A factory makes wine glasses. The manager claims that on average 2 % of the glasses are imperfect. A random sample of 200 glasses is taken and 8 of these are found to be imperfect.

Test the manager’s claim at a 1 % level of significance using a one-tailed test.

Let X denote the number of imperfect glasses in the sample     (M1)

For recognising binomial or proportion or Poisson     A1

(\(X \sim {\text{B}}(200,{\text{ }}p)\) where p-value is the probability of a glass being imperfect)

Let \({{\text{H}}_0}:p{\text{-value}} = 0.02{\text{ and }}{{\text{H}}_1}:p{\text{-value}} > 0.02\)     A1A1

EITHER

p-value = 0.0493     A2

Using the binomial distribution \(p{\text{-value}} = 0.0493 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

OR

p-value = 0.0511     A2

Using the Poisson approximation to the binomial distribution since \(p{\text{-value}} = 0.0511 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

OR

p-value = 0.0217     A2

Using the one proportion z-test since \(p{\text{-value}} = 0.0217 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

Note: Use of critical values is acceptable.

[7 marks]

Many candidates used a t-test on this question. This was possibly because the sample was large enough to approximate normality of a proportion. The need to use a one-tailed test was often missed. When using the z-test of proportions p = 0.04 was often used instead of p = 0.02 . Not many candidates used the binomial distribution.