| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Other continuous |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with expected frequencies provided. Part (i) requires routine calculation of P(5<X≤10) using the exponential distribution CDF, and part (ii) is a standard hypothesis test following a textbook procedure. The only minor complexity is combining cells to meet the expected frequency requirement, but this is a standard technique taught in S3. |
| Spec | 5.06c Fit other distributions: discrete and continuous |
| Time interval \(( x\) seconds \()\) | \(0 < x \leqslant 5\) | \(5 < x \leqslant 10\) | \(10 < x \leqslant 20\) | \(20 < x \leqslant 40\) | \(40 < x\) |
| Observed frequency | 49 | 22 | 20 | 7 | 2 |
| Time interval \(( x\) seconds \()\) | \(0 < x \leqslant 5\) | \(5 < x \leqslant 10\) | \(10 < x \leqslant 20\) | \(20 < x \leqslant 40\) | \(40 < x\) |
| Expected frequency | 39.35 | 23.87 | 23.25 | 11.70 | 1.83 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(f_c = 100 \times \int_5^{10} 0.1e^{-0.1x}dx\) | M1 | For attempting to integrate \(f(x)\) |
| \(= 100\left[-e^{-0.1x}\right]_5^{10}\) | A1 | For correct indefinite integral |
| \(= 100(e^{-0.5} - e^{-1}) = 23.87\) | M1 | For multiplying by total frequency |
| M1 | For use of correct limits | |
| A1 | 5 | For obtaining given answer correctly |
| (ii) Combining: \(\frac{f_o}{f_c}: \frac{49}{39.35}, \frac{22}{23.87}, \frac{20}{23.25}, \frac{9}{13.53}\) | M1 | For combining the last two classes |
| Test statistic is \(\frac{9.65^2}{39.35} + \frac{1.87^2}{23.87} + \frac{3.25^2}{23.25} + \frac{4.53^2}{13.53}\) | M1 | For correct calculation process |
| \(= 4.484\) | A1 | For correct value 4.48 |
| This is less than 6.251 | M1 | For comparison with the correct critical value |
| Hence there is a satisfactory fit | A1 | 5 |
**(i)** $f_c = 100 \times \int_5^{10} 0.1e^{-0.1x}dx$ | M1 | For attempting to integrate $f(x)$
$= 100\left[-e^{-0.1x}\right]_5^{10}$ | A1 | For correct indefinite integral
$= 100(e^{-0.5} - e^{-1}) = 23.87$ | M1 | For multiplying by total frequency
| M1 | For use of correct limits
| A1 | 5 | For obtaining given answer correctly
**(ii)** Combining: $\frac{f_o}{f_c}: \frac{49}{39.35}, \frac{22}{23.87}, \frac{20}{23.25}, \frac{9}{13.53}$ | M1 | For combining the last two classes
Test statistic is $\frac{9.65^2}{39.35} + \frac{1.87^2}{23.87} + \frac{3.25^2}{23.25} + \frac{4.53^2}{13.53}$ | M1 | For correct calculation process
$= 4.484$ | A1 | For correct value 4.48
This is less than 6.251 | M1 | For comparison with the correct critical value
Hence there is a satisfactory fit | A1 | 5 | For correct conclusion, in terms of the fit4 The lengths of time, in seconds, between vehicles passing a fixed observation point on a road were recorded at a time when traffic was flowing freely. The frequency distribution in Table 1 is a summary of the data from 100 observations.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c c c c c | }
\hline
Time interval $( x$ seconds $)$ & $0 < x \leqslant 5$ & $5 < x \leqslant 10$ & $10 < x \leqslant 20$ & $20 < x \leqslant 40$ & $40 < x$ \\
Observed frequency & 49 & 22 & 20 & 7 & 2 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
It is thought that the distribution of times might be modelled by the continuous random variable $X$ with probability density function given by
$$f ( x ) = \begin{cases} 0.1 e ^ { - 0.1 x } & x > 0 \\ 0 & \text { otherwise } \end{cases}$$
Using this model, the expected frequencies (correct to 2 decimal places) for the given time intervals are shown in Table 2.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c c c c c | }
\hline
Time interval $( x$ seconds $)$ & $0 < x \leqslant 5$ & $5 < x \leqslant 10$ & $10 < x \leqslant 20$ & $20 < x \leqslant 40$ & $40 < x$ \\
Expected frequency & 39.35 & 23.87 & 23.25 & 11.70 & 1.83 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}
(i) Show how the expected frequency of 23.87, corresponding to the interval $5 < x \leqslant 10$, is obtained.\\
(ii) Test, at the 10\% significance level, the goodness of fit of the model to the data.
\hfill \mbox{\textit{OCR S3 Q4 [10]}}