| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 17 |
| 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 calculations. Part (a) requires recall of binomial assumptions, (b) involves simple arithmetic using the fact frequencies sum to 150, (c) is a routine hypothesis test procedure, and (d) tests understanding of degrees of freedom when parameters are estimated. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.06b 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 | \(r\) | 8.9 | \(s\) |
| 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 | Marks | Guidance |
| Independence of each occurrence (of a fake coin) | B1 | |
| Constant probability for each occurrence (of a fake) | B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = 150 \times P(X = 2) = 150 \times \binom{20}{2} \times 0.05^2 \times 0.95^{18}\) | M1 | |
| \(r = 28.3015...\) | A1 | awrt 28.3 |
| \(s = 150 - (53.8 + 56.6 + 28.3 + 8.9) = 2.4\) | A1ft | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Bin(20, 0.05) is a suitable model; \(H_1\): Bin(20, 0.05) is not a suitable model | B1 | |
| Combining last two groups | ||
| Observed frequency: \(\geq 3\) = 19; Expected frequency = 11.3 | M1 | |
| \(\nu = 4 - 1 - 3 = 3\) | B1 | |
| Critical value, \(\chi^2(0.05) = 7.815\) (accept 9.488 if their \(\nu = 4\)) | B1ft | |
| Test statistic, \(\Sigma \frac{(O-E)^2}{E} = \frac{(43-53.8)^2}{53.8} + \frac{(62-56.6)^2}{56.6} + ...\) | M1 | |
| \(= 2.168... + 0.515... + 0.186... + 5.246...\) | ||
| \(= 8.117\) (accept 10.16 if groups not combined) | A1ft | |
| In critical region, sufficient evidence to reject \(H_0\), accept \(H_1\) | A1ft | |
| Significant evidence at 5% level to reject the manager's model | A1ft | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\nu = 4 - 2 = 2\) | B1 | |
| 4 classes due to pooling; 2 restrictions (equal total and mean/proportion) | B1 | (2) |
| (17 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Binomial distribution is a good model; \(H_1\): Binomial distribution is not a good model | B1 | |
| Critical value, \(\chi^2(0.05) = 5.991\) | B1 | |
| Test statistic is not in critical region, insufficient evidence to reject \(H_0\) | B1 | |
| Accept the assistant manager's model for the number of fake coins per bag. | B1 | (3) |
| (17 marks) |
**6(a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Independence of each occurrence (of a fake coin) | B1 | |
| Constant probability for each occurrence (of a fake) | B1 | (2) |
**6(b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = 150 \times P(X = 2) = 150 \times \binom{20}{2} \times 0.05^2 \times 0.95^{18}$ | M1 | |
| $r = 28.3015...$ | A1 | awrt 28.3 |
| $s = 150 - (53.8 + 56.6 + 28.3 + 8.9) = 2.4$ | A1ft | (3) |
**6(c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Bin(20, 0.05) is a suitable model; $H_1$: Bin(20, 0.05) is not a suitable model | B1 | |
| Combining last two groups | | |
| Observed frequency: $\geq 3$ = 19; Expected frequency = 11.3 | M1 | |
| $\nu = 4 - 1 - 3 = 3$ | B1 | |
| Critical value, $\chi^2(0.05) = 7.815$ (accept 9.488 if their $\nu = 4$) | B1ft | |
| Test statistic, $\Sigma \frac{(O-E)^2}{E} = \frac{(43-53.8)^2}{53.8} + \frac{(62-56.6)^2}{56.6} + ...$ | M1 | |
| $= 2.168... + 0.515... + 0.186... + 5.246...$ | | |
| $= 8.117$ (accept 10.16 if groups not combined) | A1ft | |
| In critical region, sufficient evidence to reject $H_0$, accept $H_1$ | A1ft | |
| Significant evidence at 5% level to reject the manager's model | A1ft | (7) |
**6(d)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\nu = 4 - 2 = 2$ | B1 | |
| 4 classes due to pooling; 2 restrictions (equal total and mean/proportion) | B1 | (2) |
| | (17 marks) | |
**6(e)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Binomial distribution is a good model; $H_1$: Binomial distribution is not a good model | B1 | |
| Critical value, $\chi^2(0.05) = 5.991$ | B1 | |
| Test statistic is not in critical region, insufficient evidence to reject $H_0$ | B1 | |
| Accept the assistant manager's model for the number of fake coins per bag. | B1 | (3) |
| | (17 marks) | |
**Guidance Notes for 6(b):**
- M1A1 for one of $r$ or $s$ correct
- A1ft for other value if using $150 - ...$ and answer must be $>0$
**Guidance Notes for 6(c):**
- 1st B1 can be in words but must include $p = 0.05$
- 3rd B1 ft on their $\nu$
- Test statistic alternative method: $\Sigma \frac{O^2}{E} - 150 = \frac{43^2}{53.8} + \frac{62^2}{56.6} + ... - 150 = 8.117...$
- 1st A1 if their groups not combined
- 2nd A1 ft if their test and critical values but must be comment in context e.g. mention of "manager's model" or "fake coins"
**Guidance Notes for 6(d):**
- 1st B1 evidence that pooling is required
- 2nd B1 must have correct reasons for restrictions.
---6. Bags of $\pounds 1$ coins are paid into a bank. Each bag contains 20 coins.
The bank manager believes that $5 \%$ of the $\pounds 1$ coins paid into the bank are fakes. He decides to use the distribution $X \sim \mathrm {~B} ( 20,0.05 )$ to model the random variable $X$, the number of fake $\pounds 1$ coins in each bag.
\begin{enumerate}[label=(\alph*)]
\item State the assumptions necessary for the binomial distribution to be an appropriate model in this case.
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.
\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 & $r$ & 8.9 & $s$ \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
\item Calculate the values of $r$ and $s$, giving your answers to 1 decimal place.
\item 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.
Question 6 parts (d) and (e) are continued on page 24
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}
\item Explain why there are 2 degrees of freedom in this case.
\item 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.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2014 Q6 [17]}}