| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 12 |
| Topic | Type I/II errors and power of test |
| Type | Carry out hypothesis test |
| Difficulty | Moderate -0.8 This is a straightforward hypothesis testing question covering standard bookwork: identifying sampling disadvantages, justifying test choice, stating Type II error probability, and executing a routine z-test. All parts require recall of standard concepts with no problem-solving or novel insight, making it easier than average A-level material. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.05a Hypothesis testing language: null, alternative, p-value, significance2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Similar for other Drs | B1, B1 | Or may not have a good spread of appts |
| Answer | Marks |
|---|---|
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Researcher suspects "more than" 10 mins | B1 | oe |
| [1] |
| Answer | Marks |
|---|---|
| \(0.99\) | B1 |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_1: \mu > 10\) | B1, B1 | One error, eg undefined \(\mu\) or 2-tail: B0B1 |
| \(\overline{X} \sim N\left(10, \frac{3.4}{\sqrt{24}}\right)\) and \(X = \frac{285}{24}\) \((= 11.875)\) | M1 | May be implied |
| Answer | Marks | Guidance |
|---|---|---|
| Reject \(H_0\) | A1, M1, M1 | or 0.003 BC |
| Sufficient evidence that mean time is \(> 10\) mins | A1 | In context. Not definite, |
| Answer | Marks |
|---|---|
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Times spent assumed normally distributed, hence sample mean also normally distributed | B1 | oe |
| [1] |
## 10(i)
Method A: eg May not be representative, e.g. may have many (or few) appointments from one Dr
Or may have many a.m. appts & few p.m.
Method B: eg Time of 1st Dr's 1st & 2nd appointments may not be typical of his later ones.
Similar for other Drs | B1, B1 | Or may not have a good spread of appts
Or other sensible
Or other sensible
| [2]
## 10(ii)
Researcher suspects "more than" 10 mins | B1 | oe
| [1]
## 10(iii)
$0.99$ | B1 |
| [1]
## 10(iv)
$H_0: \mu = 10$ where $\mu$ is pop mean appointment time
$H_1: \mu > 10$ | B1, B1 | One error, eg undefined $\mu$ or 2-tail: B0B1
$\overline{X} \sim N\left(10, \frac{3.4}{\sqrt{24}}\right)$ and $X = \frac{285}{24}$ $(= 11.875)$ | M1 | May be implied
$\text{P}(\overline{X} > 11.875) = 0.00345$ or better
Compare with 0.01
Reject $H_0$ | A1, M1, M1 | or 0.003 BC
Sufficient evidence that mean time is $> 10$ mins | A1 | In context. Not definite,
eg "Mean time is $> 10$ mins": A0
| [7]
## 10(v)
Times spent assumed normally distributed, hence sample mean also normally distributed | B1 | oe
| [1]
---10 A researcher is investigating the actual lengths of time that patients spend at their appointments with the doctors at a certain clinic. There are 12 doctors at the clinic, and each doctor has 24 appointments per day. The researcher plans to choose a sample of 24 appointments on a particular day.\\
(i) The researcher considers the following two methods for choosing the sample.
Method A: Choose a random sample of 24 appointments from the 288 on that day.\\
Method B: Choose one doctor's 1st and 2nd appointments. Choose another doctor's 3rd and 4th appointments and so on until the last doctor's 23rd and 24th appointments.
For each of A and B state a disadvantage of using this method.
Appointments are scheduled to last 10 minutes. The researcher suspects that the actual times that patients spend are more than 10 minutes on average. To test this suspicion, he uses method A , and takes a random sample of 24 appointments. He notes the actual time spent for each appointment and carries out a hypothesis test at the $1 \%$ significance level.\\
(ii) Explain why a 1-tail test is appropriate.
The population mean of the actual times that patients spend at their appointments is denoted by $\mu$ minutes.\\
(iii) Assuming that $\mu = 10$, state the probability that the conclusion of the test will be that $\mu$ is not greater than 10 .
The actual lengths of time, in minutes, that patients spend for their appointments may be assumed to have a normal distribution with standard deviation 3.4.\\[0pt]
(iv) Given that the total length of time spent for the 24 appointments is 285 minutes, carry out the test. [7]\\
(v) In part (iv) it was necessary to use the fact that the sample mean is normally distributed. Give a reason why you know that this is true in this case.
\hfill \mbox{\textit{OCR H240/02 2018 Q10 [12]}}