| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| 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 straightforward chi-squared goodness of fit test with a fully specified binomial distribution. Part (a) requires simple binomial probability calculations using B(4,0.5), and part (b) follows a standard template: calculate chi-squared statistic, find degrees of freedom, compare to critical value. No pooling needed as all expected frequencies exceed 5. This is slightly easier than average as it's a textbook application with no complications. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.06b Fit prescribed distribution: chi-squared test |
| Number of female cubs | 0 | 1 | 2 | 3 | 4 |
| Observed number of litters | 10 | 33 | 33 | 15 | 9 |
| Number of female cubs | 0 | 1 | 2 | 3 | 4 |
| Expected number of litters | 6.25 | \(r\) | \(s\) | \(r\) | 6.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r = P(X=3) \times 100\) or \(r = P(X=1) \times 100\) or \(s = P(X=2) \times 100\) | M1 | Using the Binomial model to find expected value. Allow if both probs 0.25 and 0.375 seen. May be implied by a correct value of \(r\) or \(s\). Alternatives \(r = 6.25 \times 4\) or \(s = 6.25 \times 6\) |
| \(r = \mathbf{25}\) (value may be in table) | A1 | for \(r = 25\) |
| \(s = \mathbf{37.5}\) (value may be in table) | A1 | for \(s = 37.5\) |
| SC "B1" | If M0 scored but their values of \(r\) and \(s\) satisfy \(2r + s = 87.5\) score as M0A0A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): \(B(4, 0.5)\) is a suitable model; \(H_1\): \(B(4, 0.5)\) is not a suitable model (o.e.) Condone \(B(0.5, 4)\) | B1 | Both hypotheses correct using correct notation in at least one, or written in full e.g. binomial with \(n=4\) and \(p=0.5\) |
| Calculating \(\frac{(O_i - E_i)^2}{E_i}\): values \(2.25,\ 2.56,\ 0.54,\ 4,\ 1.21\) | M1 | Calculating either \(\dfrac{(O_i-E_i)^2}{E_i}\) or \(\dfrac{O_i^2}{E_i}\) — at least 4 correct. Implied by sight of awrt 10.6 |
| \(\sum \dfrac{(O_i - E_i)^2}{E_i} = 10.56\) or \(\sum \dfrac{O_i^2}{E_i} - N = 110.56 - 100 = 10.56\ \left(= \dfrac{264}{25}\right)\) | A1 | Allow 10.6 (from correct working) |
| \(\nu = 5 - 1 = 4\) | B1 | Correct dof. May be implied by CV of 9.48 or 9.49 or better |
| \(CV = 9.488\) (Calc \(9.487729035\ldots\)) | B1ft | For 9.488 or better. Can ft their dof. NB \(\chi^2_3(5\%) = 7.815\) (allow awrt 7.815) |
| Significant so there is evidence that the researcher's model is not suitable | A1 | Indep of hypotheses but dep on 1st A1. Evaluating outcome by drawing correct inference. Compatible with comparison of 10.56 or 10.6 with their CV (must be \(> 1\)). Must say model not suitable (o.e.). No need to explicitly see \(B(4, 0.5)\) mentioned here |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = P(X=3) \times 100$ or $r = P(X=1) \times 100$ or $s = P(X=2) \times 100$ | M1 | Using the Binomial model to find expected value. Allow if both probs 0.25 and 0.375 seen. May be implied by a correct value of $r$ or $s$. Alternatives $r = 6.25 \times 4$ or $s = 6.25 \times 6$ |
| $r = \mathbf{25}$ (value may be in table) | A1 | for $r = 25$ |
| $s = \mathbf{37.5}$ (value may be in table) | A1 | for $s = 37.5$ |
| **SC "B1"** | | If M0 scored but their values of $r$ and $s$ satisfy $2r + s = 87.5$ score as **M0A0A1** |
**(3 marks)**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: $B(4, 0.5)$ is a suitable model; $H_1$: $B(4, 0.5)$ is not a suitable model (o.e.) Condone $B(0.5, 4)$ | B1 | Both hypotheses correct using correct notation in at least one, or written in full e.g. binomial with $n=4$ and $p=0.5$ |
| Calculating $\frac{(O_i - E_i)^2}{E_i}$: values $2.25,\ 2.56,\ 0.54,\ 4,\ 1.21$ | M1 | Calculating either $\dfrac{(O_i-E_i)^2}{E_i}$ or $\dfrac{O_i^2}{E_i}$ — at least 4 correct. Implied by sight of awrt 10.6 |
| $\sum \dfrac{(O_i - E_i)^2}{E_i} = 10.56$ or $\sum \dfrac{O_i^2}{E_i} - N = 110.56 - 100 = 10.56\ \left(= \dfrac{264}{25}\right)$ | A1 | Allow 10.6 (from correct working) |
| $\nu = 5 - 1 = 4$ | B1 | Correct dof. May be implied by CV of 9.48 or 9.49 or better |
| $CV = 9.488$ (Calc $9.487729035\ldots$) | B1ft | For 9.488 or better. Can ft their dof. NB $\chi^2_3(5\%) = 7.815$ (allow awrt 7.815) |
| Significant so there is evidence that the researcher's **model is not suitable** | A1 | **Indep of hypotheses but dep on 1st A1.** Evaluating outcome by drawing correct inference. Compatible with comparison of 10.56 or 10.6 with their CV (must be $> 1$). Must say **model not suitable** (o.e.). No need to explicitly see $B(4, 0.5)$ mentioned here |
**(6 marks)**
**Total: 9 marks**\begin{enumerate}
\item A researcher is investigating the number of female cubs present in litters of size 4 He believes that the number of female cubs in a litter can be modelled by $\mathrm { B } ( 4,0.5 )$ He randomly selects 100 litters each of size 4 and records the number of female cubs. The results are recorded in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Number of female cubs & 0 & 1 & 2 & 3 & 4 \\
\hline
Observed number of litters & 10 & 33 & 33 & 15 & 9 \\
\hline
\end{tabular}
\end{center}
He calculated the expected frequencies as follows
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Number of female cubs & 0 & 1 & 2 & 3 & 4 \\
\hline
Expected number of litters & 6.25 & $r$ & $s$ & $r$ & 6.25 \\
\hline
\end{tabular}
\end{center}
(a) Find the value of $r$ and the value of $s$\\
(b) Carry out a suitable test, at the $5 \%$ level of significance, to determine whether or not the number of female cubs in a litter can be modelled by $\mathrm { B } ( 4,0.5 )$ You should clearly state your hypotheses and the critical value used.
\hfill \mbox{\textit{Edexcel FS1 2022 Q1 [9]}}