| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×3 contingency table |
| Difficulty | Moderate -0.3 This is a standard chi-squared test of independence with clearly labeled steps. Part (a) requires routine calculation of expected frequencies using row/column totals, and part (b) follows the standard hypothesis testing framework with most of the calculation already provided. The question is slightly easier than average because it gives the sum for four cells, reducing computational burden, and follows a completely standard template with no novel elements. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 3 - 3 } \multicolumn{2}{c|}{} | Treatment | ||||
| \cline { 3 - 4 } \multicolumn{2}{c|}{} | Applied | Not applied | |||
\multirow{3}{*}{
| \(y < 1\) | 15 | 25 | ||
| \cline { 2 - 4 } | \(1 \leqslant y < 2\) | 35 | 61 | ||
| \cline { 2 - 4 } | \(2 \leqslant y\) | 124 | 140 | ||
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(\frac{40 \times 174}{400}\) (ii) \(\frac{96 \times 226}{400}\) | M1 | A correct method to work out either expected frequencies – or 1 correct |
| \(= 17.4\) and \(= 54.24\) | A1 | 17.4 and 54.24 (accept 54.2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between the application of the treatment and the number of years that a fruit tree remains free from this disease. \(H_1\): There is an association between the application of the treatment and the number of years that a fruit tree remains free from this disease. | B1 | For both hypotheses in terms of "association" or independence. Must mention application/treatment and years in at least one and be connected correctly to \(H_0\) and \(H_1\). [Use of link, relationship or connection is B0 but allow for last A1ft] |
| \(\sum \frac{(O-E)^2}{E} = \frac{(15-"17.4")^2}{"17.4"} + \frac{(61-"54.24")^2}{"54.24"} + 2.642\) | M1 | A correct method to find the total \(\chi^2\) value, ft their values from (a). If no method shown at least 1 of the two missing \(\chi^2\) contributions must be correct. \(\left(0.331\ldots\left(\frac{48}{145}\right)\right.\) and \(0.8425\ldots\) allow 2sf). Implied by awrt 3.82 |
| \(= 3.815\ldots\) awrt 3.82 | A1 | awrt 3.82 or awrt 3.83 |
| \([3.82 <]\ \chi^2_{2,(0.05)} = 5.991\) | B1 | Using the degrees of freedom to find the \(\chi^2\) CV for the appropriate model. awrt 5.991, allow 5.9915 |
| There is no evidence of association between the application of the treatment and the number of years that a fruit tree remains free from this disease. | A1ft | Ft "their 3.82" and their CV or \(p\)-value. Correct conclusion in context (application or treatment and years). Independent of hypotheses. Allow relationship, link, connection for association BUT do not accept correlation or contradictory statements. NB If \(p\)-value [0.148388] given instead of CV could get B1M1A1B0A1 unless they give the CV as well |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $\frac{40 \times 174}{400}$ (ii) $\frac{96 \times 226}{400}$ | M1 | A correct method to work out either expected frequencies – or 1 correct |
| $= 17.4$ and $= 54.24$ | A1 | 17.4 and 54.24 (accept 54.2) |
**(2 marks)**
---
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between the **application** of the **treatment** and the number of **years** that a fruit tree remains free from this disease. $H_1$: There is an association between the application of the treatment and the number of years that a fruit tree remains free from this disease. | B1 | For both hypotheses in terms of "association" or independence. Must mention application/treatment and years in at least one and be connected correctly to $H_0$ and $H_1$. [Use of link, relationship or connection is B0 but allow for last A1ft] |
| $\sum \frac{(O-E)^2}{E} = \frac{(15-"17.4")^2}{"17.4"} + \frac{(61-"54.24")^2}{"54.24"} + 2.642$ | M1 | A correct method to find the total $\chi^2$ value, ft their values from (a). If no method shown at least 1 of the two missing $\chi^2$ contributions must be correct. $\left(0.331\ldots\left(\frac{48}{145}\right)\right.$ and $0.8425\ldots$ allow 2sf). Implied by awrt 3.82 |
| $= 3.815\ldots$ awrt 3.82 | A1 | awrt 3.82 or awrt 3.83 |
| $[3.82 <]\ \chi^2_{2,(0.05)} = 5.991$ | B1 | Using the degrees of freedom to find the $\chi^2$ CV for the appropriate model. awrt 5.991, allow 5.9915 |
| There is no evidence of association between the **application** of the **treatment** and the number of **years** that a fruit tree remains free from this disease. | A1ft | Ft "their 3.82" and their CV or $p$-value. Correct conclusion in context (application or treatment and years). Independent of hypotheses. Allow relationship, link, connection for association BUT do not accept correlation or contradictory statements. **NB** If $p$-value [0.148388] given instead of CV could get B1M1A1B0A1 unless they give the CV as well |
**(5 marks)**
**(7 marks total)**\begin{enumerate}
\item Stuart is investigating a treatment for a disease that affects fruit trees. He has 400 fruit trees and applies the treatment to a random sample of these trees. The remainder of the trees have no treatment. He records the number of years, $y$, that each fruit tree remains free from this disease.
\end{enumerate}
The results are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 3 - 3 }
\multicolumn{2}{c|}{} & \multicolumn{2}{c|}{Treatment} \\
\cline { 3 - 4 }
\multicolumn{2}{c|}{} & Applied & Not applied \\
\hline
\multirow{3}{*}{\begin{tabular}{ l }
Number of years free \\
from this disease \\
\end{tabular}} & $y < 1$ & 15 & 25 \\
\cline { 2 - 4 }
& $1 \leqslant y < 2$ & 35 & 61 \\
\cline { 2 - 4 }
& $2 \leqslant y$ & 124 & 140 \\
\hline
\end{tabular}
\end{center}
The data are to be used to determine whether or not there is an association between the application of the treatment and the number of years that a fruit tree remains free from this disease.\\
(a) Calculate the expected frequencies for\\
(i) Applied and $y < 1$\\
(ii) Not applied and $1 \leqslant y < 2$
The value of $\sum \frac { ( O - E ) ^ { 2 } } { E }$ for the other four classes is 2.642 to 3 decimal places.\\
(b) Test, at the $5 \%$ level of significance, whether or not there is an association between the application of the treatment and the number of years a fruit tree remains free from this disease.
You should state your hypotheses, test statistic, critical value and conclusion clearly.
\hfill \mbox{\textit{Edexcel FS1 AS 2022 Q1 [7]}}