The mean weight of a certain breed of bird is believed to be 2.5 kg. In order to test this belief, it is planned to determine the weights \({x_1}{\text{ , }}{x_2}{\text{ , }}{x_3}{\text{ , }} \ldots {\text{, }}{x_{16}}\) (in kg) of sixteen of these birds and then to calculate the sample mean \({\bar x}\) . You may assume that these weights are a random sample from a normal distribution with standard deviation 0.1 kg.

(a)     State suitable hypotheses for a two-tailed test.

(b)     Find the critical region for \({\bar x}\) having a significance level of 5 %.

(c)     Given that the mean weight of birds of this breed is actually 2.6 kg, find the probability of making a Type II error.

(a)     \({H_0}:\mu = 2.5\)     A1

\({H_1}:\mu \ne 2.5\)     A1

[2 marks]

(b)     the critical values are \(2.5 \pm 1.96 \times \frac{{0.1}}{{\sqrt {16} }}\) ,     (M1)(A1)(A1)

i.e. 2.45, 2.55     (A1)

the critical region is \(\bar x < 2.45 \cup \bar x > 2.55\)     A1A1

Note: Accept \( \leqslant ,{\text{ }} \geqslant \) .

[6 marks]

(c)     \({\bar X}\) is now \({\text{N}}(2.6,{\text{ }}{0.025^2})\)     A1

a Type II error is accepting \({H_0}\) when \({H_1}\) is true     (R1)

thus we require

\({\text{P}}(2.45 < \bar X < 2.55)\)     M1A1

\( = 0.0228\,\,\,\,\,\)(Accept 0.0227)     A1

Note: If critical values of 2.451 and 2.549 are used, accept 0.0207.

[5 marks]

Total [13 marks]

In (a), some candidates incorrectly gave the hypotheses in terms of \({\bar x}\) instead of \(\mu \). In (b), many candidates found the correct critical values but then some gave the critical region as \(2.45 < \bar x < 2.55\) instead of \(\bar x < 2.45 \cup \bar x > 2.55\) Many candidates gave the critical values correct to four significant figures and therefore were given an arithmetic penalty. In (c), many candidates correctly defined a Type II error but were unable to calculate the corresponding probability.