| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2023 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Standard +0.3 This is a straightforward S3 hypothesis testing question with standard coded data manipulation. Part (a) involves routine mean/variance calculations from coded data using given formulas, (b)-(d) are textbook hypothesis test steps with critical values from tables, and (e) requires a simple justification about large sample size. All techniques are standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{y} = \frac{847}{100} = 8.47\) or \(847 + 100 \times 1000 = 100847\) | M1 | For 8.47 or \(\frac{847}{100}\) or \(847 + 100 \times 1000\) or \(847 = \sum x - 100 \times 1000\) or 100847 seen |
| \(\bar{x} = 1000 + \frac{847}{100} = 1008.47\) | A1* | cso correct solution including \(\bar{x} = ..\) and \(...= 1008.47\); allow alt notation for \(\bar{x}\) but must refer to \(x\) not \(y\); must not be just \(x\) e.g. \(E(X)\), \(\mu_x\), mean of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s_x^2 = s_y^2 = \frac{13510.09 - 100 \times \text{"8.47"}^2}{99}\) or \(s_x^2 = \frac{101707510.1 - \frac{\text{"100847"}^2}{100}}{99}\) | M1 | For a correct expression ft their 100847; allow answer of 1064 |
| \(= 64\) | A1 | Cao do not ISW; allow 64.00... |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_x = 1010\), \(H_1: \mu_x \neq 1010\) | B1 | Both hypotheses correct; must be in terms of \(\mu\); allow \(H_0: \mu_y = 10\), \(H_1: \mu_y \neq 10\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\bar{X} - 1010}{\text{"8"}/\sqrt{100}} = -1.96\) or \(\frac{\bar{X} - 1010}{\text{"8"}/\sqrt{100}} = 1.96\) | M1 B1 | For \(\pm\) standardisation with 1010 and their sd; allow equivalent e.g. \(1010 \pm n \times \text{"8"}/\sqrt{100}\); SC condone use of 1008.47 for 1010 or for \(\bar{X}\); for c.v. = \(\pm 1.96\) or better (Calculator gives 1.95996...); condone 1.6449 or better if they have a one tail hypothesis in (b) |
| \(\bar{X} = 1008.432\) and \(\bar{X} = 1011.568\), awrt 1008 and 1012 (or 1011) | A1 | For both limits 1008 or better and 1012 or better seen (condone 1011 from correct working) |
| \(\bar{X} \leq \text{"1008.432"}\) or \(\bar{X} \geq \text{"1011.568"}\) | A1ft | For selecting the correct region ft their figures (not \(z\) value); allow use of \(<\) and \(>\); allow other letters (condone \(\mu\)); allow other notation e.g. \([1012, \infty]\), \((\infty, 1008]\) etc. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 1008.47 is not in the critical region | M1 | ft their CR if the final A mark in part (c) is awarded; must refer to 1008.47; allow writing in the form "1008.432" \(< 1008.47 <\) "1011.568" etc but if in middle it must have both ends; if no clear CR it is M0A0 |
| The machine does not need to be stopped/reset | A1ft | dep on M1; correct conclusion consistent with comparing 1008.47 with their CR; if in the CR they must say it needs to be reset/stopped; if not in the CR it must say it does not need to be stopped/reset |
| SC: if CR is of the form "1008.432" \(< \bar{X} <\) "1011.568" then award M0A1 for concluding the machine does not need to be stopped/reset | SC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| It is reasonable since the sample size is (reasonably) large | B1 | Any suitable comment about the sample being large e.g. \(n\) is large |
# Question 1:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{y} = \frac{847}{100} = 8.47$ or $847 + 100 \times 1000 = 100847$ | M1 | For 8.47 or $\frac{847}{100}$ or $847 + 100 \times 1000$ or $847 = \sum x - 100 \times 1000$ or 100847 seen |
| $\bar{x} = 1000 + \frac{847}{100} = 1008.47$ | A1* | cso correct solution including $\bar{x} = ..$ and $...= 1008.47$; allow alt notation for $\bar{x}$ but must refer to $x$ not $y$; must not be just $x$ e.g. $E(X)$, $\mu_x$, mean of $x$ |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s_x^2 = s_y^2 = \frac{13510.09 - 100 \times \text{"8.47"}^2}{99}$ or $s_x^2 = \frac{101707510.1 - \frac{\text{"100847"}^2}{100}}{99}$ | M1 | For a correct expression ft their 100847; allow answer of 1064 |
