| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Poisson |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with Poisson distribution. Part (i) requires a single calculation using the standard formula with given mean; part (ii) tests understanding of degrees of freedom (cells minus parameters minus 1); part (iii) is routine hypothesis testing with given chi-squared value. All steps are standard S3 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number of particles emitted, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | \(\geqslant 11\) |
| Frequency | 57 | 203 | 383 | 525 | 532 | 408 | 273 | 139 | 45 | 27 | 10 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2608p}{p = e^{x.3}3.87}/61! \times (2608=253.82)\) | M1 | |
| A1 | ||
| \((273-253.82)^2/253.82\) | M1 | |
| \(=1.449\) | A1 | 4 Answer between 1.445 and 1.460 |
| Answer | Marks | Guidance |
|---|---|---|
| Number of cells \(- 1\) (estimated mean \() - 1\)(same totals) | B1 | 1(8) Not 11-1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): A Poisson distribution fits the data | B1 | For both hypotheses |
| \(H_1\): A Poisson distribution does not fit the data | B1 | |
| \(CV = 15.99\) | M1 | |
| \(13.0 < CV\) and do not reject \(H_0\) accept that there is insufficient evidence that a Poisson distribution does not fit data | A1 | 4(9) Their CV; Sufficient evidence that Poisson distribution fits data, OK |
**4(i)**
$\frac{2608p}{p = e^{x.3}3.87}/61! \times (2608=253.82)$ | M1 |
| A1 |
$(273-253.82)^2/253.82$ | M1 |
$=1.449$ | A1 | 4 Answer between 1.445 and 1.460
**4(ii)**
Number of cells $- 1$ (estimated mean $) - 1$(same totals) | B1 | 1(8) Not 11-1
**4(iii)**
$H_0$: A Poisson distribution fits the data | B1 | For both hypotheses
$H_1$: A Poisson distribution does not fit the data | B1 |
$CV = 15.99$ | M1 |
$13.0 < CV$ and do not reject $H_0$ accept that there is insufficient evidence that a Poisson distribution does not fit data | A1 | 4(9) Their CV; Sufficient evidence that Poisson distribution fits data, OK
---4 An experiment by Lord Rutherford at Cambridge in 1909 involved measuring the numbers of $\alpha$-particles emitted during radioactive decay. The following table shows emissions during 2608 intervals of 7.5 seconds.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Number of particles emitted, $x$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & $\geqslant 11$ \\
\hline
Frequency & 57 & 203 & 383 & 525 & 532 & 408 & 273 & 139 & 45 & 27 & 10 & 6 \\
\hline
\end{tabular}
\end{center}
It is given that the mean number of particles emitted per interval, calculated from the data, is 3.87 , correct to 3 significant figures.\\
(i) Find the contribution to the $\chi ^ { 2 }$ value of the frequency of 273 corresponding to $x = 6$ in a goodness of fit test for a Poisson distribution.\\
(ii) Given that no cells need to be combined, state why the number of degrees of freedom is 10 .\\
(iii) Given also that the calculated value of $\chi ^ { 2 }$ is 13.0 , correct to 3 significant figures, carry out the test at the 10\% significance level.
\hfill \mbox{\textit{OCR S3 2011 Q4 [9]}}