S3 June 2007 Q3
3. The time, in minutes, it takes Robert to complete the puzzle in his morning newspaper each day is normally distributed with mean 18 and standard deviation 3. After taking a holiday, Robert records the times taken to complete a random sample of 15 puzzles and he finds that the mean time is 16.5 minutes. You may assume that the holiday has not changed the standard deviation of times taken to complete the puzzle.
Stating your hypotheses clearly test, at the 5% level of significance, whether or not there has been a reduction in the mean time Robert takes to complete the puzzle. (7)
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \mu = 18, \qquad \mathrm{H}_1 : \mu \lt 18\) | B1, B1 |
| \(z = \dfrac{16.5 - 18}{3/\sqrt{15}} =,\ -1.9364\ldots\) AWRT −1.94 | M1, A1 |
| 5% one tail c.v. is \(z = (-)\,1.6449\) or probability (AWRT 0.026) (\(\pm\)) 1.6449 | B1 |
| \(-1.94 \lt -1.6449\) or significant or reject \(\mathrm{H}_0\) or in critical region | M1 |
| There is evidence that the (mean) time to complete the puzzles has reduced Or Robert is getting faster (at doing the puzzles) | A1f.t. |
| (7) | |
| (7 marks) |
Notes
1st & 2nd B1 must see \(\mu\) and 18
1st M1 for attempting test statistic, allow \(\pm\). Or attempt at critical value for \(\bar{X}\): \(\mu - z \times \dfrac{3}{\sqrt{15}}\)
1st A1 for AWRT −1.94. Allow use of \(|z| = +1.94\) to score M1A1. Or critical value = AWRT 16.7.
3rd B1 for AWRT 0.026 (i.e. correct probability only) or \(\pm\) 1.6449. (May be seen in cv formula)
2nd M1 for correct comparison or statement relating their test statistic and 1.6449 or their probability and 0.05. Ignore their hypotheses if any or assume they were correct.
2nd A1f.t. for conclusion in context which refers to “speed” or “time”. Depends only on previous M