| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (upper tail) |
| Difficulty | Standard +0.3 This is a straightforward one-sample z-test with standard calculations (mean, variance from summary statistics, test statistic, p-value) followed by routine interpretation questions about CLT, sampling, and test selection. While it's Further Maths content, the mechanics are entirely standard with no novel problem-solving required, making it slightly easier than average overall. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \frac{2163}{50} = 43.26\) | B1 | |
| \(s^2 = \frac{1}{49} \times \left(98508 - \frac{2163^2}{50}\right)\) | M1 | |
| \(s^2 = 100.7473...\) or \(s = 10.0(3729779...)\) | A1 | si |
| \(H_0: \mu = 38\), \(H_1: \mu > 38\) | B1 | |
| Under \(H_0\), \(\bar{X} \sim N\left(38, \frac{100.747...}{50}\right)\) | ||
| \(p\text{-value} = P(\bar{X} > 43.26 \mid H_0 \text{ is true})\) | M1 | Alternative: Test statistic \(= \frac{43.26-38}{10.03729779.../\sqrt{50}}\) |
| \(p\text{-value} = 0.000105\) | A1 | A1 \(p\)-value from tables \(= 0.00010\) |
| Since \(p << 0.05\) there is strong evidence to reject \(H_0\) | m1 | |
| There is strong evidence to reject the laboratory's claim that the average time taken for test results to be returned is 38 hours. | A1 | FT from their \(p\)-value and corresponding conclusion |
| Valid headline implying failure on part of laboratory. e.g. Lab lets down patients / Laboratory takes longer than claimed to process test results. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Valid explanation. e.g. Because \(n\) is large, the central limit theorem allows us to use the normal distribution. / Because \(n\) is large, the CLT allows us to assume that the distribution of the sample mean is normal. | E1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Valid explanation. e.g. Random sampling eliminates the bias that may occur from taking a batch from, say, the same day. / 50 consecutive results might all come from a time when the process is having a good, or bad, run. Randomisation avoids this. | E1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A \(t\)-test because the sample size is small. | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The assumption would be that the time taken for results to be returned is normally distributed. | E1 | (2) |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \frac{2163}{50} = 43.26$ | B1 | |
| $s^2 = \frac{1}{49} \times \left(98508 - \frac{2163^2}{50}\right)$ | M1 | |
| $s^2 = 100.7473...$ or $s = 10.0(3729779...)$ | A1 | si |
| $H_0: \mu = 38$, $H_1: \mu > 38$ | B1 | |
| Under $H_0$, $\bar{X} \sim N\left(38, \frac{100.747...}{50}\right)$ | | |
| $p\text{-value} = P(\bar{X} > 43.26 \mid H_0 \text{ is true})$ | M1 | Alternative: Test statistic $= \frac{43.26-38}{10.03729779.../\sqrt{50}}$ |
| $p\text{-value} = 0.000105$ | A1 | A1 $p$-value from tables $= 0.00010$ |
| Since $p << 0.05$ there is strong evidence to reject $H_0$ | m1 | |
| There is strong evidence to reject the laboratory's claim that the average time taken for test results to be returned is 38 hours. | A1 | FT from their $p$-value and corresponding conclusion |
| Valid headline implying failure on part of laboratory. e.g. Lab lets down patients / Laboratory takes longer than claimed to process test results. | B1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Valid explanation. e.g. Because $n$ is large, the central limit theorem allows us to use the normal distribution. / Because $n$ is large, the CLT allows us to assume that the distribution of the sample mean is normal. | E1 | (1) |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Valid explanation. e.g. Random sampling eliminates the bias that may occur from taking a batch from, say, the same day. / 50 consecutive results might all come from a time when the process is having a good, or bad, run. Randomisation avoids this. | E1 | (1) |
### Part (d)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A $t$-test because the sample size is small. | E1 | |
### Part (d)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The assumption would be that the time taken for results to be returned is normally distributed. | E1 | (2) |
---5. A laboratory carrying out screening for a certain blood disorder claims that the average time taken for test results to be returned is 38 hours. A reporter for a national newspaper suspects that the results take longer, on average, to be returned than claimed by the laboratory. The reporter finds the time, $x$ hours, for 50 randomly selected results, in order to conduct a hypothesis test. The following summary statistics were obtained.
$$\sum x = 2163 \quad \sum x ^ { 2 } = 98508$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the $p$-value for the reporter's hypothesis test, and complete the test using a $5 \%$ level of significance. Hence write a headline for the reporter to use.
\item Explain the relevance or otherwise of the Central Limit Theorem to your answer in part (a).
\item Briefly explain why a random sample is preferable to taking a batch of 50 consecutive results.
\item On another occasion, the reporter took a different random sample of 10 results.
\begin{enumerate}[label=(\roman*)]
\item State, with a reason, what type of hypothesis test the reporter should use on this occasion.
\item State one assumption required to carry out this test.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 2022 Q5 [13]}}