## 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.