# IB DP Maths Topic 7.6 Testing hypotheses for the mean of a normal population HL Paper 3

## Question

The random variable X is normally distributed with unknown mean $$\mu$$ and unknown variance $${\sigma ^2}$$ . A random sample of 10 observations on X was taken and the following 95 % confidence interval for $$\mu$$ was correctly calculated as [4.35, 4.53] .

(a)     Calculate an unbiased estimate for

(i)     $$\mu$$ ,

(ii)     $${\sigma ^2}$$ .

(b)     The value of $$\mu$$ is thought to be 4.5, so the following hypotheses are defined.${{\text{H}}_0}:\mu = 4.5;{\text{ }}{{\text{H}}_1}:\mu < 4.5$

(i)     Find the p-value of the observed sample mean.

(ii)     State your conclusion if the significance level is

(a)     1 %,

(b)     10 %.

## Markscheme

(a)     (i)     $$\bar x = \frac{{4.35 + 4.53}}{2} = 4.44$$ (estimate of $$\mu$$)     A2

(ii)     Degrees of freedom = 9     (A1)

Critical value of t = 2.262     (A1)

$$2.262 \times \frac{s}{{\sqrt {10} }} = 0.09$$     M1A1

$$s = 0.12582…$$     (A1)

$${s^2} = 0.0158$$ (estimate of $${\sigma ^2}$$)     A1

[8 marks]

(b)     (i)     Using t test     (M1)

$$t = \frac{{4.44 – 4.5}}{{\sqrt {\frac{{0.0158}}{{10}}} }} = – 1.50800$$   (Accept $$– 1.50946$$)     (A1)

p-value = 0.0829 (Accept 0.0827)     A2

(ii)     (a)     Accept $${{\text{H}}_0}$$ / Reject $${{\text{H}}_1}$$ .     R1

(b)     Reject $${{\text{H}}_0}$$ / Accept $${{\text{H}}_1}$$ .     R1

[6 marks]

Total [14 marks]

## Examiners report

Most candidates realised that the unbiased estimate of the mean was simply the central point of the confidence interval. Many candidates, however, failed to realise that, because the variance was unknown, the t-distribution was used to determine the confidence limits. In (b), although the p-value was asked for specifically, some candidates solved the problem correctly by comparing the value of their statistic with the appropriate critical values. This method was given full credit but, of course, marks were lost by their failure to give the p-value.

## Question

A teacher wants to determine whether practice sessions improve the ability to memorize digits.

He tests a group of 12 children to discover how many digits of a twelve-digit number could be repeated from memory after hearing them once. He gives them test 1, and following a series of practice sessions, he gives them test 2 one week later. The results are shown in the table below. (a)     State appropriate null and alternative hypotheses.

(b)     Test at the 5 % significance level whether or not practice sessions improve ability to memorize digits, justifying your choice of test.

## Markscheme

(a)     $${{\text{H}}_0}:d = 0;{\text{ }}{{\text{H}}_1}:d > 0$$, where d is the difference in the number of digits remembered     A1A1

[2 marks]

(b) A2

Notes: Award A2 for the correct d values.

Award A1 for one error, A0 for two or more errors.

Use the t-test because the variance is not known     M1R1

By GDC

t = 2.106…     (A2)

EITHER

p-value = 0.0295 (accept any value that rounds to this number)     A2

Since 0.0295 < 0.05 there is evidence that practice sessions improve ability to memorize digits     R1

OR

The critical value of t is 1.796     A2

Since 2.106… > 1.796 there is evidence that practice sessions improve ability to memorize digits     R1

Note: Award M1R1A1A1R1 for testing equality of means (t = –1.46, p-value = 0.08) .

[9 marks]

Total [11 marks]

## Examiners report

Although this question was reasonably well done the hypotheses were often not stated precisely and the fact that the two data sets were dependent escaped many candidates.

## 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 apple trees in a large orchard have, for several years, suffered from a disease for which the outward sign is a red discolouration on some leaves.

The fruit grower knows that the mean number of discoloured leaves per tree is 42.3. The fruit grower suspects that the disease is caused by an infection from a nearby group of cedar trees. He cuts down the cedar trees and, the following year, counts the number of discoloured leaves on a random sample of seven apple trees. The results are given in the table below. (a)     From these data calculate an unbiased estimate of the population variance.

