| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| 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 straightforward binomial model (5 coins, p=0.5). Part (a) requires simple identification of B(5, 0.5), and part (b) follows a routine procedure: calculate expected frequencies using binomial probabilities, compute chi-squared statistic, compare to critical value. The only minor complication is potentially combining cells if expected frequencies are too low, but this is a standard textbook technique. Overall slightly easier than average due to its formulaic nature. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number of heads | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 6 | 18 | 29 | 34 | 10 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(B_7(5, 0.5)\) | M1A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): \(B(5, 0.5)\) is a suitable model | B1 N | |
| \(H_1\): \(B(5, 0.5)\) is not a suitable model (not a good kv) | ||
| \(\chi_9^2 = 0.466\) | ||
| No.of beads | 0 | 1 |
| Expected | 3.125 | 15.625 |
| Actual | 6 | 18 |
| M1A1A1 | 100(freq)/for lost; 1 caret; -A1; all caret =A1; 3+a better | |
| O | E | \(\frac{(O-E)^2}{E}\) |
| O or 1 | 24 | 18.75 |
| 2 | 29 | 31.25 |
| 3 | 34 | 31.25 |
| 4 or 5 | 13 | 18.75 |
| M1A1 | oywd 0 and E; All caret 3+t; \(\sum\) oywd, amrt 3.44 M1A1 | (11) |
| \(\sum \frac{(O-E)^2}{E} = 3.6372\) | \(\sum\) oywd, amrt 3.44 | |
| \(\nu = 4-1 = 3\), \(\mu_c^2(0.1-s) = 6.251\) | B1 B1 N | |
| Insufficient evidence to reject \(H_0\) | A1 N | |
| \(B(5, 0.s)\) is a suitable model. | ||
| No evidence that colours linked | TOTAL 13 | |
| Unpaired sities awrt 5.84, DFS, \(m_c^2 = 4.236\) |
## Part (a)
$B_7(5, 0.5)$ | M1A1 | (2)
## Part (b)
$H_0$: $B(5, 0.5)$ is a suitable model | B1 N |
$H_1$: $B(5, 0.5)$ is not a suitable model (not a good kv) |
$\chi_9^2 = 0.466$ |
| No.of beads | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Expected | 3.125 | 15.625 | 31.25 | 31.25 | 15.625 | 3.125 |
| Actual | 6 | 18 | 29 | 34 | 10 | 3 |
| M1A1A1 | 100(freq)/for lost; 1 caret; -A1; all caret =A1; 3+a better
| O | E | $\frac{(O-E)^2}{E}$ |
|---|---|---|
| O or 1 | 24 | 18.75 | 1.47 |
| 2 | 29 | 31.25 | 0.162 |
| 3 | 34 | 31.25 | 0.242 |
| 4 or 5 | 13 | 18.75 | 1.763 |
| M1A1 | oywd 0 and E; All caret 3+t; $\sum$ oywd, amrt 3.44 M1A1 | (11)
$\sum \frac{(O-E)^2}{E} = 3.6372$ | | $\sum$ oywd, amrt 3.44
$\nu = 4-1 = 3$, $\mu_c^2(0.1-s) = 6.251$ | B1 B1 N |
Insufficient evidence to reject $H_0$ | A1 N |
$B(5, 0.s)$ is a suitable model. |
No evidence that colours linked | | TOTAL 13
Unpaired sities awrt 5.84, DFS, $m_c^2 = 4.236$ | |
---Five coins were tossed 100 times and the number of heads recorded. The results are shown in the table below.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Number of heads & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
Frequency & 6 & 18 & 29 & 34 & 10 & 3 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable distribution to model the number of heads when five unbiased coins are tossed. [2]
\item Test, at the 10\% level of significance, whether or not the five coins are unbiased. State your hypotheses clearly. [11]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2006 Q8 [13]}}