| Date | May 2011 | Marks available | 5 | Reference code | 11M.3sp.hl.TZ0.4 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Calculate | Question number | 4 | Adapted from | N/A |
Question
The random variable X has a Poisson distribution with unknown mean \(\mu \) . It is required to test the hypotheses
\({H_0}:\mu = 3\) against \({H_1}:\mu \ne 3\) .
Let S denote the sum of 10 randomly chosen values of X . The critical region is defined as \((S \leqslant 22) \cup (S \geqslant 38)\) .
Calculate the significance level of the test.
Given that the value of \(\mu \) is actually 2.5, determine the probability of a Type II error.
Markscheme
under \({H_0}\) , \(S{\text{ is Po}}(30)\) (A1)
EITHER
\({\text{P}}(S \leqslant 22) = {\text{0.080569}} \ldots \) A1
\({\text{P}}(S \geqslant 38) = {\text{0.089012}} \ldots \) A1
significance level = 0.080569… + 0.089012… (M1)
= 0.170 A1
OR
\({\text{P}}(S \leqslant 22) = {\text{0.080569}} \ldots \) A1
\({\text{P}}(S \leqslant 37) = {\text{0.910987}} \ldots \) A1
significance level = 1 – (0.910987…) + 0.089012… (M1)
= 0.170 A1
Note: Accept 17 % or 0.17.
Note: Award 2 marks out of the final 4 marks for correct use of the Central Limit Theorem, giving 0.144 without a continuity correction and 0.171 with a continuity correction. The first (A1) is independent.
[5 marks]
S is now Po(25) (A1)
P (Type II error) = P (accept \({H_0}|\mu = 2.5\)) (M1)
\( = {\text{P}}\left( {23 \leqslant S \leqslant 37|S{\text{ is Po}}(25)} \right)\) (M1)
Note: Only one of the above M1 marks can be implied.
= 0.990789… – 0.317533… (A1)
= 0.673 A1
Note: Award 2 marks out of the final 4 marks for correct use of the Central Limit Theorem, giving 0.647 without a continuity correction and 0.685 with a continuity correction. The first (A1) is independent.
[5 marks]
Examiners report
Solutions to this question were often disappointing with many candidates not knowing what had to be done. Even those candidates who knew what to do sometimes made errors in evaluating the probabilities, often by misinterpreting the inequality signs. Candidates who used the Central Limit Theorem to evaluate the probabilities were given only partial credit on the grounds that the answers obtained were approximate and not exact.
Solutions to this question were often disappointing with many candidates not knowing what had to be done. Even those candidates who knew what to do sometimes made errors in evaluating the probabilities, often by misinterpreting the inequality signs. Candidates who used the Central Limit Theorem to evaluate the probabilities were given only partial credit on the grounds that the answers obtained were approximate and not exact.
