S3 June 2018 Q6
6. David carries out an experiment with 4 identical dice, each with faces numbered 1 to 6. He rolls the 4 dice and counts the number of dice showing an even number on the uppermost face. He repeats this 150 times. The results are summarised in the table below.
| No. of dice showing an even number | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency | 12 | 45 | 36 | 39 | 18 |
David defines the random variable \(C\) as the number of dice showing an even number on the uppermost face when the four dice are thrown.
David claims that \(C \sim \mathrm{B}(4, 0.5)\)
John claims that \(C \sim \mathrm{B}(4, p)\)
John decides to test his claim. He calculates expected frequencies using the results of David’s experiment and obtains the following table.
| No. of dice showing an even number | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Expected frequency | 8.65 | 36.00 | \(d\) | 39.00 | \(e\) |
John obtained a test statistic of 16.9 and carries out a test at the 1% level of significance.
| Scheme | Marks | ||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\mathrm{H}_0 :\) B(4, 0.5) is a suitable model \(\mathrm{H}_1 :\) B(4, 0.5) is not a suitable model | B1 | ||||||||||||||||||||||||||||||
| M1A1 | ||||||||||||||||||||||||||||||
| \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\) or \(\chi^2 = \sum \frac{O_i^2}{E_i} - N\) | M1A1 | ||||||||||||||||||||||||||||||
| \(\chi^2 = 17.52\) or \(\chi^2 = 167.52 - 150 = 17.52\) awrt 17.5 | A1 | ||||||||||||||||||||||||||||||
| \(\nu = 4, \quad \chi^2_4(1\%) = 13.277\) | B1, B1ft | ||||||||||||||||||||||||||||||
| (Reject \(\mathrm{H}_0\),) B(4,0.5) is not a suitable model or David’s claim incorrect. | A1 | ||||||||||||||||||||||||||||||
| (9) |
Notes
(corrected from the printed mark scheme: the second statistic is printed as \(\chi = 167.52 - 150 = 17.52\))
1st B1 Accept ‘Binomial with \(p = 0.5\)’ replacing ‘B(4, 0.5)’
1st M1 for attempt at \(E_i = 150 \times \mathrm{P}(X = i)\) with at least 2 values correct.
1st A1 at least 4 \(E_i\) correct to 3sf cao. Condone truncation.
2nd M1 for at least 2 correct calculations from 4th or 5th column.
2nd A1 at least 4 correct to 3sf from 4th or 5th column. Condone truncation.
3rd A1 for a test statistic of awrt 17.5 Answer only implies 2ndM1 2ndA1 3rdA1
4th A1 for correct conclusion rejecting binomial model. Condone missing parameters here.
Award provided their test statistic >11.345
| Scheme | Marks |
|---|---|
| \(\hat{p} = \dfrac{0 \times 12 + 1 \times 45 + 2 \times 36 + 3 \times 39 + 4 \times 18}{4 \times 150} = 0.51\) | M1 A1 |
| (2) |
Notes
1st M1 for attempting \(\hat{p} = \frac{\sum fx}{600}\) with at least 2 values on the numerator correct
1st A1 for 0.51 cao
| Scheme | Marks |
|---|---|
| \(d = 150 \times 6 \times 0.51^2 \times 0.49^2 = 56.205009\) awrt 56.2 | M1, A1 |
| \(e = 150 - (8.65 + 36.00 + 39.00 + \text{"}d\text{"}) = 10.144991\) awrt 10.1 or 10.2 or \(e = 150 \times 0.51^4 = 10.1478015\) | B1ft |
| (3) |
Notes
1st M1 \(d = 150 \times 6 \times (\text{their } \hat{p})^2 \times (1 - \text{their } \hat{p})^2\)
1st A1 awrt 56.2
1st B1ft awrt 10.1 or follow from "\(d\)"
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 :\) B(4, \(p\)) is a suitable model \(\mathrm{H}_1 :\) B(4, \(p\)) is not a suitable model | B1 |
| (1) |
Notes
1st B1 accept \(\mathrm{H}_0 :\) Binomial is a suitable model \(\mathrm{H}_1 :\) Binomial is not a suitable model
| Scheme | Marks |
|---|---|
| \(\nu = 3, \quad \chi^2_3(1\%) = 11.345\) | B1B1ft |
| (16.9>11.345) Reject \(\mathrm{H}_0\) Binomial is not a suitable model or John’s claim incorrect or equivalent contextualised statement that rejects the Binomial model. | B1 |
| (3) | |
| (18 marks) |
Notes
1st B1 \(\nu = 3\), 2nd B1 11.345, follow through their \(\nu \neq\) their value in part (a)
3rd B1 Correct statement rejecting \(\mathrm{H}_0\)