| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Poisson |
| Difficulty | Standard +0.3 Part (i) is a standard chi-squared goodness of fit test for a Poisson distribution with most calculations already provided—students just need to complete the table, calculate the test statistic, find critical value, and conclude. Part (ii) is a routine Wilcoxon signed-rank test application. Both are textbook exercises requiring careful execution but no novel insight, making this slightly easier than average for S3 level. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.06c Fit other distributions: discrete and continuous |
| Number of particles | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | \(10 +\) |
| Frequency | 4 | 7 | 10 | 20 | 17 | 15 | 10 | 9 | 5 | 3 | 0 |
| Number of particles |
| Observed frequency |
| Expected frequency |
| Contribution to \(X ^ { 2 }\) |
| 5 | 6 | 7 | \(8 +\) |
| 15 | 10 | 9 | 8 |
| 16.33 | 11.44 | 6.86 | 6.39 |
| 0.1083 | 0.1813 | 0.6676 | 0.4056 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Probabilities correct to 3 d.p. or better | M1 | |
| \(\times 100\) for expected frequencies | M1 | |
| All expected frequencies correct | A1 | |
| Merge first two cells | M1 | |
| Calculation of \(X^2\) | M1 | |
| \(X^2 = 1.3128 + 0.7843 + 0.1183 + 0.3063 + 0.1083 + 0.1813 + 0.6676 + 0.4056 = 3.884(5)\) | A1 | c.a.o. For ungrouped cells \(X^2 = 6.816\) |
| \(H_0\): The Poisson model fits the data | B1 | Ignore any reference to the parameter |
| \(H_1\): The Poisson model does not fit the data | B1 | Do not accept "data fit model" |
| Refer to \(\chi^2_6\) | M1 | Allow correct df (= cells \(-2\)) from wrongly grouped table |
| Upper 10% point is 10.64 | A1 | No ft from here if wrong. \(\chi^2_7 = 12.02\), \(P(X^2 > 3.884) = 0.6924\) |
| Not significant | A1 | ft only c's test statistic |
| Evidence suggests model fits the data | A1 | Do not accept "data fit model" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: m = 15\), \(H_1: m > 15\) where \(m\) is the population median diameter (in \(\mu\)m) | B1 B1 | Both. Accept in words. Must include "population" |
| Given \(W_- = 53\) (\(\therefore W_+ = 157\)) | ||
| Refer to Wilcoxon paired/single sample statistic for \(n = 20\) | M1 | No ft from here if wrong |
| Lower 5% point is 60 (or upper is 150 if \(W_+\) used) | A1 | 1-tail test. No ft from here if wrong |
| Result is significant | A1 | ft only c's test statistic |
| Evidence suggests median diameter appears to be more than 15 \(\mu\)m | A1 | ft only c's test statistic. Context must include "average" o.e. |
# Question 2:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Probabilities correct to 3 d.p. or better | M1 | |
| $\times 100$ for expected frequencies | M1 | |
| All expected frequencies correct | A1 | |
| Merge first two cells | M1 | |
| Calculation of $X^2$ | M1 | |
| $X^2 = 1.3128 + 0.7843 + 0.1183 + 0.3063 + 0.1083 + 0.1813 + 0.6676 + 0.4056 = 3.884(5)$ | A1 | c.a.o. For ungrouped cells $X^2 = 6.816$ |
| $H_0$: The Poisson model fits the data | B1 | Ignore any reference to the parameter |
| $H_1$: The Poisson model does not fit the data | B1 | Do not accept "data fit model" |
| Refer to $\chi^2_6$ | M1 | Allow correct df (= cells $-2$) from wrongly grouped table |
| Upper 10% point is 10.64 | A1 | No ft from here if wrong. $\chi^2_7 = 12.02$, $P(X^2 > 3.884) = 0.6924$ |
| Not significant | A1 | ft only c's test statistic |
| Evidence suggests model fits the data | A1 | Do not accept "data fit model" |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: m = 15$, $H_1: m > 15$ where $m$ is the population median diameter (in $\mu$m) | B1 B1 | Both. Accept in words. Must include "population" |
| Given $W_- = 53$ ($\therefore W_+ = 157$) | | |
| Refer to Wilcoxon paired/single sample statistic for $n = 20$ | M1 | No ft from here if wrong |
| Lower 5% point is 60 (or upper is 150 if $W_+$ used) | A1 | 1-tail test. No ft from here if wrong |
| Result is significant | A1 | ft only c's test statistic |
| Evidence suggests median diameter appears to be more than 15 $\mu$m | A1 | ft only c's test statistic. Context must include "average" o.e. |
---2 Scientists researching into the chemical composition of dust in space collect specimens using a specially designed spacecraft. The craft collects the particles of dust in trays that are made up of a large array of cells containing aerogel. The aerogel traps the particles that penetrate into the cells.\\
(i) For a random sample of 100 cells, the number of particles of dust in each cell was counted, giving the following results.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Number of particles & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & $10 +$ \\
\hline
Frequency & 4 & 7 & 10 & 20 & 17 & 15 & 10 & 9 & 5 & 3 & 0 \\
\hline
\end{tabular}
\end{center}
It is thought that the number of particles collected in each cell can be modelled using the distribution Poisson(4.2) since 4.2 is the sample mean for these data.
Some of the calculations for a $\chi ^ { 2 }$ test are shown below. The cells for 8,9 and $10 +$ particles have been combined.
\begin{center}
\begin{tabular}{ | l | }
\hline
Number of particles \\
\hline
Observed frequency \\
\hline
Expected frequency \\
\hline
Contribution to $X ^ { 2 }$ \\
\hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
5 & 6 & 7 & $8 +$ \\
\hline
15 & 10 & 9 & 8 \\
\hline
16.33 & 11.44 & 6.86 & 6.39 \\
\hline
0.1083 & 0.1813 & 0.6676 & 0.4056 \\
\hline
\end{tabular}
\end{center}
Complete the calculations and carry out the test using a $10 \%$ significance level to see whether the number of particles per cell may be modelled in this way.\\
(ii) The diameters of the dust particles are believed to be distributed symmetrically about a median of 15 micrometres $( \mu \mathrm { m } )$. For a random sample of 20 particles, the sum of the signed ranks of the diameters of the particles smaller than $15 \mu \mathrm {~m} \left( W _ { - } \right)$is found to be 53 . Test at the $5 \%$ level of significance whether the median diameter appears to be more than $15 \mu \mathrm {~m}$.
\hfill \mbox{\textit{OCR MEI S3 2011 Q2 [18]}}