| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | January |
| 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 Part (i) is a straightforward integration of a linear pdf over an interval of width 1, requiring only basic calculus. Part (ii) is a standard chi-squared goodness of fit test with all steps clearly signposted—calculate expected frequencies from the given probabilities, compute the test statistic, and compare to critical value. This is a routine S3 question testing core syllabus content with no novel problem-solving required. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| \(x\) | \(0 \leqslant x < 1\) | \(1 \leqslant x < 2\) | \(2 \leqslant x < 3\) | \(3 \leqslant x < 4\) | \(4 \leqslant x < 5\) |
| Frequency | 16 | 12 | 18 | 30 | 24 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_u^{u+1} \frac{2x+3}{40} dx = \left[\frac{x^2+3x}{40}\right]_u^{u+1}\) | \(M1 A1\) | Or correct use of F(x) or (2x+3)²/160 |
| \(= (u+2)/20\) AG | \(A1\) | |
| \([3]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((H_0:f(x)\) fits data, \(H_1:f(x)\) does not fit data\()\) | \(B2\) | B1 for 2 correct E-values, B2 for all correct |
| E-values: 10 15 20 25 30 | \(M1\) | M1 for 2 correct \(\chi^2\) values ft |
| \(= 36/10 + 9/15 + 4/20 + 25/25 + 36/30\) | \(A1\) | CAO |
| \(= 6.6\) | ||
| Compare with 7.779 and do not reject \(H_0\) | \(B1 M1\) | B1 for 7.779 Allow M1 for consistent conclusion from wrong CV. |
| There is insufficient evidence (at the 10%SL) that \(f(x)\) does not fit the data | \(A1ft\) | Dep 7.779 used. Allow f(x) fits data NOT Data fits f(x) |
| \([7]\) |
### (i)
$\int_u^{u+1} \frac{2x+3}{40} dx = \left[\frac{x^2+3x}{40}\right]_u^{u+1}$ | $M1 A1$ | Or correct use of F(x) or (2x+3)²/160
$= (u+2)/20$ AG | $A1$ |
| $[3]$ |
### (ii)
$(H_0:f(x)$ fits data, $H_1:f(x)$ does not fit data$)$ | $B2$ | B1 for 2 correct E-values, B2 for all correct
E-values: 10 15 20 25 30 | $M1$ | M1 for 2 correct $\chi^2$ values ft
$= 36/10 + 9/15 + 4/20 + 25/25 + 36/30$ | $A1$ | CAO
$= 6.6$ | |
Compare with 7.779 and do not reject $H_0$ | $B1 M1$ | B1 for 7.779 Allow M1 for consistent conclusion from wrong CV.
There is insufficient evidence (at the 10%SL) that $f(x)$ does not fit the data | $A1ft$ | Dep 7.779 used. Allow f(x) fits data NOT Data fits f(x)
| $[7]$ |A statistician suggested that the weekly sales $X$ thousand litres at a petrol station could be modelled by the following probability density function.
$$\text{f}(x) = \begin{cases}
\frac{1}{40}(2x + 3) & 0 \leqslant x < 5, \\
0 & \text{otherwise.}
\end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Show that, using this model, P$(a < X < a + 1) = \frac{a + 2}{20}$ for $0 \leqslant a < 4$. [3]
\end{enumerate}
Sales in 100 randomly chosen weeks gave the following grouped frequency table.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & $0 \leqslant x < 1$ & $1 \leqslant x < 2$ & $2 \leqslant x < 3$ & $3 \leqslant x < 4$ & $4 \leqslant x < 5$ \\
\hline
Frequency & 16 & 12 & 18 & 30 & 24 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Carry out a goodness of fit test at the $10\%$ significance level of whether f$(x)$ fits the data. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR S3 2012 Q5 [10]}}