A2 October 2021 Q1
1. Kelly throws a tetrahedral die \(n\) times and records the number on which it lands for each throw.
She calculates the expected frequency for each number to be 43 if the die was unbiased.
The table below shows three of the frequencies Kelly records but the fourth one is missing.
| Number | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Frequency | 47 | 34 | 36 | \(x\) |
Kelly wishes to test, at the 5% level of significance, whether or not there is evidence that the tetrahedral die is unbiased.
| Scheme | Marks | AO |
|---|---|---|
| \(x = 4 \times 43 - 47 - 34 - 36 = 55\)* | B1* | 3.4 |
| (1) |
Notes
B1*: Using the uniform model to show the missing observed value eg \(x = \dfrac{43 - 0.25 \times (47 + 34 + 36)}{0.25} = 55\)
| Scheme | Marks | AO |
|---|---|---|
| \(\nu = 4 - 1 = 3\) since the only constraint is that the totals agree | B1 | 2.4 |
| (1) |
Notes
B1: \(4 - 1 = 3\) (may be in words) and explanation of what the constraint is
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0\): The die is unbiased \(\mathrm{H}_1\): The die is biased | B1 | 2.1 |
| Test Statistic \(= \dfrac{(47 - 43)^2}{43} + \dfrac{(34 - 43)^2}{43} + \dfrac{(36 - 43)^2}{43} + \dfrac{(55 - 43)^2}{43}\) | M1 | 1.1b |
| \(= 6.744\ldots\) | A1 | 1.1b |
| \(\chi^2_{(3, 0.05)} = 7.815\) | B1 | 1.1b |
| Not in the critical region since \(7.815 \gt \text{“}6.74\ldots\text{”}\) therefore insufficient evidence to reject \(\mathrm{H}_0\) Inconclusive test - consistent with the die being unbiased. | A1 | 3.5a |
| (5) | ||
| (7 marks) |
Notes
B1: Both hypotheses correct. eg The data fits a discrete uniform distribution
M1: Attempting to find \(\sum\dfrac{(O - E)^2}{E}\) or \(\sum\dfrac{O^2}{E} - N\) May be implied by awrt 6.74 or \(p\) value of 0.0805…
A1: awrt 6.74 or \(\dfrac{290}{43}\) oe May be implied by \(p\) value of 0.0805…
B1: awrt 7.82 (Calc 7.8147…)
A1: Drawing correct inference in context. Need the word die or tetrahedral