(b)     Stating null and alternative hypotheses, carry out an appropriate test at the 10 % level to justify the cutting down of the cedar trees.

## Markscheme

(a)     $$n = 7,{\text{ sample mean }} = 35$$     (A1)

$$s_{n – 1}^2 = \frac{{\sum {{{(x – 35)}^2}} }}{6} = 322$$     (M1)A1

[3 marks]

(b)     null hypothesis $${{\text{H}}_0}:\mu = 42.3$$     A1

alternative hypothesis $${{\text{H}}_1}:\mu < 42.3$$     A1

using one-sided t-test

$$\left| {{t_{{\text{calc}}}}} \right| = \sqrt 7 \frac{{42.3 – 35}}{{\sqrt {322} }} = 1.076$$     (M1)(A1)

with 6 degrees of freedom , $${t_{{\text{crit}}}} = 1.440 > 1.076$$

$${\text{(or }}p{\text{-value }} = 0.162 > 0.1)$$     A1

we conclude that there is no justification for cutting down the cedar trees     R1     N0

Note: FT on their t or p-value.

[6 marks]

Total [9 marks]

## Examiners report

This question was generally well attempted as an example of the t-test. Very few used the Z statistic, and many found p-values.

## Question

(a)     After a chemical spillage at sea, a scientist measures the amount, x units, of the chemical in the water at 15 randomly chosen sites. The results are summarised in the form $$\sum {x = 18}$$ and $$\sum {{x^2} = 28.94}$$. Before the spillage occurred the mean level of the chemical in the water was 1.1. Test at the 5 % significance level the hypothesis that there has been an increase in the amount of the chemical in the water.

(b)     Six months later the scientist returns and finds that the mean amount of the chemical in the water at the 15 randomly chosen sites is 1.18. Assuming that this sample came from a normal population with variance 0.0256, find a 90 % confidence interval for the mean level of the chemical.

## Markscheme

(a)     $$\bar x = \frac{{\sum x }}{n} = 1.2$$     (A1)

$$s_{n – 1}^2 = 0.524 \ldots$$     (A1)

it is a one tailed test

$${{\text{H}}_0}:\mu = 1.1,{\text{ }}{{\text{H}}_1}:\mu > 1.1$$     A1

EITHER

$$t = \frac{{1.2 – 1.1}}{{\sqrt {\frac{{0.524 \ldots }}{{15}}} }} = 0.535$$     (M1) A1

$$v = 14$$     (A1)

$${t_{crit}} = 1.761$$     A1

since $$0.535 < {t_{crit}}$$ we accept $${{\text{H}}_0}$$ that there is no increase in the amount of the chemical     R1

OR

$$p = 0.301$$     A4

since $$p > 0.05$$ we accept $${{\text{H}}_0}$$ that there is no increase in the amount of the chemical     R1

[8 marks]

(b)     90 % confidence interval $$= 1.18 \pm 1.645\sqrt {\frac{{0.0256}}{{15}}}$$     (M1)A1A1A1

$$= [1.11,{\text{ }}1.25]$$     A1     N5

[5 marks]

Total [13 marks]

## Examiners report

This question also proved accessible to a majority of candidates with many wholly correct or nearly wholly correct answers seen. A few candidates did not recognise that part (a) was a t-distribution and part (b) was a Normal distribution, but most recognised the difference. Many candidates received an accuracy penalty on this question for not giving the final answer to part (b) to 3 significant figures.

## 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

Anna cycles to her new school. She records the times taken for the first ten days with the following results (in minutes).

12.4 13.7 12.5 13.4 13.8 12.3 14.0 12.8 12.6 13.5

Assume that these times are a random sample from the $${\text{N}}(\mu ,{\text{ }}{\sigma ^2})$$ distribution.

(a)     Determine unbiased estimates for $$\mu$$ and $${\sigma ^2}$$.

(b)     Calculate a 95 % confidence interval for $$\mu$$.

(c)     Before Anna calculated the confidence interval she thought that the value of $$\mu$$ would be 12.5. In order to check this, she sets up the null hypothesis $${{\text{H}}_0}:\mu = 12.5$$.

