| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2020 |
| Session | June |
| Marks | 1 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Larger contingency table (4+ categories) |
| Difficulty | Moderate -0.8 This is a straightforward recall question requiring only the formula for degrees of freedom in a chi-squared test: (r-1)(c-1) = (3-1)(5-1) = 8. It requires no problem-solving, just direct application of a standard formula to count rows and columns. |
| Spec | 5.06a Chi-squared: contingency tables |
| Number of times late | |||||
| 0 | 1 | 2 | 3 | 4 | |
| A | 8.12 | 14 | 15.12 | 14 | 4.76 |
| Class B | 8.99 | 15.5 | 16.74 | 15.5 | 5.27 |
| C | 11.89 | 20.5 | 22.14 | 20.5 | 6.97 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | Circles correct answer. | 1.1b |
| Total | 1 | |
| Q | Marking Instructions | AO |
Question 2:
2 | Circles correct answer. | 1.1b | B1 | 6
Total | 1
Q | Marking Instructions | AO | Marks | Typical SolutionA $\chi^2$ test is carried out in a school to test for association between the class a student belongs to and the number of times they are late to school in a week.
The contingency table below gives the expected values for the test.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\multicolumn{6}{|c|}{Number of times late} \\
\hline
& 0 & 1 & 2 & 3 & 4 \\
\hline
A & 8.12 & 14 & 15.12 & 14 & 4.76 \\
\hline
Class B & 8.99 & 15.5 & 16.74 & 15.5 & 5.27 \\
\hline
C & 11.89 & 20.5 & 22.14 & 20.5 & 6.97 \\
\hline
\end{tabular}
\end{center}
Find a possible value for the degrees of freedom for the test.
Circle your answer.
[1 mark]
6 \quad 8 \quad 12 \quad 15
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2020 Q2 [1]}}