| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2016 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample t-test |
| Difficulty | Standard +0.3 This is a straightforward paired t-test question with standard bookwork components. Part (i) requires identifying experimental design flaws (routine critical thinking), part (ii) is a standard hypothesis test calculation with given data, part (iii) asks for conditions (pure recall), and part (iv) is a routine confidence interval calculation. While it requires multiple techniques, each step follows standard procedures with no novel insight needed, making it slightly easier than average for an S3 question. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Employee | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Old system | 40.5 | 42.9 | 52.8 | 51.7 | 77.2 | 66.7 | 65.2 | 49.2 | 55.6 | 58.3 |
| New system | 39.2 | 40.7 | 50.6 | 50.7 | 71.4 | 70.5 | 71.1 | 47.7 | 52.1 | 55.5 |
| Answer | Marks | Guidance |
|---|---|---|
| A) One set of claim forms could be more difficult to process. | B1 | Allow suitable alternatives |
| B) A form would be more familiar on second processing. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Values of \(d\): 1.3, 2.2, 2.2, 1.0, 5.8, -3.8, -5.9, 1.5, 3.5, 2.8 | M1 | calculating differences (at least 3 correct) |
| \(\bar{d} = 1.06\), \(s = 3.4378\) | A1 | both. Allow \(s^2 = 11.818\). Do not allow \(s_n = 3.2613\) or \(s_n^2 = 10.636\) |
| \(H_0: \mu_D = 0\) | B1 | hypotheses (allow \(<0\) if consistent) |
| \(H_1: \mu_D > 0\) | B1 | definition including difference, mean, and context. Allow other symbols only if they are defined as the population mean difference. Hypotheses in words must include 'population'. Not 'difference of means' |
| \(t = \frac{1.06 - 0}{3.4378/\sqrt{10}} = 0.975\) | M1 | including \(\sqrt{10}\). FT their \(s\) or \(s_n\) |
| 9 degrees of freedom | A1 | cao |
| Critical value \(= 1.833\) | B1 | no FT if wrong |
| \(0.975 < 1.833\) so cannot reject \(H_0\) | B1 | no FT if wrong |
| Insufficient evidence to suggest that the mean time for processing forms has reduced using the new system | M1 | sensible comparison using their test statistic if the previous M1 awarded |
| A1 [10] | including mean and context FT; not assertive but accept 'Evidence suggests mean time has reduced' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differences should be: Normally distributed in population | B1 | or 'underlying distribution of differences is normal' |
| With unknown variance | B1 | NB: candidates may say the sample should be small. This is incorrect, but should be ignored for the purposes of marking. Also ignore 'paired' |
| Sample (of differences) must be random | B1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.06 \pm \frac{3.25(3.4378)}{\sqrt{10}}\) | M1 | for their \(\bar{d}\) and \(s/\sqrt{10}\) in correct position |
| B1 | for 3.25 | |
| \(= (-2.473, 4.593)\) | A1 [3] | cao. SC: Answers from calculator with no working — 3sf or 4sf gets 3/3, \(>\)4sf gets 1/3. \((-4.593, 2.473)\) gets 2/3 |
# Question 4:
## Part i:
A) One set of claim forms could be more difficult to process. | B1 | Allow suitable alternatives
B) A form would be more familiar on second processing. | B1 |
## Question 4ii:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Values of $d$: 1.3, 2.2, 2.2, 1.0, 5.8, -3.8, -5.9, 1.5, 3.5, 2.8 | M1 | calculating differences (at least 3 correct) |
| $\bar{d} = 1.06$, $s = 3.4378$ | A1 | both. Allow $s^2 = 11.818$. Do not allow $s_n = 3.2613$ or $s_n^2 = 10.636$ |
| $H_0: \mu_D = 0$ | B1 | hypotheses (allow $<0$ if consistent) |
| $H_1: \mu_D > 0$ | B1 | definition including difference, mean, and context. Allow other symbols only if they are defined as the population mean difference. Hypotheses in words must include 'population'. Not 'difference of means' |
| $t = \frac{1.06 - 0}{3.4378/\sqrt{10}} = 0.975$ | M1 | including $\sqrt{10}$. FT their $s$ or $s_n$ |
| 9 degrees of freedom | A1 | cao |
| Critical value $= 1.833$ | B1 | no FT if wrong |
| $0.975 < 1.833$ so cannot reject $H_0$ | B1 | no FT if wrong |
| Insufficient evidence to suggest that the mean time for processing forms has reduced using the new system | M1 | sensible comparison using their test statistic if the previous M1 awarded |
| | A1 [10] | including mean and context FT; not assertive but accept 'Evidence suggests mean time has reduced' |
---
## Question 4iii:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differences should be: Normally distributed in population | B1 | or 'underlying distribution of differences is normal' |
| With unknown variance | B1 | NB: candidates may say the sample should be small. This is incorrect, but should be ignored for the purposes of marking. Also ignore 'paired' |
| Sample (of differences) must be random | B1 [3] | |
---
## Question 4iv:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.06 \pm \frac{3.25(3.4378)}{\sqrt{10}}$ | M1 | for their $\bar{d}$ and $s/\sqrt{10}$ in correct position |
| | B1 | for 3.25 |
| $= (-2.473, 4.593)$ | A1 [3] | cao. SC: Answers from calculator with no working — 3sf or 4sf gets 3/3, $>$4sf gets 1/3. $(-4.593, 2.473)$ gets 2/3 |4 An insurance company is investigating a new system designed to reduce the average time taken to process claim forms. The company has decided to use 10 experienced employees to process claims using the old system and the new system.
Two procedures for comparing the systems are proposed.\\
Procedure $A$ There are two sets of claim forms, set 1 and set 2. Each contains the same number of forms. Each employee processes set 1 on the old system and set 2 on the new system. The times taken are compared.
Procedure $B$ There is just one set of claim forms which each employee processes firstly on the old system and then on the new system. The times taken are compared.\\
(i) State one weakness of each of these procedures.
In fact a third procedure which avoids these two weaknesses is adopted. In this procedure each employee is given a randomly selected set of claim forms. Each set contains the same number of forms. The employees each process their set of claim forms on both systems. The times taken, in minutes, are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
Employee & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
Old system & 40.5 & 42.9 & 52.8 & 51.7 & 77.2 & 66.7 & 65.2 & 49.2 & 55.6 & 58.3 \\
\hline
New system & 39.2 & 40.7 & 50.6 & 50.7 & 71.4 & 70.5 & 71.1 & 47.7 & 52.1 & 55.5 \\
\hline
\end{tabular}
\end{center}
(ii) Carry out a paired $t$ test at the $5 \%$ level of significance to investigate whether the mean length of time taken to process a set of forms has reduced using the new system.\\
(iii) State fully the usual conditions for a paired $t$ test.\\
(iv) Construct a $99 \%$ confidence interval for the mean reduction in time taken to process a set of forms using the new system.
\hfill \mbox{\textit{OCR MEI S3 2016 Q4 [18]}}