| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| 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 uniformity with clearly defined hypotheses, equal expected frequencies (150 each), and a standard 4-category setup. The calculation is routine with no conceptual challenges—students simply compute (O-E)²/E for each cell, sum to get the test statistic, compare to χ²₃ critical value, and conclude. Slightly above average difficulty only because it requires proper hypothesis formulation and interpretation, but remains a textbook application of the chi-squared test. |
| Spec | 5.06a Chi-squared: contingency tables |
| Period | 1 | 2 | 3 | 4 |
| Number of accidents | 138 | 127 | 165 | 170 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: p_1 = p_2 = p_3 = p_4\) | Indication of equality of proportions | |
| (\(H_1\): They are not all equal) | B1 | |
| Expected values under \(H_0 = 150\) | B1 | |
| \(X^2 = (12^2 + 23^2 + 15^2 + 20^2)/150\) | M1 | At least one correct term |
| \(= 8.653\) | A1 | |
| Critical value with 3 d.f. \(= 7.815\) | B1 | |
| (\(X^2 > 7.185\) so) reject \(H_0\) and accept that proportions are different | B1\(\sqrt{}\) | 6 |
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: p_1 = p_2 = p_3 = p_4$ | | Indication of equality of proportions |
| ($H_1$: They are not all equal) | B1 | |
| Expected values under $H_0 = 150$ | B1 | |
| $X^2 = (12^2 + 23^2 + 15^2 + 20^2)/150$ | M1 | At least one correct term |
| $= 8.653$ | | A1 | Accept art $8.65$ or $8.66$ |
| Critical value with 3 d.f. $= 7.815$ | B1 | |
| ($X^2 > 7.185$ so) reject $H_0$ and accept that proportions are different | B1$\sqrt{}$ | **6** | ft critical value |
---2 The manager of a factory with a large number of employees investigated when accidents to employees occurred during 8-hour shifts. An analysis was made of 600 randomly chosen accidents that occurred over a year. The following table shows the numbers of accidents occurring in the four consecutive 2-hour periods of the 8-hour shifts.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Period & 1 & 2 & 3 & 4 \\
\hline
Number of accidents & 138 & 127 & 165 & 170 \\
\hline
\end{tabular}
\end{center}
Test, at the $5 \%$ significance level, whether the proportions of all accidents that occur in the four time periods differ.
\hfill \mbox{\textit{OCR S3 2006 Q2 [6]}}