S2 January 2005 Q4
4. In an experiment, there are 250 trials and each trial results in a success or a failure.
(a) Write down two other conditions needed to make this into a binomial experiment. (2)
It is claimed that 10% of students can tell the difference between two brands of baked beans. In a random sample of 250 students, 40 of them were able to distinguish the difference between the two brands.
(b) Using a normal approximation, test at the 1% level of significance whether or not the claim is justified. Use a one-tailed test. (6)
(c) Comment on the acceptability of the assumptions you needed to carry out the test. (2)
| Scheme | Marks |
|---|---|
| Probability of success/failure is constant | B1 |
| Trials are independent | B1 |
| (2) |
| Scheme | Marks |
|---|---|
| Let \(p\) represent proportion of students who can distinguish between brands \(\mathrm{H}_0: p = 0.1;\ \mathrm{H}_1: p \gt 0.1\) | B1 |
| \(\alpha = 0.01;\ \text{CR}: z \gt 2.3263\) | B1 |
| \(np = 25;\ npq = 22.5\) | B1 |
| \(z = \dfrac{39.5 - 25}{\sqrt{22.5}} = 3.0568\ldots\) | M1 A1 |
| Reject \(\mathrm{H}_0\): claim cannot be accepted | A1ft |
| (6) |
Notes
1st B1 both
2nd B1 2.3263
3rd B1 both; can be implied
M1 standardisation with \(\pm 0.5\) & their \(\sqrt{npq}\)
A1 awrt 3.06
A1ft based on clear evidence from \(z\) or \(p\)
Alternative (b)
| Scheme | Marks |
|---|---|
| \(z = 3.06 \Rightarrow p = 0.9989 \gt 0.99\) or \(p = 0.0011 \lt 0.01\) | B1 |
B1 equivalent to 2.3263
| Scheme | Marks |
|---|---|
| e.g. \(np, nq\) both \(\gt 5\) – true so acceptable \(p\) close to 0.5 – not true, assumption not met success/failure not clear cut necessarily independence – one student influences another | B1 B1 |
| (2) | |
| (10 marks) |