| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with a binomial distribution. Part (a) requires calculating the sample mean to estimate p (routine calculation: total germinations/total seeds = 315/750). Part (b) follows a textbook procedure: calculate expected frequencies using B(5,0.42), combine cells with expected frequencies <5, compute the test statistic, and compare to critical value. While it requires multiple steps and careful arithmetic, it involves no novel insight—just systematic application of a well-practiced technique. Slightly easier than average due to its procedural nature. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number of seeds that germinate | 0 | 1 | 2 | 3 | 4 | 5 |
| Number of pots | 12 | 40 | 43 | 35 | 16 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x}=\frac{40+86+105+64+20}{150}=\frac{315}{150}=2.1\), \(\quad p=\frac{2.1}{5}=0.42\) | B1 | Must see either 315 or 2.1. AG |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Number of seeds that germinate | 0 | 1 |
| Number of pots | 12 | 40 |
| Expected frequency | 9.846 | 35.6475 |
| B1 | Calculate expected frequencies (must be seen) at least 2 correct to at least 2 decimal places. | |
| B1 | At least 4 correct to at least 2 decimal places. | |
| Combine last two columns | M1 | 20, 15.50, may be implied by answer \(3.90-3.91\). |
| Chi-squared contributions: \(0.4712\quad 0.5314\quad 1.4416\quad 0.1521\quad 1.3088\) | M1 | At least 2 correct, may be implied by answer \(3.90-3.91\). |
| Test statistic \(= 3.905\) | A1 | accept \(3.90-3.910\) |
| \(H_0\): Binomial B(5, 0.42) fits the data; \(H_1\): Binomial B(5, 0.42) does not fit the data | B1 | Allow 'Binomial' for 'B(5, 0.42)'. Allow 'Number of seeds that germinate can be modelled by B(5, 0.42)' |
| Critical value is 6.251 | B1 | Must come from combined columns. Allow 7.779. |
| \('3.905' < '6.251'\) Accept \(H_0\) | M1 | Reject \(H_1\), not significant. |
| Insufficient evidence to suggest that B(5, 0.42) is not a good fit (to the data) | A1 | Correct work only, including hypotheses, level of uncertainty in language. |
| 9 |
## Question 3(a):
$\bar{x}=\frac{40+86+105+64+20}{150}=\frac{315}{150}=2.1$, $\quad p=\frac{2.1}{5}=0.42$ | B1 | Must see either 315 or 2.1. AG
| 1 |
## Question 3(b):
| Number of seeds that germinate | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Number of pots | 12 | 40 | 43 | 35 | 16 | 4 |
| Expected frequency | 9.846 | 35.6475 | 51.627 | 37.3845 | 13.536 | 1.9605 |
B1 | Calculate expected frequencies (must be seen) at least 2 correct to at least 2 decimal places.
B1 | At least 4 correct to at least 2 decimal places.
Combine last two columns | M1 | 20, 15.50, may be implied by answer $3.90-3.91$.
Chi-squared contributions: $0.4712\quad 0.5314\quad 1.4416\quad 0.1521\quad 1.3088$ | M1 | At least 2 correct, may be implied by answer $3.90-3.91$.
Test statistic $= 3.905$ | A1 | accept $3.90-3.910$
$H_0$: Binomial B(5, 0.42) fits the data; $H_1$: Binomial B(5, 0.42) does not fit the data | B1 | Allow 'Binomial' for 'B(5, 0.42)'. Allow 'Number of seeds that germinate can be modelled by B(5, 0.42)'
Critical value is 6.251 | B1 | Must come from combined columns. Allow 7.779.
$'3.905' < '6.251'$ Accept $H_0$ | M1 | Reject $H_1$, not significant.
Insufficient evidence to suggest that B(5, 0.42) is not a good fit (to the data) | A1 | Correct work only, **including hypotheses**, level of uncertainty in language.
| 9 |
---3 Rosie sows 5 seeds in each of 150 plant pots. The number of seeds that germinate is recorded for each pot. The results are summarised in the following table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
Number of seeds that germinate & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
Number of pots & 12 & 40 & 43 & 35 & 16 & 4 \\
\hline
\end{tabular}
\end{center}
Rosie suggests that the number of seeds that germinate follows the binomial distribution $\mathrm { B } ( 5 , p )$.
\begin{enumerate}[label=(\alph*)]
\item Use Rosie's results to show that $p = 0.42$.
\item Carry out a goodness of fit test, at the $10 \%$ significance level, to test whether the distribution $\mathrm { B } ( 5,0.42 )$ is a good fit for the data.\\
\includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-06_2720_38_109_2010}\\
\includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-07_2726_35_97_20}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q3 [10]}}