| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample t-test (unknown variances) |
| Difficulty | Standard +0.3 This is a standard two-sample t-test with large sample sizes (n₁=605, n₂=45), making it straightforward. Part (a) follows a routine hypothesis testing procedure with clear numerical values. Part (b) requires recall of standard assumptions. Part (c) is a simple variance update calculation using the formula for combining samples. The large sample size for full-time employees makes the test robust and the calculations mechanical. This is slightly easier than average because it's a textbook application with no conceptual challenges or unusual features. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| \(n\) | \(\bar { t }\) | \(s ^ { 2 }\) | |
| Full-time employees | 605 | 5.6 | 9 |
| Part-time employees | 45 | 7.0 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_p - \mu_f = 1\) | B1 | Correct; must be in terms of \(\mu\); condone other letters if defined |
| \(H_1: \mu_p - \mu_f > 1\) | B1 | Correct; must be in terms of \(\mu\) |
| s.e. \(= \sqrt{\dfrac{9}{605} + \dfrac{4}{45}} = \sqrt{0.10376...} = [0.322...]\) | M1 | Correct method for s.e.; may be seen within standardisation formula |
| \(z = \dfrac{7.0 - 5.6 - 1}{\sqrt{\dfrac{9}{605} + \dfrac{4}{45}}}\) | dM1 | Dependent on previous M; correct method for \(z\) |
| \(= 1.24175...\) awrt \(1.24\) | A1 | — |
| CV 5% one-tailed \(= \pm 1.6449\) | B1 | awrt \(\pm 1.6449\) or \(p\)-value awrt \(0.107 > 0.05\) |
| Not significant, do not reject \(H_0\) | dM1 | All previous method marks awarded; correct conclusion |
| Insufficient evidence that full-time staff are more than one minute faster than part-time staff / manager's claim not supported | A1ft | Dependent on all previous method marks; words in bold required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assume both samples are normal or both large enough for CLT | B1 | One correct assumption |
| Assume \(s^2 = \sigma^2\) for both samples / individual results are independent | B1 | Second correct assumption |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{a} = \dfrac{45 \times 7 + 8}{46} = 7.0217...\) | M1 | Correct method |
| \(\sum a^2 = 44 \times 4 + 45 \times 7^2 + 8^2 = 2445\) | M1 | Correct method |
| \(s^2 = \dfrac{2445 - 46 \times (7.0217...)^2}{45}\) | M1 | Correct method for \(s^2\) |
| \(= 3.93285...\) awrt \(3.93\) | A1 | — |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_p - \mu_f = 1$ | B1 | Correct; must be in terms of $\mu$; condone other letters if defined |
| $H_1: \mu_p - \mu_f > 1$ | B1 | Correct; must be in terms of $\mu$ |
| s.e. $= \sqrt{\dfrac{9}{605} + \dfrac{4}{45}} = \sqrt{0.10376...} = [0.322...]$ | M1 | Correct method for s.e.; may be seen within standardisation formula |
| $z = \dfrac{7.0 - 5.6 - 1}{\sqrt{\dfrac{9}{605} + \dfrac{4}{45}}}$ | dM1 | Dependent on previous M; correct method for $z$ |
| $= 1.24175...$ awrt $1.24$ | A1 | — |
| CV 5% one-tailed $= \pm 1.6449$ | B1 | awrt $\pm 1.6449$ or $p$-value awrt $0.107 > 0.05$ |
| Not significant, do not reject $H_0$ | dM1 | All previous method marks awarded; correct conclusion |
| Insufficient evidence that full-time staff are more than one minute faster than part-time staff / manager's claim not supported | A1ft | Dependent on all previous method marks; words in bold required |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume **both** samples are **normal** or both large enough for CLT | B1 | One correct assumption |
| Assume $s^2 = \sigma^2$ for **both** samples / individual results are independent | B1 | Second correct assumption |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{a} = \dfrac{45 \times 7 + 8}{46} = 7.0217...$ | M1 | Correct method |
| $\sum a^2 = 44 \times 4 + 45 \times 7^2 + 8^2 = 2445$ | M1 | Correct method |
| $s^2 = \dfrac{2445 - 46 \times (7.0217...)^2}{45}$ | M1 | Correct method for $s^2$ |
| $= 3.93285...$ awrt $3.93$ | A1 | — |\begin{enumerate}
\item A manager of a large company is investigating the time it takes the company's employees to complete a task.
\end{enumerate}
The manager believes that the mean time for full-time employees to complete the task is more than a minute quicker than the mean time for part-time employees to complete the task.
The manager collects a random sample of 605 full-time employees and 45 part-time employees and records the times, $t$ minutes, it takes each employee to complete the task.
The results are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& $n$ & $\bar { t }$ & $s ^ { 2 }$ \\
\hline
Full-time employees & 605 & 5.6 & 9 \\
\hline
Part-time employees & 45 & 7.0 & 4 \\
\hline
\end{tabular}
\end{center}
(a) Test, at the $5 \%$ level of significance, the manager's claim.
You should state your hypotheses, test statistic, critical value and conclusion clearly.\\
(b) State two assumptions you have made in carrying out the test in part (a)
The company increases the size of the sample of part-time employees to 46 The time taken to complete the task by the extra employee is 8 minutes.\\
(c) Find an unbiased estimate of the variance for the sample of 46 part-time employees.
\hfill \mbox{\textit{Edexcel S3 2024 Q5 [14]}}