| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Cell combining required |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with straightforward hypotheses, routine expected frequency calculation (row total × column total / grand total), standard degrees of freedom formula (r-1)(c-1), and table lookup for critical value comparison. All steps are textbook procedures requiring only recall and basic arithmetic, making it slightly easier than average for A-level. |
| Spec | 5.06a Chi-squared: contingency tables |
| \backslashbox{Age \(\boldsymbol { a }\) years}{Activity} | Badminton | Bowls | Snooker |
| \(a < 20\) | 9 | 3 | 3 |
| \(20 \leqslant a < 40\) | 10 | 10 | 14 |
| \(40 \leqslant a < 50\) | 16 | 15 | 5 |
| \(50 \leqslant a < 60\) | 15 | 13 | 11 |
| \(a \geqslant 60\) | 4 | 19 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \(H_0\): There is no association between age and activity; \(H_1\): There is an association between age and activity | B1 |
| (b) | \(\frac{26 \times 36}{150} = 6.24\) | B1 |
| (c) | Since expected value in \(a < 20\) and snooker \(= 3.6 < 5\) we amalgamate two rows. Table is now \(4 \times 3\) so degrees of freedom \(= (4-1) \times (3-1) = 6\) (*) | B1, B1* |
| (d) | Critical value \(\chi_c^2(5\%) = 12.592\). [Significant result]: so there is evidence to support manager's belief | B1, B1ft |
| (6 marks) |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $H_0$: There is no association between age and activity; $H_1$: There is an association between age and activity | B1 | B1 for both hypotheses in terms of "association" or "independence". Must mention age and activity (or sport). Use of "relationship" or "link" here is B0 but allow for last B1ft |
| (b) | $\frac{26 \times 36}{150} = 6.24$ | B1 | B1 for 6.24 |
| (c) | Since expected value in $a < 20$ and snooker $= 3.6 < 5$ we amalgamate two rows. Table is now $4 \times 3$ so degrees of freedom $= (4-1) \times (3-1) = 6$ (*) | B1, B1* | 1st B1 for a reason to get a $4 \times 3$ table based on amalgamation of rows. Must mention $a < 20$ and snooker and see 3.6 and be combining rows (not columns). 2nd B1* for 6 degrees of freedom. [ $8 - 2 = 6$ is B0 ] |
| (d) | Critical value $\chi_c^2(5\%) = 12.592$. [Significant result]: so there is evidence to support manager's belief | B1, B1ft | 1st B1 for correct critical value (allow 12.6 or 12.59 or awrt 12.592). NB $p$-value $= 0.0032839...$ so allow awrt 0.00328. 2nd B1ft for a correct comparison and conclusion (fit their cv). [Independent of hypotheses]. e.g. there is an "association" or "relationship" or "link" between age and activity is OK BUT there is a "correlation" between age and activity is B0. Do not accept contradictory contextual statements e.g. "manager's belief supported, there is no association between age and activity" |
| | | **(6 marks)** | |\begin{enumerate}
\item A leisure club offers a choice of one of three activities to its 150 members on a Tuesday evening. The manager believes that there may be an association between the choice of activity and the age of the member and collected the following data.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
\backslashbox{Age $\boldsymbol { a }$ years}{Activity} & Badminton & Bowls & Snooker \\
\hline
$a < 20$ & 9 & 3 & 3 \\
\hline
$20 \leqslant a < 40$ & 10 & 10 & 14 \\
\hline
$40 \leqslant a < 50$ & 16 & 15 & 5 \\
\hline
$50 \leqslant a < 60$ & 15 & 13 & 11 \\
\hline
$a \geqslant 60$ & 4 & 19 & 3 \\
\hline
\end{tabular}
\end{center}
(a) Write down suitable hypotheses for a test of the manager's belief.
The manager calculated expected frequencies to use in the test.\\
(b) Calculate the expected frequency of members aged 60 or over who choose snooker, used by the manager.\\
(c) Explain why there are 6 degrees of freedom used in this test.
The test statistic used to test the manager's belief is 19.583\\
(d) Using a 5\% level of significance, complete the test of the manager's belief.
\hfill \mbox{\textit{Edexcel FS1 AS 2019 Q1 [6]}}