| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Uniform |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness-of-fit test for uniform distribution with equal expected frequencies. Students need to state hypotheses, calculate expected frequencies (96/6=16), compute chi-squared statistic, find critical value, and conclude. While it requires multiple steps, it's a standard S3 application with no conceptual challenges beyond textbook exercises. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Sector | \(0° -\) | \(45° -\) | \(90° -\) | \(180° -\) | \(270° -\) | \(315° - 360°\) |
| No. of Shots | 18 | 19 | 15 | 20 | 9 | 15 |
| Answer | Marks |
|---|---|
| Method 1: exp. freq. \(0-45 = \frac{45}{360} \times 96 = 12\) etc. giving exp. freqs. 12, 12, 24, 24, 12, 12 | M1 A1 |
| Hypothesis setup: \(H_0\): continuous uniform distribution is a suitable model; \(H_1\): continuous uniform distribution is not a suitable model | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| O | E | (O−E) |
| 18 | 12 | 6 |
| 19 | 12 | 7 |
| 15 | 24 | −9 |
| 20 | 24 | −4 |
| 9 | 12 | −3 |
| 15 | 12 | 3 |
| Test statistic: \(\sum \frac{(O-E)^2}{E} = 12.625\) | M1 A2 | |
| Critical value and conclusion: \(\nu = 6 - 1 = 5\); \(\chi^2_{crit}(5\%) = 11.070\); \(12.625 > 11.070\) \(\therefore\) reject \(H_0\); continuous uniform distribution is not a suitable model | M1 A1 A1 | (9 marks total) |
**Method 1:** exp. freq. $0-45 = \frac{45}{360} \times 96 = 12$ etc. giving exp. freqs. 12, 12, 24, 24, 12, 12 | M1 A1 |
**Hypothesis setup:** $H_0$: continuous uniform distribution is a suitable model; $H_1$: continuous uniform distribution is not a suitable model | B1 |
**Chi-squared calculation table:**
| O | E | (O−E) | $\frac{(O-E)^2}{E}$ |
|---|---|-------|---------|
| 18 | 12 | 6 | 3 |
| 19 | 12 | 7 | 4.0833 |
| 15 | 24 | −9 | 3.375 |
| 20 | 24 | −4 | 0.6667 |
| 9 | 12 | −3 | 0.75 |
| 15 | 12 | 3 | 0.75 |
**Test statistic:** $\sum \frac{(O-E)^2}{E} = 12.625$ | M1 A2 |
**Critical value and conclusion:** $\nu = 6 - 1 = 5$; $\chi^2_{crit}(5\%) = 11.070$; $12.625 > 11.070$ $\therefore$ reject $H_0$; continuous uniform distribution is not a suitable model | M1 A1 A1 | (9 marks total)
---Commentators on a game of cricket say that a certain batsman is "playing shots all round the ground". A sports statistician wishes to analyse this claim and records the direction of shots played by the batsman during the course of his innings. She divides the $360°$ around the batsman into six sectors, measuring the angle of each shot clockwise from the line between the wickets, and obtains the following results:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Sector & $0° -$ & $45° -$ & $90° -$ & $180° -$ & $270° -$ & $315° - 360°$ \\
\hline
No. of Shots & 18 & 19 & 15 & 20 & 9 & 15 \\
\hline
\end{tabular}
\end{center}
Stating your hypotheses clearly and using a 5% level of significance test whether or not these data can be modelled by a continuous uniform distribution. [9]
\hfill \mbox{\textit{Edexcel S3 Q2 [9]}}