| Date | May 2013 | Marks available | 8 | Reference code | 13M.3sp.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Calculate and Determine | Question number | 3 | Adapted from | N/A |
Question
The number of machine breakdowns occurring in a day in a certain factory may be assumed to follow a Poisson distribution with mean \(\mu \). The value of \(\mu \) is known, from past experience, to be 1.2. In an attempt to reduce the value of \(\mu \), all the machines are fitted with new control units. To investigate whether or not this reduces the value of \(\mu \), the total number of breakdowns, x, occurring during a 30-day period following the installation of these new units is recorded.
State suitable hypotheses for this investigation.
It is decided to define the critical region by \(x \leqslant 25\).
(i) Calculate the significance level.
(ii) Assuming that the value of \(\mu \) was actually reduced to 0.75, determine the probability of a Type II error.
Markscheme
\({{\text{H}}_0}:\mu = 1.2\); \({{\text{H}}_1}:\mu < 1.2\) A1
Note: Accept “ \({{\text{H}}_0}:\) (\(30\)-day) mean \( = 36\); \({{\text{H}}_1}:\) (\(30\)-day) mean \( = 36\) ”.
[1 mark]
(i) let X denote the number of breakdowns in 30 days
then under \({{\text{H}}_0}\) , \(E(X) = 36\) (A1)
\({\text{sig level}} = {\text{P}}(X \leqslant 25|{\text{mean}} = 36)\) (M1)(A1)
= 0.0345 (3.45%) A1
Note: Accept any answer that rounds to 0.035 (3.5%) .
Note: Do not accept the use of a normal approximation.
(ii) under \({{\text{H}}_1}\), \(E(X) = 22.5\) (A1)
\(P{\text{(Type II error)}} = P(X \geqslant 26|{\text{mean}} = 22.5)\) (M1)(A1)
= 0.257 A1
Note: Accept any answer that rounds to 0.26.
Note: Do not accept the use of a normal approximation.
[8 marks]
Examiners report
This question was well answered by many candidates. The most common error was to attempt to use a normal approximation to find approximate probabilities instead of the Poisson distribution to find the exact probabilities. Some candidates appeared not to be familiar with the term ‘Type II error probability’ which made (b)(ii) inaccessible. Another fairly common error was to believe that the complement of \(x \leqslant 25\) is \(x \geqslant 25\).
This question was well answered by many candidates. The most common error was to attempt to use a normal approximation to find approximate probabilities instead of the Poisson distribution to find the exact probabilities. Some candidates appeared not to be familiar with the term ‘Type II error probability’ which made (b)(ii) inaccessible. Another fairly common error was to believe that the complement of \(x \leqslant 25\) is \(x \geqslant 25\).
