(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?

(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]

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.