| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2023 |
| Session | January |
| Marks | 14 |
| 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 straightforward binomial calculations. Part (a) requires routine application of the test with given p=0.5, part (b) is simple mean calculation, and part (c) applies the test again with adjusted degrees of freedom. All steps are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.06c Fit other distributions: discrete and continuous |
| Number of girls | 0 | 1 | 2 | 3 | 4 |
| Frequency | 15 | 68 | 69 | 38 | 10 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: B(4, 0.5)\) is a suitable model; \(H_1: B(4, 0.5)\) is not a suitable model | B1 | Must mention B(4,0.5) at least once. May be in words: Binomial, probability \(p=0.5\) and reference to 4 children or \(n=4\). Condone B(0.5,4) |
| Expected frequencies: 12.5, 50, 75, 50, 12.5 | M1 A1 | M1 for correct method to find at least one expected frequency e.g. \(0.5^4 \times 200 = 12.5\). A1 all 5 correct, must be seen |
| \(\sum\frac{(O-E)^2}{E} = \frac{(15-12.5)^2}{12.5}+\ldots+\frac{(10-12.5)^2}{12.5}\) or \(\sum\frac{O^2}{E}-N = \frac{15^2}{12.5}+\ldots+\frac{10^2}{12.5}-200\) | M1 | At least 2 correct expressions/values seen (include \(-200\) if needed) |
| \(= 10.84\) (or 10.8) | A1 | Allow 10.8 |
| \(\nu = 4\) | B1 | This mark can be implied by a correct critical value of 9.488 |
| \(\chi^2_4(0.05) = 9.488 \Rightarrow CR \geqslant 9.488\) | B1 | 9.488 ft their degrees of freedom; for \(\nu=3\) it is 7.815 |
| Sufficient evidence to say that the research students claim is not supported | A1ft | Dep on 2nd M1. Need claim or student or binomial. ft their CV and test statistic. If test statistic \(>\) CV must say not supported; if test statistic \(<\) CV must say supported |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0\times15+1\times68+2\times69+3\times38+4\times10] = 360\) | M1 | Correct method for total number of girls. At least 3 non-zero terms correct. Useful figures \([0]+68+138+114+40\). Implied by 360 or 1.8 |
| \(\frac{360}{200\times4} = 0.45\) | A1* | cso; allow for 360/800 or 1.8/4 or \(1.8=4p\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): Binomial is a suitable model; \(H_1\): Binomial is not a suitable model | B1 | Must mention binomial at least once. Condone inclusion of B(4,0.45)/B(0.45,4) |
| \(\nu = 3\) | B1 | Implied by correct critical value of 7.815. Condone their \(\nu\) in part(a)\(-1\) |
| \(\chi^2_3(0.05) = 7.815 \Rightarrow CR \geqslant 7.815\) | B1ft | 7.815 ft their degrees of freedom if they have their \(\nu\) in part(a)\(-1\) |
| No significant evidence to say that the binomial is not a reasonable model | B1ft | Ft their CV only. Independent of hypotheses. Correct conclusion based on awrt 2.47 and their \(\chi^2\) critical value. Ignore any parameter given. Do not allow contradicting statements |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: B(4, 0.5)$ is a suitable model; $H_1: B(4, 0.5)$ is not a suitable model | B1 | Must mention B(4,0.5) at least once. May be in words: Binomial, probability $p=0.5$ and reference to 4 children or $n=4$. Condone B(0.5,4) |
| Expected frequencies: 12.5, 50, 75, 50, 12.5 | M1 A1 | M1 for correct method to find at least one expected frequency e.g. $0.5^4 \times 200 = 12.5$. A1 all 5 correct, must be seen |
| $\sum\frac{(O-E)^2}{E} = \frac{(15-12.5)^2}{12.5}+\ldots+\frac{(10-12.5)^2}{12.5}$ or $\sum\frac{O^2}{E}-N = \frac{15^2}{12.5}+\ldots+\frac{10^2}{12.5}-200$ | M1 | At least 2 correct expressions/values seen (include $-200$ if needed) |
| $= 10.84$ (or 10.8) | A1 | Allow 10.8 |
| $\nu = 4$ | B1 | This mark can be implied by a correct critical value of 9.488 |
| $\chi^2_4(0.05) = 9.488 \Rightarrow CR \geqslant 9.488$ | B1 | 9.488 ft their degrees of freedom; for $\nu=3$ it is 7.815 |
| Sufficient evidence to say that the research students claim is not supported | A1ft | Dep on 2nd M1. Need claim or student or binomial. ft their CV and test statistic. If test statistic $>$ CV must say not supported; if test statistic $<$ CV must say supported |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0\times15+1\times68+2\times69+3\times38+4\times10] = 360$ | M1 | Correct method for total number of girls. At least 3 non-zero terms correct. Useful figures $[0]+68+138+114+40$. Implied by 360 or 1.8 |
| $\frac{360}{200\times4} = 0.45$ | A1* | cso; allow for 360/800 or 1.8/4 or $1.8=4p$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: Binomial is a suitable model; $H_1$: Binomial is not a suitable model | B1 | Must mention binomial at least once. Condone inclusion of B(4,0.45)/B(0.45,4) |
| $\nu = 3$ | B1 | Implied by correct critical value of 7.815. Condone their $\nu$ in part(a)$-1$ |
| $\chi^2_3(0.05) = 7.815 \Rightarrow CR \geqslant 7.815$ | B1ft | 7.815 ft their degrees of freedom if they have their $\nu$ in part(a)$-1$ |
| No significant evidence to say that the binomial is not a reasonable model | B1ft | Ft their CV only. Independent of hypotheses. Correct conclusion based on awrt 2.47 and their $\chi^2$ critical value. Ignore any parameter given. Do not allow contradicting statements |
---4 A research student is investigating the number of children who are girls in families with 4 children.
The table below shows her results for 200 such families.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Number of girls & 0 & 1 & 2 & 3 & 4 \\
\hline
Frequency & 15 & 68 & 69 & 38 & 10 \\
\hline
\end{tabular}
\end{center}
The research student suggests that a binomial distribution with $p = \frac { 1 } { 2 }$ could be a suitable model for the number of children who are girls in a family of 4 children.
\begin{enumerate}[label=(\alph*)]
\item Using her results and a $5 \%$ significance level, test the research student's claim. You should state your hypotheses, expected frequencies, test statistic and the critical value used.
The research student decides to refine the model and retains the idea of using a binomial distribution but does not specify the probability that the child is a girl.
\item Use the data in the table to show that the probability that a child is a girl is 0.45
The research student uses the probability from part (b) to calculate a new set of expected frequencies, none of which are less than 5\\
The statistic $\sum \frac { ( O - E ) ^ { 2 } } { E }$ is evaluated and found to be 2.47
\item Test, at the $5 \%$ significance level, whether using a binomial distribution is suitable to model the number of children who are girls in a family of 4 children. You should state your hypotheses and the critical value used.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2023 Q4 [14]}}