S2 June 2007 Q6
6. Linda regularly takes a taxi to work five times a week. Over a long period of time she finds the taxi is late once a week. The taxi firm changes her driver and Linda thinks the taxi is late more often. In the first week, with the new driver, the taxi is late 3 times.
You may assume that the number of times a taxi is late in a week has a Binomial distribution.
Test, at the 5% level of significance, whether or not there is evidence of an increase in the proportion of times the taxi is late. State your hypotheses clearly. (7)
One tail test Method 1
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : p = 0.2\) | B1 |
| \(\mathrm{H}_1 : p \gt 0.2\) | B1 |
| \(X \sim \mathrm{B}(5, 0.2)\) | M1 |
| \(\mathrm{P}(X \geqslant 3) = 1 - \mathrm{P}(X \leqslant 2)\) \(\left[\mathrm{P}(X \geqslant 3) = 1 - 0.9421 = 0.0579\right.\) \(\qquad\qquad\ \ = 1 - 0.9421\) \(\left.\mathrm{P}(X \geqslant 4) = 1 - 0.9933 = 0.0067\right]\) | M1 |
| \(= 0.0579\) CR \(X \geqslant 4\) | A1 |
| \(0.0579 \gt 0.05\) \(3 \leqslant 4\) or 3 is not in critical region or 3 is not significant | M1 |
| (Do not reject \(\mathrm{H}_0\).) There is insufficient evidence at the 5% significance level that there is an increase in the number of times the taxi/driver is late. Or Linda’s claim is not justified | B1 |
| (7) | |
| (7 marks) |
Notes
1st M1 may be implied
2nd M1 att \(\mathrm{P}(X \geqslant 3)\) | \(\mathrm{P}(X \geqslant 4)\)
A1 awrt 0.0579
One tail test: Method 2
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : p = 0.2\) | B1 |
| \(\mathrm{H}_1 : p \gt 0.2\) | B1 |
| \(X \sim \mathrm{B}(5, 0.2)\) | M1 |
| \(\mathrm{P}(X \lt 3) =\) \(\left[\mathrm{P}(X \lt 3) = 0.9421\right]\), \(\mathrm{P}(X \lt 4) = 0.9933\) \(0.9421\) CR \(X \geqslant 4\) | M1 A1 |
| \(0.9421 \lt 0.95\) \(3 \leqslant 4\) or 3 is not in critical region or 3 is not significant | M1 |
| (Do not reject \(\mathrm{H}_0\).) There is insufficient evidence at the 5% significance level that there is an increase in the number of times the taxi/driver is late. Or Linda’s claim is not justified | B1 |
| (7) |
1st M1 may be implied
2nd M1 att \(\mathrm{P}(X \lt 3)\) | \(\mathrm{P}(X \lt 4)\)
A1 awrt 0.942
Two tail test: Method 1
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : p = 0.2\) | B1 |
| \(\mathrm{H}_1 : p \ne 0.2\) | B0 |
| \(X \sim \mathrm{B}(5, 0.2)\) | M1 |
| \(\mathrm{P}(X \geqslant 3) = 1 - \mathrm{P}(X \leqslant 2)\) \(\left[\mathrm{P}(X \geqslant 3) = 1 - 0.9421 = 0.0579\right.\) \(\qquad\qquad\ \ = 1 - 0.9421\) \(\left.\mathrm{P}(X \geqslant 4) = 1 - 0.9933 = 0.0067\right]\) | M1 |
| \(= 0.0579\) CR \(X \geqslant 4\) | A1 |
| \(0.0579 \gt 0.025\) \(3 \leqslant 4\) or 3 is not in critical region or 3 is not significant | M1 |
| (Do not reject \(\mathrm{H}_0\).) There is insufficient evidence at the 5% significance level that there is an increase in the number of times the taxi/driver is late. Or Linda’s claim is not justified | B1 |
| (7) |
1st M1 may be implied
2nd M1 att \(\mathrm{P}(X \geqslant 3)\) | \(\mathrm{P}(X \geqslant 4)\)
A1 awrt 0.0579
Two tail test: Method 2
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : p = 0.2\) | B1 |
| \(\mathrm{H}_1 : p \ne 0.2\) | B0 |
| \(X \sim \mathrm{B}(5, 0.2)\) | M1 |
| \(\mathrm{P}(X \lt 3) =\) \(\left[\mathrm{P}(X \lt 3) = 0.9421\right]\), \(\mathrm{P}(X \lt 4) = 0.9933\) \(0.9421\) CR \(X \geqslant 4\) | M1 A1 |
| \(0.9421 \lt 0.975\) \(3 \leqslant 4\) or 3 is not in critical region or 3 is not significant | M1 |
| Do not reject \(\mathrm{H}_0\). There is insufficient evidence at the 5% significance level that there is an increase in the number of times the taxi/driver is late. Or Linda’s claim is not justified | B1 |
| (7) |
1st M1 may be implied
2nd M1 att \(\mathrm{P}(X \lt 3)\) | \(\mathrm{P}(X \lt 4)\)
A1 awrt 0.942
Special Case
If they use a probability of \(\dfrac{1}{7}\) throughout the question they may gain B1 B1 M0 M1 A0 M1 B1.
NB they must attempt to work out the probabilities using \(\dfrac{1}{7}\)