Question
A coin was tossed 200 times and 115 of these tosses resulted in ‘heads’. Use a two-tailed test with significance level 1 % to investigate whether or not the coin is biased.
▶️Answer/Explanation
Markscheme
The number of’ ‘heads’ X is B(200, p) (M1)
\({{\text{H}}_0}:p = 0.5;{\text{ }}{{\text{H}}_1}:p \ne 0.5\) A1A1
Note: Award A1A0 for the statement “ \({{\text{H}}_0}:\) coin is fair; \({{\text{H}}_1}:\) coin is biased”.
EITHER
\({\text{P}}(\left. {X \geqslant 115} \right|{{\text{H}}_0}) = 0.0200\) (M1)(A1)
p-value = 0.0400 A1
This is greater than 0.01. R1
There is insufficient evidence to conclude that the coin is biased (or the coin is not biased). R1
OR
(Using a proportion test on a GDC) p-value = 0.0339 N3
This is greater than 0.01. R1
There is insufficient evidence to conclude that the coin is biased (or the coin is not biased). R1
OR
Under \({{\text{H}}_0}X\) is approximately N(100, 50) (M1)
\(z = \frac{{115 – 100}}{{\sqrt {50} }} = 2.12\) (M1)A1
(Accept 2.05 with continuity correction)
This is less than 2.58 R1
There is insufficient evidence to conclude that the coin is biased (or the coin is not biased). R1
OR
99 % confidence limits for p are \(\frac{{115}}{{200}} \pm 2.576\sqrt {\frac{{115}}{{200}} \times \frac{{85}}{{200}} \times \frac{1}{{200}}} \) (M1)A1
giving [0.485, 0.665] A1
This interval contains 0.5 R1
There is insufficient evidence to conclude that the coin is biased (or the coin is not biased). R1
[8 marks]
Examiners report
This question was well answered in general with several correct methods seen. The most popular method was to use a GDC to carry out a proportion test which is equivalent to using a normal approximation. Relatively few candidates calculated an exact p-value using the binomial distribution. Candidates who found a 95% confidence interval for p, the probability of obtaining a head, and noted that this contained 0.5 were given full credit.
Question
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.
▶️Answer/Explanation
Markscheme
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]
Examiners report
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.
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.