| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Moderate -0.3 This is a straightforward one-sample z-test with clearly stated hypotheses (H₀: μ=30 vs H₁: μ≠30), standard test statistic calculation z = (28.2-30)/(8.5/√75), and comparison to critical values at 10% significance. Parts (b) and (c) test standard bookwork about CLT justifying normal approximation and the assumption that population variance equals sample variance. While it requires understanding of hypothesis testing framework, all steps are routine applications of S3 content with no problem-solving insight needed. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.01a Permutations and combinations: evaluate probabilities5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu = 30\), \(H_1: \mu \neq 30\) | B1 | Both hypotheses correct |
| \(z = \dfrac{28.2 - 30}{\frac{8.5}{\sqrt{75}}} = -1.833936...\) | M1 | For standardising with 28.2, 30 and \(\frac{8.5}{\sqrt{75}}\) (or awrt 0.981). Allow use of \(8.5 \times \sqrt{\frac{74}{75}}\) (= awrt 8.44) |
| awrt \(-1.83\) | A1 | |
| Two tailed c.v.'s \(Z = \pm 1.6449\), or CR: \(Z \leq -1.6449\) or \(Z \geq 1.6449\), or p-value = awrt 0.033 or awrt 0.034 \(< 0.05\) | B1 | Critical value of \(-1.6449\) (compatible with sign of test statistic) or correct probability comparison |
| Conclude: mean age of gym customers is not 30 years / manager's claim is not correct | A1 | Dependent on M1; correct contextualised rejection of \(H_0\). Contradictory statements score A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-1.6449 = \dfrac{\bar{X}_c - 30}{\frac{8.5}{\sqrt{75}}}\), so \(\bar{X}_c = 28.38883812...\) | M1/A1 | M1: For \(\dfrac{c-30}{\frac{8.5}{\sqrt{75}}} = -1.6449/-1.645/-1.64/-1.65\); A1: \(\bar{X}_c =\) awrt 28.4 |
| Critical value of \(-1.6449\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{X}\) is (approximately) normally distributed | B1 | Allow in words e.g. "sample mean is normally distributed" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assumed \(s^2 = \sigma^2\) or variance of sample = variance of population | B1 | Also allow \(s = \sigma\) or standard deviation of sample = standard deviation of population |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 30$, $H_1: \mu \neq 30$ | B1 | Both hypotheses correct |
| $z = \dfrac{28.2 - 30}{\frac{8.5}{\sqrt{75}}} = -1.833936...$ | M1 | For standardising with 28.2, 30 and $\frac{8.5}{\sqrt{75}}$ (or awrt 0.981). Allow use of $8.5 \times \sqrt{\frac{74}{75}}$ (= awrt 8.44) |
| awrt $-1.83$ | A1 | |
| Two tailed c.v.'s $Z = \pm 1.6449$, or CR: $Z \leq -1.6449$ or $Z \geq 1.6449$, or p-value = awrt 0.033 or awrt 0.034 $< 0.05$ | B1 | Critical value of $-1.6449$ (compatible with sign of test statistic) **or** correct probability comparison |
| Conclude: mean age of gym customers is not 30 years / manager's claim is not correct | A1 | Dependent on M1; correct contextualised rejection of $H_0$. Contradictory statements score A0 |
**Alternative method (M1A1B1):** Let $\bar{X}_c$ be critical value of sample mean.
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-1.6449 = \dfrac{\bar{X}_c - 30}{\frac{8.5}{\sqrt{75}}}$, so $\bar{X}_c = 28.38883812...$ | M1/A1 | M1: For $\dfrac{c-30}{\frac{8.5}{\sqrt{75}}} = -1.6449/-1.645/-1.64/-1.65$; A1: $\bar{X}_c =$ awrt 28.4 |
| Critical value of $-1.6449$ | B1 | |
**Note:** One tailed test scores maximum B0M1A1B1A0. Allow $\pm 1.2816$ to score 2nd B1.
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{X}$ is (approximately) normally distributed | B1 | Allow in words e.g. "sample mean is normally distributed" |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assumed $s^2 = \sigma^2$ or variance of sample = variance of population | B1 | Also allow $s = \sigma$ or standard deviation of sample = standard deviation of population |
---3. The manager of a gym claimed that the mean age of its customers is 30 years. A random sample of 75 customers is taken and their ages have a mean of 28.2 years and a standard deviation, $s$, of 8.5 years.
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly and using a 10\% level of significance, test whether or not the manager's claim is supported by the data.
\item Explain the relevance of the Central Limit Theorem to your calculation in part (a).
\item State an additional assumption needed to carry out the test in part (a).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2017 Q3 [7]}}