| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Larger contingency table (4+ categories) |
| Difficulty | Standard +0.3 This is a standard chi-squared test question from Further Statistics 1, requiring two routine tests (independence and goodness-of-fit) with clearly structured data. While it involves multiple parts and careful calculation, the procedures are textbook applications with no novel insight required. The binomial distribution in part (b) is straightforward. Slightly easier than average A-level due to the mechanical nature of the tests. |
| Spec | 5.02b Expectation and variance: discrete random variables5.06a Chi-squared: contingency tables5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| \multirow{2}{*}{} | Number of heads | |||
| 0 | 1 | 2 | 3 | |
| Left hand | 7 | 29 | 42 | 22 |
| Right hand | 13 | 35 | 36 | 16 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between the hand and the number of heads. \(H_1\): There is an association between the hand and the number of heads. | B1 | For both hypotheses correct with at least one in context |
| Expected frequencies: \(E = \frac{100 \times 20}{200}=10,\ \frac{100 \times 64}{200}=32,\ \frac{100 \times 78}{200}=39,\ \frac{100 \times 38}{200}=19\) | M1, A1 | M1: attempt at \(\frac{\text{row total} \times \text{col total}}{\text{grand total}}\); A1: all correct |
| \(\chi^2 = \sum\frac{(O-E)^2}{E} = \frac{(7-10)^2}{10} + \frac{(13-10)^2}{10} + \ldots + \frac{(16-19)^2}{19}\) | M1 | Applying \(\sum\frac{(O-E)^2}{E}\) ft their values |
| \(= 3.7714\ldots\) awrt 3.77 | A1 | awrt 3.77 |
| Degrees of freedom \(= (4-1)\times(2-1) = 3\); \(\chi^2_{3,0.05} = 7.815\) | M1 | For using degrees of freedom to set up \(\chi^2\) model |
| (Do not reject \(H_0\).) There is not enough evidence to suggest an association between the hand flipping the coin and the number of heads. | A1 | Correct conclusion in context with all other marks scored |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(B(3, 0.5)\) | B1 | Allow a complete probability distribution with labels: \(P(X=0)=0.125,\ P(X=1)=0.375,\ P(X=2)=0.375,\ P(X=3)=0.125\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): \(B(3, 0.5)\) is a suitable model. \(H_1\): \(B(3, 0.5)\) is not a suitable model. | B1ft | Must have binomial and (3, 0.5) or ft their distribution from (b) |
| Expected frequencies: \(200\times P(X=0)=25,\ 200\times P(X=1)=75,\ 200\times P(X=2)=75,\ 200\times P(X=3)=25\) | M1, A1 | M1: attempt using distribution from (b); A1: all correct |
| \(\chi^2 = \frac{(20-25)^2}{25} + \frac{(64-75)^2}{75} + \frac{(78-75)^2}{75} + \frac{(38-25)^2}{25}\) | M1 | Applying \(\sum\frac{(O-E)^2}{E}\) ft their values |
| \(= 9.493\ldots\) awrt 9.49 | A1 | awrt 9.49 |
| \([\text{df}=3]\ \chi^2_{3,0.1} = 6.251\) | M1 | For using degrees of freedom to set up \(\chi^2\) model |
| (Reject \(H_0\)) \(B(3, 0.5)\) is not a suitable model for the number of heads. | A1 | Correct conclusion in context with all other marks scored |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between the hand and the number of heads. $H_1$: There is an association between the hand and the number of heads. | B1 | For both hypotheses correct with at least one in context |
| Expected frequencies: $E = \frac{100 \times 20}{200}=10,\ \frac{100 \times 64}{200}=32,\ \frac{100 \times 78}{200}=39,\ \frac{100 \times 38}{200}=19$ | M1, A1 | M1: attempt at $\frac{\text{row total} \times \text{col total}}{\text{grand total}}$; A1: all correct |
| $\chi^2 = \sum\frac{(O-E)^2}{E} = \frac{(7-10)^2}{10} + \frac{(13-10)^2}{10} + \ldots + \frac{(16-19)^2}{19}$ | M1 | Applying $\sum\frac{(O-E)^2}{E}$ ft their values |
| $= 3.7714\ldots$ awrt **3.77** | A1 | awrt 3.77 |
| Degrees of freedom $= (4-1)\times(2-1) = 3$; $\chi^2_{3,0.05} = 7.815$ | M1 | For using degrees of freedom to set up $\chi^2$ model |
| (Do not reject $H_0$.) There is not enough evidence to suggest an association between the hand flipping the coin and the number of heads. | A1 | Correct conclusion in context with all other marks scored |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B(3, 0.5)$ | B1 | Allow a complete probability distribution with labels: $P(X=0)=0.125,\ P(X=1)=0.375,\ P(X=2)=0.375,\ P(X=3)=0.125$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: $B(3, 0.5)$ is a suitable model. $H_1$: $B(3, 0.5)$ is not a suitable model. | B1ft | Must have binomial and (3, 0.5) or ft their distribution from (b) |
| Expected frequencies: $200\times P(X=0)=25,\ 200\times P(X=1)=75,\ 200\times P(X=2)=75,\ 200\times P(X=3)=25$ | M1, A1 | M1: attempt using distribution from (b); A1: all correct |
| $\chi^2 = \frac{(20-25)^2}{25} + \frac{(64-75)^2}{75} + \frac{(78-75)^2}{75} + \frac{(38-25)^2}{25}$ | M1 | Applying $\sum\frac{(O-E)^2}{E}$ ft their values |
| $= 9.493\ldots$ awrt **9.49** | A1 | awrt 9.49 |
| $[\text{df}=3]\ \chi^2_{3,0.1} = 6.251$ | M1 | For using degrees of freedom to set up $\chi^2$ model |
| (Reject $H_0$) $B(3, 0.5)$ is **not** a suitable model for the number of heads. | A1 | Correct conclusion in context with all other marks scored |
---\begin{enumerate}
\item In an experiment, James flips a coin 3 times and records the number of heads. He carries out the experiment 100 times with his left hand and 100 times with his right hand.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multirow{2}{*}{} & \multicolumn{4}{|c|}{Number of heads} \\
\hline
& 0 & 1 & 2 & 3 \\
\hline
Left hand & 7 & 29 & 42 & 22 \\
\hline
Right hand & 13 & 35 & 36 & 16 \\
\hline
\end{tabular}
\end{center}
(a) Test, at the $5 \%$ level of significance, whether or not there is an association between the hand he flips the coin with and the number of heads.
You should state your hypotheses, the degrees of freedom and the critical value used for this test.\\
(b) Assuming the coin is unbiased, write down the distribution of the number of heads in 3 flips.\\
(c) Carry out a $\chi ^ { 2 }$ test, at the $10 \%$ level of significance, to test whether or not the distribution you wrote down in part (b) is a suitable model for the number of heads obtained in the 200 trials of James' experiment.
You should state your hypotheses, the degrees of freedom and the critical value used for this test.
\hfill \mbox{\textit{Edexcel FS1 AS 2020 Q2 [15]}}