt \(34.8\) | M1, A1 | For selecting appropriate model; for test statistic awrt \(34.8\) |
| \(\chi_{24}^2\ (10\%)\) 2-tail CR: \(\chi_{24}^2 < 13.848\) or \(\chi_{24}^2 > 36.415\) | B1 | For at least one correct critical value |
| \(34.8\) is not significant so insufficient evidence that \(\sigma \neq 0.175\) | A1 | For a correct conclusion confirming that assuming st. dev \(= 0.175\) is OK |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{"1.28"} \pm z\times\frac{0.175}{\sqrt{25}}\) | M1 | For use of correct formula with \(1.6 < z < 2\) (ft \(\bar{x}\) if found in (a)) |
| \(z = 1.96\) | B1 | For \(z=1.96\) or better used |
| \(= (1.211\ldots,\ 1.349\ldots)\) awrt (1.21, 1.35) | A1 | For an interval awrt \((1.21, 1.35)\) |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| From CI: $\bar{x} = \frac{1.193+1.367}{2} = 1.28$, or width $= 1.367-1.193 = 0.174$ | B1 | For finding mean from CI or calculation of width of CI |
| $1.367 = \text{"1.28"} \pm 2.064 \times \frac{s}{\sqrt{25}}$ or $\text{"0.174"} = 2\times 2.064\times\frac{s}{\sqrt{25}}$ | M1; A1 | M1 for using the given $t$ model; A1 for correct use of $t_{24}=2.064$ |
| $\Rightarrow s = 0.210755\ldots$ | A1 | For $s=0.21$ or better |
| $H_0: \sigma = 0.175$, $H_1: \sigma \neq 0.175$ | B1 | For correct hypotheses in terms of $\sigma$ |
| $\chi_{24}^2 = \frac{24s^2}{\sigma^2} = 34.8092\ldots$ awrt $34.8$ | M1, A1 | For selecting appropriate model; for test statistic awrt $34.8$ |
| $\chi_{24}^2\ (10\%)$ 2-tail CR: $\chi_{24}^2 < 13.848$ or $\chi_{24}^2 > 36.415$ | B1 | For at least one correct critical value |
| $34.8$ is not significant so insufficient evidence that $\sigma \neq 0.175$ | A1 | For a correct conclusion confirming that assuming st. dev $= 0.175$ is OK |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{"1.28"} \pm z\times\frac{0.175}{\sqrt{25}}$ | M1 | For use of correct formula with $1.6 < z < 2$ (ft $\bar{x}$ if found in (a)) |
| $z = 1.96$ | B1 | For $z=1.96$ or better used |
| $= (1.211\ldots,\ 1.349\ldots)$ awrt **(1.21, 1.35)** | A1 | For an interval awrt $(1.21, 1.35)$ |
---6 A new employee, Kim, joins an existing employee, Jiang, to work in the quality control department of a company producing steel rods.\\
Each day a random sample of rods is taken, their lengths measured and a $95 \%$ confidence interval for the mean length of the rods, in metres, is calculated. It is assumed that the lengths of the rods produced are normally distributed.
Kim took a random sample of 25 rods and used the $t$ distribution to obtain a $95 \%$ confidence interval of $( 1.193,1.367 )$ for the mean length of the rods. Jiang commented that this interval was a little wider than usual and explained that they usually assume that the standard deviation does not change and can be taken as 0.175 metres.\\
(a) Test, at the $10 \%$ level of significance, whether or not Kim's sample suggests that the standard deviation is different from 0.175 metres. State your hypotheses clearly.
Using Kim's sample and the normal distribution with a standard deviation of 0.175 metres, (b) find a 95\% confidence interval for the mean length of the rods.
\hfill \mbox{\textit{Edexcel FS2 2020 Q6 [12]}}