| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample z-test large samples |
| Difficulty | Standard +0.3 This is a standard two-sample t-test question with clearly provided summary statistics. Part (a) requires a two-tailed test with routine hypothesis setup and calculation, while part (b) involves interpreting a given test statistic for a one-tailed test. The question is slightly easier than average because the data is pre-summarized, the procedure is well-rehearsed in S3, and no unusual complications arise—though it does require understanding the distinction between one-tailed and two-tailed tests. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Sample Size, \(n\) | Mean, \(\bar { x }\) | Standard Deviation, \(s\) | |
| Mr Alan | 29 | 80 | 10 |
| Ms Burns | 26 | 74 | 15 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \mu_A = \mu_B\); \(H_1: \mu_A \neq \mu_B\) | B1 | |
| \(z = \frac{\pm(80-74)}{\sqrt{\frac{100}{29} + \frac{225}{26}}}\) | M1A1 | M1 for attempt at s.e. (condone one number wrong) and using s.e. in correct formula; A1 for correct expression |
| \(z = \pm 1.7247...\) awrt \(\pm 1.72\) | A1 | |
| \(1.7247 > 1.6449\) so reject \(H_0\) | dM1 | Dependent on 1st M1; correct statement linking test statistic and cv |
| There is evidence of a difference in the (mean) scores of their students | A1 | Must mention "scores" and "students/groups/classes"; A0 for one-tailed comment |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(z=1.6\), test not significant so no evidence of a difference | B1 | |
| For Mr A's claim: \(H_0: \mu_A = \mu_B\), \(H_1: \mu_A > \mu_B\), critical value \(z = 1.2816\) | B1, B1 | Alternative hypothesis should be clearly defined |
| Both \(z\) values significant, Mr Alan's claim supported | B1 |
# Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu_A = \mu_B$; $H_1: \mu_A \neq \mu_B$ | B1 | |
| $z = \frac{\pm(80-74)}{\sqrt{\frac{100}{29} + \frac{225}{26}}}$ | M1A1 | M1 for attempt at s.e. (condone one number wrong) and using s.e. in correct formula; A1 for correct expression |
| $z = \pm 1.7247...$ awrt $\pm 1.72$ | A1 | |
| $1.7247 > 1.6449$ so reject $H_0$ | dM1 | Dependent on 1st M1; correct statement linking test statistic and cv |
| There is evidence of a difference in the (mean) scores of their students | A1 | Must mention "scores" and "students/groups/classes"; A0 for one-tailed comment |
---
# Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $z=1.6$, test not significant so no evidence of a difference | B1 | |
| For Mr A's claim: $H_0: \mu_A = \mu_B$, $H_1: \mu_A > \mu_B$, critical value $z = 1.2816$ | B1, B1 | Alternative hypothesis should be clearly defined |
| Both $z$ values significant, Mr Alan's claim supported | B1 | |
---5. Mr Alan and Ms Burns are two Mathematics teachers teaching mixed ability groups of students in a large college. At the end of the college year all students took the same examination. A random sample of 29 of Mr Alan's students and a random sample of 26 of Ms Burns' students are chosen. The results are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& Sample Size, $n$ & Mean, $\bar { x }$ & Standard Deviation, $s$ \\
\hline
Mr Alan & 29 & 80 & 10 \\
\hline
Ms Burns & 26 & 74 & 15 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, test, at the $10 \%$ level of significance whether there is evidence that there is a difference in the mean scores of their students.
Ms Burns thinks the comparison was unfair as the examination was set by Mr Alan. She looks up a different set of examination results for these students and, although Mr Alan's sample has a higher mean, she calculates the test statistic for this new set of results to be 1.6
However, Mr Alan now claims that the mean marks of his students are higher than the mean marks of Ms Burns' students.
\item Test Mr Alan's claim, stating the hypotheses and critical values you would use. Use a $10 \%$ level of significance.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2012 Q5 [9]}}