| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Test statistic given, complete the test |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with clearly structured data requiring routine calculation of expected frequencies, test statistic, and comparison with critical values. Part (b) simply requires looking up the same critical value. While it involves multiple computational steps (11 marks), it follows a completely standard procedure taught in S3 with no conceptual challenges or novel insights required. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 2 - 3 } \multicolumn{1}{c|}{} | Pro-Europe | Eurosceptic |
| \(18 - 34\) years | 43 | 21 |
| \(35 - 54\) years | 30 | 36 |
| 55 years or over | 27 | 43 |
| Answer | Marks |
|---|---|
| giving expected freqs. 32 32 / 33 33 / 35 35 | M1 A2 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_1:\) association between age and attitude to Europe | B1 | |
| \(O\) | \(E\) | \((O-E)\) |
| 43 | 32 | 11 |
| 21 | 32 | \(\neg 11\) |
| 30 | 33 | \(\neg 3\) |
| 36 | 33 | 3 |
| 27 | 35 | \(\neg 8\) |
| 43 | 35 | 8 |
| \(\therefore \sum \frac{(O-E)^2}{E} = 11.765\) | M1 A2 |
| Answer | Marks |
|---|---|
| there is an association between age and attitude to Europe | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| there is no association amongst those who voted, get different result | M1 A1 | (13) |
(a) expected freq. $18\text{-}34/$Pro $= \frac{100 \times 64}{200} = 32$
$35\text{-}54/$Pro $= \frac{100 \times 66}{200} = 33$
giving expected freqs. 32 32 / 33 33 / 35 35 | M1 A2 A1 |
$H_0:$ no association between age and attitude to Europe
$H_1:$ association between age and attitude to Europe | B1 |
| $O$ | $E$ | $(O-E)$ | $\frac{(O-E)^2}{E}$ |
|---|---|---|---|
| 43 | 32 | 11 | 3.7813 |
| 21 | 32 | $\neg 11$ | 3.7813 |
| 30 | 33 | $\neg 3$ | 0.2727 |
| 36 | 33 | 3 | 0.2727 |
| 27 | 35 | $\neg 8$ | 1.8286 |
| 43 | 35 | 8 | 1.8286 |
$\therefore \sum \frac{(O-E)^2}{E} = 11.765$ | M1 A2 |
$v = 2$, $\chi^2_{\text{crit}}(5\%) = 5.991$
$11.765 > 5.991 \therefore$ significant
there is an association between age and attitude to Europe | M1 A1 |
(b) $v = 2$, $\chi^2_{\text{crit}}(5\%) = 5.991$
$4.872 < 5.991 \therefore$ not significant
there is no association amongst those who voted, get different result | M1 A1 | (13)
---5. A Policy Unit wished to find out whether attitudes to the European Union varied with age. It conducted a survey asking 200 individuals to which of three age groups they belonged and whether they regarded themselves as generally pro-Europe or Eurosceptic. The results are shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | }
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & Pro-Europe & Eurosceptic \\
\hline
$18 - 34$ years & 43 & 21 \\
\hline
$35 - 54$ years & 30 & 36 \\
\hline
55 years or over & 27 & 43 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, test at the $5 \%$ level of significance whether attitudes to Europe are associated with age.\\
(11 marks)\\
The survey also asked people if they voted at the last election. When the above test was repeated using only the results from those who had voted a value of 4.872 was calculated for $\sum \frac { ( O - E ) ^ { 2 } } { E }$. No classes were combined.
\item Find if this value leads to a different result.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q5 [13]}}