OCR S3 2014 June — Question 2 6 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2014
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared goodness of fit
TypeChi-squared goodness of fit: Given ratios
DifficultyStandard +0.3 This is a straightforward chi-squared goodness of fit test with given expected ratios. Students need to calculate expected frequencies from the 9:3:3:1 ratio, compute the test statistic using the standard formula, and compare to critical values. It's slightly easier than average because the ratio is given explicitly, requiring only routine application of the chi-squared test procedure with no conceptual complications.
Spec5.06b Fit prescribed distribution: chi-squared test

2 In a study of the inheritance of skin colouration in corn snakes, a researcher found 865 snakes with black and orange bodies, 320 snakes with black bodies, 335 snakes with orange bodies and 112 snakes with bodies of other colours. Theory predicts that snakes of these colours should occur in the ratios \(9 : 3 : 3 : 1\). Test, at the \(5 \%\) significance level, whether these experimental results are compatible with theory.

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0\): The data can be modelled by the theory; \(H_1\): The data can't be modelled by the theory.B1 For both. Allow compatible.
Expected values 918, 306, 306, 102B1 Can be implied by 7.43
\(TS = \frac{(865-918)^2}{918} + \ldots\)M1
\(= 7.43\)A1
\(TS < 7.815\), do not reject \(H_0\)M1 ft TS. \(p > 0.05\) do not reject \(H_0\)
There is insufficient evidence to conclude that the data can't be modelled by the theoryA1 ft TS. \(p = 0.05939\) and conclusion
[6]
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: The data can be modelled by the theory; $H_1$: The data can't be modelled by the theory. | B1 | For both. Allow compatible. |
| Expected values 918, 306, 306, 102 | B1 | Can be implied by 7.43 |
| $TS = \frac{(865-918)^2}{918} + \ldots$ | M1 | |
| $= 7.43$ | A1 | |
| $TS < 7.815$, do not reject $H_0$ | M1 | ft TS. $p > 0.05$ do not reject $H_0$ |
| There is insufficient evidence to conclude that the data can't be modelled by the theory | A1 | ft TS. $p = 0.05939$ and conclusion |
| **[6]** | | |

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2 In a study of the inheritance of skin colouration in corn snakes, a researcher found 865 snakes with black and orange bodies, 320 snakes with black bodies, 335 snakes with orange bodies and 112 snakes with bodies of other colours. Theory predicts that snakes of these colours should occur in the ratios $9 : 3 : 3 : 1$. Test, at the $5 \%$ significance level, whether these experimental results are compatible with theory.

\hfill \mbox{\textit{OCR S3 2014 Q2 [6]}}