| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2018 |
| Session | September |
| Marks | 8 |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Given ratios |
| Difficulty | Standard +0.3 This is a straightforward chi-squared goodness of fit test with given ratios (1:2:3:4). Students must calculate expected frequencies from the ratio, compute the test statistic, and compare to critical values. It's slightly easier than average because the ratio is explicitly stated, the sample size is small (simple arithmetic), and it's a standard textbook application requiring no novel insight. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Edge number | 1 | 2 | 3 | 4 |
| Frequency | 3 | 7 | 4 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): data consistent with proportions \(1 : 2 : 3 : 4\) | B1 | 1.1a |
| \(H_1\): data not consistent with the proportions | ||
| Expected frequencies \(2, 4, 6, 8\) | B1 | 3.1b |
| Combine \(2 \& 4\) | M1 | 1.1a |
| Calculate \(\Sigma(O-E)^2/E\) | M1* | 1.1a |
| \(2.666\ldots + 0.666\ldots + 0.5 = 3.833\ldots\) | A1 | 1.1 |
| \(\nu = 2\) so CV = 4.605 | B1ft | 1.1 |
| \(3.833 < 4.605\) so do not reject \(H_0\) | depM1ft | 2.2b |
| Insufficient evidence that data is inconsistent with intended proportions | A1ft [8] | 3.5a |
$H_0$: data consistent with proportions $1 : 2 : 3 : 4$ | B1 | 1.1a | Must refer to proportions but needn't state them
$H_1$: data not consistent with the proportions | | |
Expected frequencies $2, 4, 6, 8$ | B1 | 3.1b |
Combine $2 \& 4$ | M1 | 1.1a |
Calculate $\Sigma(O-E)^2/E$ | M1* | 1.1a | Can be implied by 3.83
$2.666\ldots + 0.666\ldots + 0.5 = 3.833\ldots$ | A1 | 1.1 | $\chi^2$, awrt 3.83
$\nu = 2$ so CV = 4.605 | B1ft | 1.1 | Correct $\chi^2$ CV, ft only on not combined
$3.833 < 4.605$ so do not reject $H_0$ | depM1ft | 2.2b | Correct comparison
Insufficient evidence that data is inconsistent with intended proportions | A1ft [8] | 3.5a | Contextualised, not too definite | SC not combined: B1B1 M0M1 A0B1 M1A1
---5 Hal designs a 4-edged spinner with edges labelled 1, 2, 3 and 4. He intends that the probability that the spinner will land on any edge should be proportional to the number on that edge. He spins the spinner 20 times and on each spin he records the number of the edge on which it lands. The results are shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
Edge number & 1 & 2 & 3 & 4 \\
\hline
Frequency & 3 & 7 & 4 & 6 \\
\hline
\end{tabular}
\end{center}
Test at the $10 \%$ significance level whether the results are consistent with the intended probabilities.
\hfill \mbox{\textit{OCR Further Statistics 2018 Q5 [8]}}