| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-sample t-test with summary statistics |
| Difficulty | Standard +0.3 This is a standard two-sample z-test with large samples (n=300, 400) where students apply a routine procedure: state hypotheses, calculate test statistic using given means and standard deviations, compare to critical value, and conclude. Part (b) requires recalling that CLT justifies normality for large samples. While it involves multiple steps, it's a textbook application with no novel insight required, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| \(N\) | mean | s.d. | |
| Low income group | 300 | £6.40 | £6.69 |
| High income group | 400 | £7.42 | £8.13 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0 : \mu_L = \mu_H; H_1 : \mu_L \neq \mu_H\) | B1 B1 | |
| \(s.e. = \sqrt{\frac{8.13^2}{400} + \frac{6.69^2}{300}}\) | M1 | Substitute into s.e. |
| \(= 0.5607\) | A1 | Complete correct expression |
| AWRT 0.561 | ||
| \(\alpha = 0.05 \Rightarrow C.R: z < -1.96\) or \(z > 1.96\) | B1 | \(\pm 1.96\) |
| Test statistic: \(z = \frac{6.40 - 7.42}{0.5607} = -1.819\) | M1 | \(\left(\bar{x}_L - \bar{x}_H\right) / \) their s.e. |
| AWRT \(\pm 1.82\) | ||
| A1 | ||
| Since \(-1.819\) is not in the critical region then there is no evidence to reject \(H_0\) and thus it can be concluded that there is no difference in mean expenditure on tobacco | A1 ft | |
| (9 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| C. L. Theorem enables use of \(\bar{L}\) ~ Normal and \(\bar{H}\) ~ Normal. | B1 | \(\bar{L}\) or \(\bar{H}\) |
| B1 | Normal | |
| (2 marks) |
## (a)
$H_0 : \mu_L = \mu_H; H_1 : \mu_L \neq \mu_H$ | B1 B1 |
$s.e. = \sqrt{\frac{8.13^2}{400} + \frac{6.69^2}{300}}$ | M1 | Substitute into s.e.
$= 0.5607$ | A1 | Complete correct expression
| | AWRT 0.561
$\alpha = 0.05 \Rightarrow C.R: z < -1.96$ or $z > 1.96$ | B1 | $\pm 1.96$
Test statistic: $z = \frac{6.40 - 7.42}{0.5607} = -1.819$ | M1 | $\left(\bar{x}_L - \bar{x}_H\right) / $ their s.e.
| | AWRT $\pm 1.82$
| | A1 |
Since $-1.819$ is not in the critical region then there is no evidence to reject $H_0$ and thus it can be concluded that there is no difference in mean expenditure on tobacco | A1 ft |
| (9 marks) |
## (b)
C. L. Theorem enables use of $\bar{L}$ ~ Normal and $\bar{H}$ ~ Normal. | B1 | $\bar{L}$ or $\bar{H}$
| | B1 | Normal
| (2 marks) |A sociologist was studying the smoking habits of adults. A random sample of 300 adult smokers from a low income group and an independent random sample of 400 adult smokers from a high income group were asked what their weekly expenditure on tobacco was. The results are summarised below.
\begin{center}
\begin{tabular}{|l|c|c|c|}
\hline
& $N$ & mean & s.d. \\
\hline
Low income group & 300 & £6.40 & £6.69 \\
\hline
High income group & 400 & £7.42 & £8.13 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Using a 5\% significance level, test whether or not the two groups differ in the mean amounts spent on tobacco. [9]
\item Explain briefly the importance of the central limit theorem in this example. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q6 [11]}}