(i)     Use the above data to calculate the value of an appropriate test statistic. Find the corresponding p-value using a two-tailed test.

(ii)     Interpret your p-value at the 1 % level of significance, justifying your conclusion.

## Markscheme

(a)     estimate of $$\mu = 13.1$$     A1

estimate of $${\sigma ^2} = 0.416$$     A1

[2 marks]

(b)     using a GDC (or otherwise), the 95% confidence interval is     (M1)

[12.6, 13.6]     A1A1

Note: Accept open or closed intervals.

[3 marks]

(c)     (i)     $$t = \frac{{13.1 – 12.5}}{{0.6446 \ldots /\sqrt {10} }} = 2.94$$     (M1)A1

$$v = 9$$     (A1)

p-value $$= 2 \times {\text{P}}(T > 2.9433 \ldots )$$     (M1)

$$= 0.0164\,\,\,\,\,$$(accept 0.0165)     A1

(ii)     we accept the null hypothesis (the mean travel time is 12.5 minutes)     A1

because 0.0164 (or 0.0165) > 0.01     R1

Note: Allow follow through on their p-value.

[7 marks]

Total [12 marks]

## Examiners report

This was well answered by many candidates. In (a), some candidates chose the wrong standard deviation from their calculator and often failed to square their result to obtain the unbiased variance estimate. Candidates should realise that it is the smaller of the two values (ie the one obtained by dividing by (n – 1)) that is required. The most common error was to use the normal distribution instead of the t-distribution. The signpost towards the t-distribution is the fact that the variance had to be estimated in (a). Accuracy penalties were often given for failure to round the confidence limits, the t-statistic or the p-value to three significant figures.

## 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

Two species of plant, $$A$$ and $$B$$, are identical in appearance though it is known that the mean length of leaves from a plant of species $$A$$ is $$5.2$$ cm, whereas the mean length of leaves from a plant of species $$B$$ is $$4.6$$ cm. Both lengths can be modelled by normal distributions with standard deviation $$1.2$$ cm.

In order to test whether a particular plant is from species $$A$$ or species $$B$$, $$16$$ leaves are collected at random from the plant. The length, $$x$$, of each leaf is measured and the mean length evaluated. A one-tailed test of the sample mean, $$\bar X$$, is then performed at the $$5\%$$ level, with the hypotheses: $${H_0}:\mu = 5.2$$ and $${H_1}:\mu < 5.2$$.

Find the critical region for this test.


a.

It is now known that in the area in which the plant was found $$90\%$$ of all the plants are of species $$A$$ and $$10\%$$ are of species $$B$$.

Find the probability that $$\bar X$$ will fall within the critical region of the test.


c.

If, having done the test, the sample mean is found to lie within the critical region, find the probability that the leaves came from a plant of species $$A$$.


d.

## Markscheme

$$\bar X \sim N\left( {5.2,{\text{ }}\frac{{{{1.2}^2}}}{{16}}} \right)$$     (M1)

critical value is $$5.2 – 1.64485 \ldots \times \frac{{1.2}}{4} = 4.70654 \ldots$$     (A1)

critical region is $$] – \infty ,{\text{ }}4.71]$$     A1

Note:     Allow follow through for the final A1 from their critical value.

Note:     Follow through previous values in (b), (c) and (d).

[3 marks]

a.

$$0.9 \times 0.05 + 0.1 \times (1 – 0.361 \ldots ) = 0.108875997 \ldots = 0.109$$     M1A1

Note:     Award M1 for a weighted average of probabilities with weights $$0.1,0.9$$.

[2 marks]

c.

attempt to use conditional probability formula     M1

$$\frac{{0.9 \times 0.05}}{{0.108875997 \ldots }}$$     (A1)

$$= 0.41334 \ldots = 0.413$$     A1

[3 marks]

Total [10 marks]

d.

## Examiners report

Solutions to this question were generally disappointing.

In (a), the standard error of the mean was often taken to be $$\sigma (1.2)$$ instead of $$\frac{\sigma }{{\sqrt n }}(0.3)$$ and the solution sometimes ended with the critical value without the critical region being given.

a.

In (c), the question was often misunderstood with candidates finding the weighted mean of the two means, ie $$0.9 \times 5.2 + 0.1 \times 4.6 = 5.14$$ instead of the weighted mean of two probabilities.

c.

