| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| 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 a continuous distribution. Part (a) is routine integration of a simple pdf, and part (b) follows a standard template: calculate expected frequencies using the given formula, compute chi-squared statistic, compare to critical value. All steps are mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Delay ( \(\boldsymbol { t }\) hours) | Frequency |
| \(0 \leqslant t < 1\) | 10 |
| \(1 \leqslant t < 2\) | 13 |
| \(2 \leqslant t < 3\) | 24 |
| \(3 \leqslant t < 4\) | 35 |
| \(4 \leqslant t < 5\) | 68 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_a^{a+1}\frac{2}{25}t\,dt = \frac{2}{25}\left[\frac{t^2}{2}\right]_a^{a+1}\) | M1 | For correct integration, ignore limits or finding area of trapezium |
| \(\frac{1}{25}\left((a+1)^2-a^2\right)\) or \(\frac{1}{25}(a+1)^2-\frac{1}{25}a^2\) | M1 | For substitution of limits |
| \(\frac{1}{25}(a^2+2a+1-a^2) = \frac{1}{25}(2a+1)\) | A1* | Answer given; at least one correct line of working from method mark to final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): The data could be modelled by the p.d.f.; \(H_1\): The data could not be modelled by the p.d.f. | B1 | Both hypotheses correct. Allow \(H_0\): \(f(t)\) is a suitable model |
| Expected frequencies: 6, 18, 30, 42, 54 | M1 A1 | M1 for correct method to find at least one expected frequency e.g. \(\frac{1}{25}\times 150\); A1 all 5 correct |
| \(\sum\frac{(O-E)^2}{E}=\frac{(10-6)^2}{6}+\cdots+\frac{(68-54)^2}{54}\) | M1 | At least 2 correct expressions/values ft their expected frequencies |
| \(=10.05\ldots\) awrt 10.1 | A1 | awrt 10.1 |
| \(\nu=4\) | B1 | Implied by correct critical value 9.488 |
| \(\chi^2_4(0.05)=9.488 \Rightarrow\) CR \(\geqslant 9.488\) | B1ft | 9.488 or better ft their DoF |
| In CR; sufficient evidence to reject \(H_0\); data does not fit the given p.d.f. | dA1 | Dependent on 2nd M1; if no hypotheses or wrong way round, A0 |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_a^{a+1}\frac{2}{25}t\,dt = \frac{2}{25}\left[\frac{t^2}{2}\right]_a^{a+1}$ | M1 | For correct integration, ignore limits or finding area of trapezium |
| $\frac{1}{25}\left((a+1)^2-a^2\right)$ or $\frac{1}{25}(a+1)^2-\frac{1}{25}a^2$ | M1 | For substitution of limits |
| $\frac{1}{25}(a^2+2a+1-a^2) = \frac{1}{25}(2a+1)$ | A1* | Answer given; at least one correct line of working from method mark to final answer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: The data could be modelled by the p.d.f.; $H_1$: The data could not be modelled by the p.d.f. | B1 | Both hypotheses correct. Allow $H_0$: $f(t)$ is a suitable model |
| Expected frequencies: 6, 18, 30, 42, 54 | M1 A1 | M1 for correct method to find at least one expected frequency e.g. $\frac{1}{25}\times 150$; A1 all 5 correct |
| $\sum\frac{(O-E)^2}{E}=\frac{(10-6)^2}{6}+\cdots+\frac{(68-54)^2}{54}$ | M1 | At least 2 correct expressions/values ft their expected frequencies |
| $=10.05\ldots$ awrt 10.1 | A1 | awrt 10.1 |
| $\nu=4$ | B1 | Implied by correct critical value 9.488 |
| $\chi^2_4(0.05)=9.488 \Rightarrow$ CR $\geqslant 9.488$ | B1ft | 9.488 or better ft their DoF |
| In CR; sufficient evidence to reject $H_0$; data does not fit the given p.d.f. | dA1 | Dependent on 2nd M1; if no hypotheses or wrong way round, A0 |
---\begin{enumerate}
\item It is suggested that the delay, in hours, of certain flights from a particular country may be modelled by the continuous random variable, $T$, with probability density function
\end{enumerate}
$$f ( t ) = \left\{ \begin{array} { c l }
\frac { 2 } { 25 } t & 0 \leqslant t < 5 \\
0 & \text { otherwise }
\end{array} \right.$$
(a) Show that for $0 \leqslant a \leqslant 4$
$$P ( a \leqslant T < a + 1 ) = \frac { 1 } { 25 } ( 2 a + 1 )$$
A random sample of 150 of these flights is taken. The delays are summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Delay ( $\boldsymbol { t }$ hours) & Frequency \\
\hline
$0 \leqslant t < 1$ & 10 \\
\hline
$1 \leqslant t < 2$ & 13 \\
\hline
$2 \leqslant t < 3$ & 24 \\
\hline
$3 \leqslant t < 4$ & 35 \\
\hline
$4 \leqslant t < 5$ & 68 \\
\hline
\end{tabular}
\end{center}
(b) Test, at the $5 \%$ significance level, whether the given probability density function is a suitable model for these delays.\\
You should state your hypotheses, expected frequencies, test statistic and the critical value used.
\hfill \mbox{\textit{Edexcel S3 2023 Q4 [11]}}