A teacher has forgotten his computer password. He knows that it is either six of the letter J followed by two of the letter R (i.e. JJJJJJRR) or three of the letter J followed by four of the letter R (i.e. JJJRRRR). The computer is able to tell him at random just two of the letters in his password.

The teacher decides to use the following rule to attempt to find his password.

If the computer gives him a J and a J, he will accept the null hypothesis that his password is JJJJJJRR.

Otherwise he will accept the alternative hypothesis that his password is JJJRRRR.

(a)     Define a Type I error.

(b)     Find the probability that the teacher makes a Type I error.

(c)     Define a Type II error.

(d)     Find the probability that the teacher makes a Type II error.

(a)     a Type I error is when \({{\text{H}}_0}\) is rejected, when \({{\text{H}}_0}\) is actually true     A1

[1 mark]

(b)     \({\text{P(}}{{\text{H}}_0}{\text{ rejected}}|{{\text{H}}_0}{\text{ true)}} = {\text{P(at least one R}}|{\text{6 J and 2 R)}}\)     M1

EITHER

\({\text{P(no R}}|{{\text{H}}_0}{\text{ true)}} = \frac{6}{8} \times \frac{5}{7} = \frac{{15}}{{28}}\)     (A1)

OR

let X count the number of R’s given by the computer under \({{\text{H}}_0},{\text{ }}X \sim {\text{Hyp(}}2,{\text{ }}2,{\text{ }}8)\)

\({\text{P}}(X = 0) = \frac{{\left( {\begin{array}{*{20}{c}}
  2 \\
  0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  6 \\
  2
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
  8 \\
  2
\end{array}} \right)}} = \frac{{15}}{{28}}\)     (A1)

THEN

\({\text{P(at least one R}}|{{\text{H}}_0}{\text{ true)}} = 1 - \frac{{15}}{{28}}\)     (M1)

\({\text{P(Type I error)}} = \frac{{13}}{{28}}\,\,\,\,\,( = 0.464)\)     A1

[4 marks]

(c)     a Type II error is when \({{\text{H}}_0}\) is accepted, when \({{\text{H}}_0}\) is actually false     A1

[1 mark]

(d)     \({\text{P(}}{{\text{H}}_0}{\text{ accepted}}|{{\text{H}}_0}{\text{ false)}} = {\text{P(2 J}}|{\text{3 J and 4 R)}}\)     M1

EITHER

\({\text{P(2 J}}|{{\text{H}}_0}{\text{ false)}} = \frac{3}{7} \times \frac{2}{6} = \frac{1}{7}\)     (A1)

OR

let Y count the number of R’s given by the computer.

\({{\text{H}}_0}\) false implies \(Y \sim {\text{Hyp(}}2,{\text{ }}4,{\text{ }}7)\)

\({\text{P}}(Y = 0) = \frac{{\left( {\begin{array}{*{20}{c}}
  4 \\
  0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  3 \\
  2
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
  7 \\
  2
\end{array}} \right)}} = \frac{1}{7}\)     (A1)
THEN

\({\text{P}}({\text{Type II error)}} = \frac{1}{7}( = 0.143)\)     A1

[3 marks]

Total [9 marks]

Poorer candidates just gained the 2 marks for saying what a Type I and Type II error were and could not then apply the definitions to obtain the conditional probabilities required. It was clear from some crossings out that even the 2 definition continue to cause confusion. Good, clear-thinking candidates were able to do the question correctly.