Question
A population is known to have a normal distribution with a variance of 3 and an unknown mean \(\mu \) . It is proposed to test the hypotheses \({{\text{H}}_0}:\mu = 13,{\text{ }}{{\text{H}}_1}:\mu > 13\) using the mean of a sample of size 2.
(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?
▶️Answer/Explanation
Markscheme
(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]
Examiners report
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.
Question
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.
▶️Answer/Explanation
Markscheme
(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]
Examiners report
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.
Question
The random variable X has a Poisson distribution with mean \(\mu \). The value of \(\mu \) is known to be either 1 or 2 so the following hypotheses are set up.
\[{{\text{H}}_0}:\mu = 1;{\text{ }}{{\text{H}}_1}:\mu = 2\]
A random sample \({x_1},{\text{ }}{x_2},{\text{ }} \ldots ,{\text{ }}{x_{10}}\) of 10 observations is taken from the distribution of X and the following critical region is defined.
\[\sum\limits_{i = 1}^{10} {{x_i} \geqslant 15} \]
Determine the probability of
(a) a Type I error;
(b) a Type II error.
▶️Answer/Explanation
Markscheme
(a) let \(T = \sum\limits_{i = 1}^{10} {{X_i}} \) so that T is Po(10) under \({{\text{H}}_0}\) (M1)
\({\text{P(Type I error)}} = {\text{P }}T \geqslant 15|\mu = 1\) M1A1
\( = 0.0835\) A2 N3
Note: Candidates who write the first line and only the correct answer award (M1)M0A0A2.
[5 marks]
(b) let \(T = \sum\limits_{i = 1}^{10} {{X_i}} \) so that T is Po(20) under \({{\text{H}}_1}\) (M1)
\({\text{P(Type II error)}} = {\text{P }}T \leqslant 14|\mu = 2\) M1A1
\( = 0.105\) A2 N3
Note: Candidates who write the first line and only the correct answer award (M1)M0A0A2.
Note: Award 5 marks to a candidate who confuses Type I and Type II errors and has both answers correct.
[5 marks]
Total [10 marks]
Examiners report
This question caused problems for many candidates and the solutions were often disappointing. Some candidates seemed to be unaware of the meaning of Type I and Type II errors. Others were unable to calculate the probabilities even when they knew what they represented. Candidates who used a normal approximation to obtain the probabilities were not given full credit – there seems little point in using an approximation when the exact value could be found.
Question
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.
▶️Answer/Explanation
Markscheme
(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]
Examiners report
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.