| $= 64$ | A1 | Cao do not ISW; allow 64.00... |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_x = 1010$, $H_1: \mu_x \neq 1010$ | B1 | Both hypotheses correct; must be in terms of $\mu$; allow $H_0: \mu_y = 10$, $H_1: \mu_y \neq 10$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\bar{X} - 1010}{\text{"8"}/\sqrt{100}} = -1.96$ or $\frac{\bar{X} - 1010}{\text{"8"}/\sqrt{100}} = 1.96$ | M1 B1 | For $\pm$ standardisation with 1010 and their sd; allow equivalent e.g. $1010 \pm n \times \text{"8"}/\sqrt{100}$; SC condone use of 1008.47 for 1010 or for $\bar{X}$; for c.v. = $\pm 1.96$ or better (Calculator gives 1.95996...); condone 1.6449 or better if they have a one tail hypothesis in (b) |
| $\bar{X} = 1008.432$ and $\bar{X} = 1011.568$, awrt 1008 and 1012 (or 1011) | A1 | For **both** limits 1008 or better **and** 1012 or better seen (condone 1011 from correct working) |
| $\bar{X} \leq \text{"1008.432"}$ or $\bar{X} \geq \text{"1011.568"}$ | A1ft | For selecting the correct region ft their figures (not $z$ value); allow use of $<$ and $>$; allow other letters (condone $\mu$); allow other notation e.g. $[1012, \infty]$, $(\infty, 1008]$ etc. |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| 1008.47 is not in the critical region | M1 | ft their CR if the final A mark in part (c) is awarded; must refer to 1008.47; allow writing in the form "1008.432" $< 1008.47 <$ "1011.568" etc but if in middle it must have both ends; if no clear CR it is M0A0 |
| The machine does not need to be stopped/reset | A1ft | dep on M1; correct conclusion consistent with comparing 1008.47 with their CR; if in the CR they must say it needs to be reset/stopped; if not in the CR it must say it does not need to be stopped/reset |
| SC: if CR is of the form "1008.432" $< \bar{X} <$ "1011.568" then award M0A1 for concluding the machine does not need to be stopped/reset | SC | |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| It is reasonable since the sample size is (reasonably) large | B1 | Any suitable comment about the sample being large e.g. $n$ is large |
---1 A machine fills bottles with mineral water.\\
The machine is checked every day to ensure that it is working correctly. On a particular day a random sample of 100 bottles is taken. The volume of water, $x$ millilitres, for each bottle is measured and each measurement is coded using
$$y = x - 1000$$
The results are summarised below
$$\sum y = 847 \quad \sum y ^ { 2 } = 13510.09$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the value of the unbiased estimate of the mean of $x$ is 1008.47
\item Calculate the unbiased estimate of the variance of $x$
The machine was initially set so that the volume of water in a bottle had a mean value of 1010 millilitres.
Later, a test at the $5 \%$ significance level is used to determine whether or not the mean volume of water in a bottle has changed. If it has changed then the machine is stopped and reset.
\end{enumerate}\item Write down suitable null and alternative hypotheses for a 2-tailed test.
\item Find the critical region for $\bar { X }$ in the above test.
\item Using your answer to part (a) and your critical region found in part (c), comment on whether or not the machine needs to be stopped and reset.\\
Give a reason for your answer.
\item Explain why the use of $\sigma ^ { 2 } = s ^ { 2 }$ is reasonable in this situation.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2023 Q1 [12]}}