| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Expected frequencies partially provided |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with all scaffolding provided: 7 of 9 expected frequencies given, partial test statistic calculated, and clear instructions. Students only need to find 2 missing expected frequencies, complete the calculation, compare to critical value, and state conclusion. Slightly easier than average due to extensive scaffolding. |
| Spec | 5.06a Chi-squared: contingency tables |
| \multirow{2}{*}{} | Treatment | |||
| No action | Plant sprayed once | Plant sprayed every day | ||
| \multirow{3}{*}{Outcome} | Plant died within a month | 15 | 16 | 25 |
| Plant survived for 1-6 months | 8 | 25 | 10 | |
| Plant survived beyond 6 months | 7 | 14 | 5 | |
| \multirow{2}{*}{} | Treatment | |||
| No action | Plant sprayed once | Plant sprayed every day | ||
| \multirow{3}{*}{Outcome} | Plant died within a month | 17.92 | ||
| Plant survived for 1-6 months | 10.32 | 18.92 | 13.76 | |
| Plant survived beyond 6 months | 6.24 | 11.44 | 8.32 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between the treatment of the plants and their survival/outcome. \(H_1\): There is an association between the treatment of the plants and their survival/outcome | B1 | For correct hypotheses at least one in context. Allow independent/not independent. Do not accept correlation |
| Expected frequencies: \(13.44, 24.64, 17.92, 10.32, 18.92, 13.76, 6.24, 11.44, 8.32\) | M1, A1 | M1: attempt at \(\frac{\text{(Row Total)(Column Total)}}{\text{(Grand Total)}}\). A1: awrt \(13.44\) and \(24.64\) (may be implied by correct \(\chi^2\)) |
| \(\chi^2=\sum\frac{(O-E)^2}{E}=\frac{(15-"13.44")^2}{"13.44"}+\frac{(16-"24.64")^2}{"24.64"}+8.29\) | M1 | For applying \(\sum\frac{(O-E)^2}{E}\) ft their expected values. If no method shown, at least 1 of the two missing \(\chi^2\) contributions must be correct |
| awrt \(\mathbf{11.5}\) | A1 | |
| Degrees of freedom \((3-1)(3-1)=4\), \(\chi^2_{4,0.025}=11.143\) | M1 | For using degrees of freedom to set up \(\chi^2\) model critical value, implied by CV \(11.143\) or better |
| Reject \(H_0\). There is an association between the treatment of the plants and their survival/outcome | dA1ft | Dependent on 2nd and 3rd M marks. Correct conclusion ft their \(\sum\frac{(O-E)^2}{E}\). Must reference association between treatment and survival/outcome. Do not allow contradicting statements |
| (7 marks) |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between the treatment of the plants and their survival/outcome. $H_1$: There is an association between the treatment of the plants and their survival/outcome | B1 | For correct hypotheses at least one in context. Allow independent/not independent. Do not accept correlation |
| Expected frequencies: $13.44, 24.64, 17.92, 10.32, 18.92, 13.76, 6.24, 11.44, 8.32$ | M1, A1 | M1: attempt at $\frac{\text{(Row Total)(Column Total)}}{\text{(Grand Total)}}$. A1: awrt $13.44$ and $24.64$ (may be implied by correct $\chi^2$) |
| $\chi^2=\sum\frac{(O-E)^2}{E}=\frac{(15-"13.44")^2}{"13.44"}+\frac{(16-"24.64")^2}{"24.64"}+8.29$ | M1 | For applying $\sum\frac{(O-E)^2}{E}$ ft their expected values. If no method shown, at least 1 of the two missing $\chi^2$ contributions must be correct |
| awrt $\mathbf{11.5}$ | A1 | |
| Degrees of freedom $(3-1)(3-1)=4$, $\chi^2_{4,0.025}=11.143$ | M1 | For using degrees of freedom to set up $\chi^2$ model critical value, implied by CV $11.143$ or better |
| Reject $H_0$. There is an association between the treatment of the plants and their survival/outcome | dA1ft | Dependent on 2nd and 3rd M marks. Correct conclusion ft their $\sum\frac{(O-E)^2}{E}$. Must reference **association** between **treatment** and **survival**/outcome. Do not allow contradicting statements |
| | **(7 marks)** | |\begin{enumerate}
\item Abram carried out a survey of two treatments for a plant fungus. The contingency table below shows the results of a survey of a random sample of 125 plants with the fungus.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Treatment} \\
\hline
& & No action & Plant sprayed once & Plant sprayed every day \\
\hline
\multirow{3}{*}{Outcome} & Plant died within a month & 15 & 16 & 25 \\
\hline
& Plant survived for 1-6 months & 8 & 25 & 10 \\
\hline
& Plant survived beyond 6 months & 7 & 14 & 5 \\
\hline
\end{tabular}
\end{center}
Abram calculates expected frequencies to carry out a suitable test. Seven of these are given in the partly-completed table below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Treatment} \\
\hline
& & No action & Plant sprayed once & Plant sprayed every day \\
\hline
\multirow{3}{*}{Outcome} & Plant died within a month & & & 17.92 \\
\hline
& Plant survived for 1-6 months & 10.32 & 18.92 & 13.76 \\
\hline
& Plant survived beyond 6 months & 6.24 & 11.44 & 8.32 \\
\hline
\end{tabular}
\end{center}
The value of $\sum \frac { ( O - E ) ^ { 2 } } { E }$ for the 7 given values is 8.29\\
Test at the $2.5 \%$ level of significance, whether or not there is an association between the treatment of the plants and their survival. State your hypotheses and conclusion clearly.
\hfill \mbox{\textit{Edexcel FS1 AS 2018 Q4 [7]}}