# IB DP Maths Topic 7.6 Type I and II errors, including calculations of their probabilities HL Paper 3

## 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?

## 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.

## 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.

## 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.

## 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.

## Question

A baker produces loaves of bread that he claims weigh on average 800 g each. Many customers believe the average weight of his loaves is less than this. A food inspector visits the bakery and weighs a random sample of 10 loaves, with the following results, in grams:

783, 802, 804, 785, 810, 805, 789, 781, 800, 791.

Assume that these results are taken from a normal distribution.

Determine unbiased estimates for the mean and variance of the distribution.


a.

In spite of these results the baker insists that his claim is correct.

Stating appropriate hypotheses, test the baker’s claim at the 10 % level of significance.


b.

## Markscheme

unbiased estimate of the mean: 795 (grams)     A1

unbiased estimate of the variance: 108 $$(gram{s^2})$$     (M1)A1

[3 marks]

a.

null hypothesis $${H_0}:\mu = 800$$     A1

alternative hypothesis $${H_1}:\mu < 800$$     A1

using 1-tailed t-test     (M1)

EITHER

p = 0.0812…     A3

OR

with 9 degrees of freedom     (A1)

$${t_{calc}} = \frac{{\sqrt {10} (795 – 800)}}{{\sqrt {108} }} = – 1.521$$     A1

$${t_{crit}} = – 1.383$$     A1

Note: Accept 2sf intermediate results.

THEN

so the baker’s claim is rejected     R1

Note: Accept “reject $${H_0}$$ ” provided $${H_0}$$ has been correctly stated.

Note: FT for the final R1.

[7 marks]

b.

## Examiners report

A successful question for many candidates. A few candidates did not read the question and adopted a 2-tailed test.

a.

A successful question for many candidates. A few candidates did not read the question and adopted a 2-tailed test.

b.

## Question

A smartphone’s battery life is defined as the number of hours a fully charged battery can be used before the smartphone stops working. A company claims that the battery life of a model of smartphone is, on average, 9.5 hours. To test this claim, an experiment is conducted on a random sample of 20 smartphones of this model. For each smartphone, the battery life, $$b$$ hours, is measured and the sample mean, $${\bar b}$$, calculated. It can be assumed the battery lives are normally distributed with standard deviation 0.4 hours.

It is then found that this model of smartphone has an average battery life of 9.8 hours.

State suitable hypotheses for a two-tailed test.


a.

Find the critical region for testing $${\bar b}$$ at the 5 % significance level.


b.

Find the probability of making a Type II error.


c.

Another model of smartphone whose battery life may be assumed to be normally distributed with mean μ hours and standard deviation 1.2 hours is tested. A researcher measures the battery life of six of these smartphones and calculates a confidence interval of [10.2, 11.4] for μ.

Calculate the confidence level of this interval.


d.

## Markscheme

Note: In question 3, accept answers that round correctly to 2 significant figures.

$${{\text{H}}_0}\,{\text{:}}\,\mu = 9.5{\text{;}}\,\,{{\text{H}}_1}\,{\text{:}}\,\mu \ne 9.5$$     A1

[1 mark]

a.

Note: In question 3, accept answers that round correctly to 2 significant figures.

the critical values are $$9.5 \pm 1.95996 \ldots \times \frac{{0.4}}{{\sqrt {20} }}$$     (M1)(A1)

i.e. 9.3247…, 9.6753…

the critical region is $${\bar b}$$ < 9.32, $${\bar b}$$ > 9.68     A1A1

Note: Award A1 for correct inequalities, A1 for correct values.

Note: Award M0 if t-distribution used, note that t(19)97.5 = 2.093 …

[4 marks]

b.

Note: In question 3, accept answers that round correctly to 2 significant figures.

$$\bar B \sim {\text{N}}\left( {9.8,\,{{\left( {\frac{{0.4}}{{\sqrt {20} }}} \right)}^2}} \right)$$     (A1)

$${\text{P}}\left( {9.3247 \ldots < \bar B < 9.6753 \ldots } \right)$$     (M1)

=0.0816     A1

Note: FT the critical values from (b). Note that critical values of 9.32 and 9.68 give 0.0899.

[3 marks]

c.

Note: In question 3, accept answers that round correctly to 2 significant figures.

METHOD 1

$$X \sim {\text{N}}\left( {{\text{10}}{\text{.8,}}\,\frac{{{{1.2}^2}}}{6}} \right)$$     (M1)(A1)

P(10.2 < X < 11.4) = 0.7793…     (A1)

confidence level is 77.9%    A1

Note: Accept 78%.

METHOD 2

$$11.4 – 10.2 = 2z \times \frac{{1.2}}{{\sqrt 6 }}$$      (M1)

$$z = 1.224 \ldots$$     (A1)

P(−1.224… < Z < 1.224…) = 0.7793…      (A1)

confidence level is 77.9%      A1

Note: Accept 78%.

[4 marks]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.