| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Standard +0.8 This is a standard chi-squared goodness of fit test with binomial distribution, requiring calculation of missing expected frequencies, proper hypothesis formulation, and understanding of degrees of freedom when parameters are estimated. Part (b) requires conceptual understanding of why df = cells - 1 - estimated parameters. While methodical, it's above average difficulty due to the parameter estimation aspect and multi-part structure typical of Further Statistics. |
| Spec | 5.02b Expectation and variance: discrete random variables5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number of fake coins in each bag | 0 | 1 | 2 | 3 | 4 or more |
| Observed frequency | 43 | 62 | 26 | 13 | 6 |
| Expected frequency | 53.8 | 56.6 | 8.9 |
| Number of fake coins in each bag | 0 | 1 | 2 | 3 | 4 or more |
| Observed frequency | 43 | 62 | 26 | 13 | 6 |
| Expected frequency | 44.5 | 55.7 | 33.2 | 12.5 | 4.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expected value for \(2 = 150 \times P(X=2)\) | M1 | Using binomial model \(150 \times p^2 \times (1-p)^{18}\); may be implied by 28.3 |
| \(= 28.3015\ldots\) | A1 | awrt 28.3 |
| Expected value for 4 or more \(= 150 - (53.8 + 56.6 + 28.3 + 8.9) = 2.4\) | A1ft | awrt 2.4 or ft their "28.3" |
| \(H_0\): \(\text{Bin}(20, 0.05)\) is a suitable model; \(H_1\): \(\text{Bin}(20, 0.05)\) is not a suitable model | B1 | Both hypotheses correct using correct notation or written in full |
| Combining last two groups; \(\geq 3\), Observed \(= 19\), Expected \(= 11.3\) | M1 | For recognising need to combine groups |
| \(\nu = 4 - 1 = 3\) | B1 | Number of degrees of freedom \(= 3\); may be implied by correct CV |
| Critical value \(\chi^2(0.05) = 7.815\) | B1 | awrt 7.82 |
| Test statistic \(= \frac{(43-53.8)^2}{53.8} + \frac{(62-56.6)^2}{56.6} + \ldots\) | M1 | Attempting to find \(\sum\frac{(O_i - E_i)^2}{E_i}\) or \(\sum\frac{O_i^2}{E_i} - N\); may be implied by awrt 8.12 |
| \(= 8.117\) | A1 | awrt 8.12 |
| In critical region; sufficient evidence to reject \(H_0\), accept \(H_1\); significant evidence at 5% level to reject manager's model | A1 | Evaluating outcome of model by drawing correct inference in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\nu = 4 - 2 = 2\); 4 classes due to pooling | B1 | Explaining why there are 4 classes |
| 2 restrictions (equal total and mean/proportion) | B1 | Explanation of why 2 is subtracted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): Binomial distribution is a good model; \(H_1\): Binomial distribution is not a good model | B1 | Correct hypotheses for the refined model |
| Critical value \(\chi^2(0.05) = 5.991\); test statistic not in critical region; insufficient evidence to reject \(H_0\); Binomial distribution is a good model | B1 | CV awrt 5.99 and drawing correct inference for the refined model |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expected value for $2 = 150 \times P(X=2)$ | M1 | Using binomial model $150 \times p^2 \times (1-p)^{18}$; may be implied by 28.3 |
| $= 28.3015\ldots$ | A1 | awrt 28.3 |
| Expected value for 4 or more $= 150 - (53.8 + 56.6 + 28.3 + 8.9) = 2.4$ | A1ft | awrt 2.4 or ft their "28.3" |
| $H_0$: $\text{Bin}(20, 0.05)$ is a suitable model; $H_1$: $\text{Bin}(20, 0.05)$ is not a suitable model | B1 | Both hypotheses correct using correct notation or written in full |
| Combining last two groups; $\geq 3$, Observed $= 19$, Expected $= 11.3$ | M1 | For recognising need to combine groups |
| $\nu = 4 - 1 = 3$ | B1 | Number of degrees of freedom $= 3$; may be implied by correct CV |
| Critical value $\chi^2(0.05) = 7.815$ | B1 | awrt 7.82 |
| Test statistic $= \frac{(43-53.8)^2}{53.8} + \frac{(62-56.6)^2}{56.6} + \ldots$ | M1 | Attempting to find $\sum\frac{(O_i - E_i)^2}{E_i}$ or $\sum\frac{O_i^2}{E_i} - N$; may be implied by awrt 8.12 |
| $= 8.117$ | A1 | awrt 8.12 |
| In critical region; sufficient evidence to reject $H_0$, accept $H_1$; significant evidence at 5% level to reject manager's model | A1 | Evaluating outcome of model by drawing correct inference in context |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\nu = 4 - 2 = 2$; 4 classes due to pooling | B1 | Explaining why there are 4 classes |
| 2 restrictions (equal total and mean/proportion) | B1 | Explanation of why 2 is subtracted |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: Binomial distribution is a good model; $H_1$: Binomial distribution is not a good model | B1 | Correct hypotheses for the refined model |
| Critical value $\chi^2(0.05) = 5.991$; test statistic not in critical region; insufficient evidence to reject $H_0$; Binomial distribution is a good model | B1 | CV awrt 5.99 and drawing correct inference for the refined model |
---\begin{enumerate}
\item Bags of $\pounds 1$ coins are paid into a bank. Each bag contains 20 coins.
\end{enumerate}
The bank manager believes that $5 \%$ of the $\pounds 1$ coins paid into the bank are fakes. He decides to use the distribution $X \sim B ( 20,0.05 )$ to model the random variable $X$, the number of fake $\pounds 1$ coins in each bag.
The bank manager checks a random sample of 150 bags of $\pounds 1$ coins and records the number of fake coins found in each bag. His results are summarised in Table 1. He then calculates some of the expected frequencies, correct to 1 decimal place.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Number of fake coins in each bag & 0 & 1 & 2 & 3 & 4 or more \\
\hline
Observed frequency & 43 & 62 & 26 & 13 & 6 \\
\hline
Expected frequency & 53.8 & 56.6 & & 8.9 & \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
(a) Carry out a hypothesis test, at the $5 \%$ significance level, to see if the data supports the bank manager's statistical model. State your hypotheses clearly.
The assistant manager thinks that a binomial distribution is a good model but suggests that the proportion of fake coins is higher than $5 \%$. She calculates the actual proportion of fake coins in the sample and uses this value to carry out a new hypothesis test on the data. Her expected frequencies are shown in Table 2.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Number of fake coins in each bag & 0 & 1 & 2 & 3 & 4 or more \\
\hline
Observed frequency & 43 & 62 & 26 & 13 & 6 \\
\hline
Expected frequency & 44.5 & 55.7 & 33.2 & 12.5 & 4.1 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}
(b) Explain why there are 2 degrees of freedom in this case.\\
(c) Given that she obtains a $\chi ^ { 2 }$ test statistic of 2.67 , test the assistant manager's hypothesis that the binomial distribution is a good model for the number of fake coins in each bag. Use a $5 \%$ level of significance and state your hypotheses clearly.\\
\hfill \mbox{\textit{Edexcel FS1 Q3 [14]}}