| Date | May 2008 | Marks available | 16 | Reference code | 08M.3sp.hl.TZ2.5 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ2 |
| Command term | Calculate and Find | Question number | 5 | Adapted from | N/A |
Question
A population is known to have a normal distribution with a variance of 3 and an unknown mean \(\mu \) . It is proposed to test the hypotheses \({{\text{H}}_0}:\mu = 13,{\text{ }}{{\text{H}}_1}:\mu > 13\) using the mean of a sample of size 2.
(a) Find the appropriate critical regions corresponding to a significance level of
(i) 0.05;
(ii) 0.01.
(b) Given that the true population mean is 15.2, calculate the probability of making a Type II error when the level of significance is
(i) 0.05;
(ii) 0.01.
(c) How is the change in the probability of a Type I error related to the change in the probability of a Type II error?
Markscheme
(a) With \({{\text{H}}_0},{\text{ }}\bar X \sim {\text{N}}\left( {13,\frac{3}{2}} \right) = {\text{N(13, 1.5)}}\) (M1)(A1)
(i) 5 % for N(0,1) is 1.645
so \(\frac{{\bar x - 13}}{{\sqrt {1.5} }} = 1.645\) (M1)(A1)
\(\bar x = 13 + 1.645\sqrt {1.5} \)
\( = 15.0\,\,\,\,\,{\text{(3 s.f.)}}\) A1 N0
\({\text{[15.0, }}\infty {\text{[}}\)
(ii) 1% for N(0, 1) is 2.326
so \(\frac{{\bar x - 13}}{{\sqrt {1.5} }} = 2.326\) (M1)(A1)
\(\bar x = 13 + 2.326\sqrt {1.5} \)
\( = 15.8\,\,\,\,\,{\text{(3 s.f., accept 15.9)}}\) A1 N0
\({\text{[15.8, }}\infty {\text{[}}\)
[8 marks]
(b) (i) \(\beta = {\text{P}}(\bar X < 15.0147)\) M1
\( = 0.440\) A2
(ii) \(\beta = {\text{P}}(\bar X < 15.8488)\) M1
\( = 0.702\) A2
[6 marks]
(c) The probability of a Type II error increases when the probability of a Type I error decreases. R2
[2 marks]
Total [16 marks]
Examiners report
This question proved to be the most difficult. The range of solutions ranged from very good to very poor. Many students thought that \(P(TypeI) = 1 - P(TypeII)\) when in fact \(1 - P(TypeII)\) is the power of the test.
