| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Uniform |
| Difficulty | Standard +0.8 This S3 chi-squared test requires recognizing that unequal class widths necessitate calculating expected frequencies proportional to interval width (not just 228/6), then performing the standard test procedure. The unequal intervals add conceptual complexity beyond routine goodness-of-fit questions, requiring careful thought about what 'uniform distribution' means geometrically. |
| Spec | 5.06c Fit other distributions: discrete and continuous |
| \(0 - 1\) | \(1 - 2\) | \(2 - 4\) | \(4 - 6\) | \(6 - 9\) | \(9 - 12\) | ||
| Number of items | 22 | 15 | 44 | 37 | 52 | 58 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
Calculate widths \(b-a\) and probabilities \(P(a\leq X| M1 |
For calculation of at least 3 widths and attempting proportions/probs, or for 1:2:3 ratio |
|
| Correct probabilities: \(\frac{1}{12}, \frac{1}{12}, \frac{1}{6}, \frac{1}{6}, \frac{1}{4}, \frac{1}{4}\) | A1 | For correct probabilities |
| Expected frequencies: \(19, 19, 38, 38, 57, 57\) | A1 | All correct expected frequencies |
| \(\sum\frac{(O_i-E_i)^2}{E_i} = \frac{313}{114} = 2.75\) | M1, dM1, A1 | At least 3 correct expressions; dependent on 2nd M1; correct sum awrt \(2.75\) |
| \(H_0\): continuous uniform distribution is a good fit; \(H_1\): continuous uniform distribution is not a good fit | B1 | |
| \(v = 6-1 = 5\) | B1 | |
| \(\chi^2_5(0.05) = 11.070\) | B1ft | ft their \(v\) |
| \(2.75 < 11.070\), insufficient evidence to reject \(H_0\) | M1 | Correct statement based on test statistic \((>1)\) and cv \((>3.8)\) |
| Continuous uniform distribution is a suitable model | A1 | Correct comment suggesting continuous uniform model is suitable; no ft |
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Calculate widths $b-a$ and probabilities $P(a\leq X<b)$ for uniform distribution | M1 | For calculation of at least 3 widths and attempting proportions/probs, or for 1:2:3 ratio |
| Correct probabilities: $\frac{1}{12}, \frac{1}{12}, \frac{1}{6}, \frac{1}{6}, \frac{1}{4}, \frac{1}{4}$ | A1 | For correct probabilities |
| Expected frequencies: $19, 19, 38, 38, 57, 57$ | A1 | All correct expected frequencies |
| $\sum\frac{(O_i-E_i)^2}{E_i} = \frac{313}{114} = 2.75$ | M1, dM1, A1 | At least 3 correct expressions; dependent on 2nd M1; correct sum awrt $2.75$ |
| $H_0$: continuous uniform distribution is a good fit; $H_1$: continuous uniform distribution is not a good fit | B1 | |
| $v = 6-1 = 5$ | B1 | |
| $\chi^2_5(0.05) = 11.070$ | B1ft | ft their $v$ |
| $2.75 < 11.070$, insufficient evidence to reject $H_0$ | M1 | Correct statement based on test statistic $(>1)$ and cv $(>3.8)$ |
| Continuous uniform distribution is a suitable model | A1 | Correct comment suggesting continuous uniform model is suitable; no ft |6. A total of 228 items are collected from an archaeological site. The distance from the centre of the site is recorded for each item. The results are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ c }
Distance from the \\
centre of the site (m) \\
\end{tabular} & $0 - 1$ & $1 - 2$ & $2 - 4$ & $4 - 6$ & $6 - 9$ & $9 - 12$ \\
\hline
Number of items & 22 & 15 & 44 & 37 & 52 & 58 \\
\hline
\end{tabular}
\end{center}
Test, at the $5 \%$ level of significance, whether or not the data can be modelled by a continuous uniform distribution. State your hypotheses clearly.\\
\hfill \mbox{\textit{Edexcel S3 Q6 [12]}}