| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 3×3 contingency table |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with a 3×3 contingency table. Part (a) requires routine calculation of expected frequencies using (row total × column total)/grand total for diagonal cells only. Part (b) is straightforward comparison with critical value from tables. Part (c) asks for contextual interpretation (children winning prizes suggests dependence), which is accessible. Slightly easier than average due to being given the test statistic and only calculating 3 expected frequencies rather than all 9. |
| Spec | 5.06a Chi-squared: contingency tables |
| \backslashbox{2nd colour}{1st colour} | Red | Blue | Yellow | Total |
| Red | 31 | 11 | 18 | 60 |
| Blue | 8 | 10 | 9 | 27 |
| Yellow | 21 | 9 | 33 | 63 |
| Total | 60 | 30 | 60 | 150 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(RR) = \frac{60 \times 60}{150}\) or \(E(BB) = \frac{30 \times 27}{150}\) or \(E(YY) = \frac{60 \times 63}{150}\) | M1 | For at least one correct calculation seen |
| \(E(RR) = 24\) and \(E(BB) = 5.4\) and \(E(YY) = 25.2\) | A1 | For all 3 expected frequencies correct, must be exact |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([E(BB) > 5\) so no need for pooling\(]\) \(v = (3-1)(3-1) = 4\) | B1 | For 4 degrees of freedom |
| Critical value: \(\chi_4^2(5\%) = 9.488\) | B1ft | For awrt 9.488 or ft their degrees of freedom for 5% cv |
| (significant) evidence that the colour of balls is not independent | B1ft | For correct contextual conclusion mentioning "colour" and "independent"; must be consistent with their cv; allow "associated" instead of "not independent" |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(O_i > E_i\) for pairings of the same colour and some children may be cheating when selecting the 2nd ball so choice not independent | B1 | Stating \(O_i > E_i\) and a suitable reason such as cheating/looking, or the first ball being replaced by the child so more likely to be picked again |
# Question 2:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $E(RR) = \frac{60 \times 60}{150}$ or $E(BB) = \frac{30 \times 27}{150}$ or $E(YY) = \frac{60 \times 63}{150}$ | M1 | For at least one correct calculation seen |
| $E(RR) = 24$ and $E(BB) = 5.4$ and $E(YY) = 25.2$ | A1 | For all 3 expected frequencies correct, must be exact |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $[E(BB) > 5$ so no need for pooling$]$ $v = (3-1)(3-1) = 4$ | B1 | For 4 degrees of freedom |
| Critical value: $\chi_4^2(5\%) = 9.488$ | B1ft | For awrt 9.488 or ft their degrees of freedom for 5% cv |
| (significant) evidence that the colour of balls is **not** independent | B1ft | For correct contextual conclusion mentioning "colour" and "independent"; must be consistent with their cv; allow "associated" instead of "not independent" |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $O_i > E_i$ for pairings of the same colour **and** some children may be cheating when selecting the 2nd ball so choice not independent | B1 | Stating $O_i > E_i$ **and** a suitable reason such as cheating/looking, or the first ball being replaced by the child so more likely to be picked again |
---\begin{enumerate}
\item A bag contains a large number of balls, all of the same size and weight. The balls are coloured Red, Blue or Yellow.
\end{enumerate}
Jasmine asks each child in a group of 150 children to close their eyes, select a ball from the bag and show it to her. The child then replaces the ball and repeats the process a second time.
If both balls are the same colour the child receives a prize.\\
The results are given in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\backslashbox{2nd colour}{1st colour} & Red & Blue & Yellow & Total \\
\hline
Red & 31 & 11 & 18 & 60 \\
\hline
Blue & 8 & 10 & 9 & 27 \\
\hline
Yellow & 21 & 9 & 33 & 63 \\
\hline
Total & 60 & 30 & 60 & 150 \\
\hline
\end{tabular}
\end{center}
Jasmine carries out a test, at the $5 \%$ level of significance, to see whether or not the colour of the 2nd ball is independent of the colour of the 1st ball.\\
(a) Calculate the expected frequencies for the cases where both balls are the same colour.
The test statistic Jasmine obtained was 12.712 to three decimal places.\\
(b) Use this value to complete the test, stating the critical value and conclusion clearly.
With reference to your calculations in part (a) and the nature of the experiment, (c) give a plausible reason why Jasmine may have obtained her conclusion in part (b).
\hfill \mbox{\textit{Edexcel FS1 AS 2023 Q2 [6]}}