Without having the solution to (c), part (d) was inaccessible to most of the candidates so that very few correct solutions were seen.

d.

## Question

Eleven students who had under-performed in a philosophy practice examination were given extra tuition before their final examination. The differences between their final examination marks and their practice examination marks were

$10,{\text{ }} – 1,{\text{ }}6,{\text{ }}7,{\text{ }} – 5,{\text{ }} – 5,{\text{ }}2,{\text{ }} – 3,{\text{ }}8,{\text{ }}9,{\text{ }} – 2.$

Assume that these differences form a random sample from a normal distribution with mean $$\mu$$ and variance $${\sigma ^2}$$.

Determine unbiased estimates of $$\mu$$ and $${\sigma ^2}$$.


a.

(i)     State suitable hypotheses to test the claim that extra tuition improves examination marks.

(ii)     Calculate the $$p$$-value of the sample.

(iii)     Determine whether or not the above claim is supported at the $$5\%$$ significance level.


b.

## Markscheme

unbiased estimate of $$\mu$$ is $$2.36(36 \ldots )\;\;\;(26/11)$$     (M1)A1

unbiased estimate of $${\sigma ^2}$$ is $$33.65(45 \ldots ) = ({5.801^2})\;\;\;(1851/55)$$     (M1)A1

Note:     Accept any answer that rounds correctly to $$3$$ significant figures.

Note:     Award M1A0 for any unbiased estimate of $${\sigma ^2}$$ that rounds to $$5.80$$.

[4 marks]

a.

(i)     $${{\text{H}}_0}:\mu = 0;{\text{ }}{{\text{H}}_1}:\mu > 0$$     A1A1

Note:     Award A1A0 if an inappropriate symbol is used for the mean, eg, $$r$$, $${\rm{\bar d}}$$.

(ii)     attempt to use t-test     (M1)

$$t = 1.35$$     (A1)

$${\text{DF}} = 10$$     (A1)

$$p$$-value $$= 0.103$$     A1

Note:     Accept any answer that rounds correctly to $$3$$ significant figures.

(iii)     $$0.103 > 0.05$$     A1

there is insufficient evidence at the $$5\%$$ level to support the claim (that extra tuition improves examination marks)

OR

the claim (that extra tuition improves examination marks) is not supported at the $$5\%$$ level (or equivalent statement)     R1

Note:     Follow through the candidate’s $$p$$-value.

Note:     Do not award R1 for Accept $${{\text{H}}_0}$$ or Reject $${{\text{H}}_1}$$.

[8 marks]

Total [12 marks]

b.

## Examiners report

Almost every candidate gave the correct estimate of the mean but some chose the wrong variance from their calculators to estimate $${\sigma ^2}$$. In (b)(i), the hypotheses were sometimes incorrectly written, usually with an incorrect symbol instead of $$\mu$$, for example $$d$$, $$\bar x$$ and ‘mean’ were seen. Many candidates failed to make efficient use of their calculators in (b)(ii). The intention of the question was that candidates should simply input the data into their calculators and use the software to give the p-value. Instead, many candidates found the p-value by first evaluating $$t$$ using the appropriate formula. This was a time consuming process and it gave opportunity for error. In (b)(iii), candidates were expected to refer to the claim so that the answers ‘Accept $${H_0}$$’ or ‘Reject $${H_1}$$’ were not accepted.

a.

Almost every candidate gave the correct estimate of the mean but some chose the wrong variance from their calculators to estimate $${\sigma ^2}$$. In (b)(i), the hypotheses were sometimes incorrectly written, usually with an incorrect symbol instead of $$\mu$$, for example $$d$$, $$\bar x$$ and ‘mean’ were seen. Many candidates failed to make efficient use of their calculators in (b)(ii). The intention of the question was that candidates should simply input the data into their calculators and use the software to give the $$p$$-value. Instead, many candidates found the $$p$$-value by first evaluating $$t$$ using the appropriate formula. This was a time consuming process and it gave opportunity for error. In (b)(iii), candidates were expected to refer to the claim so that the answers ‘Accept $${H_0}$$’ or ‘Reject $${H_1}$$’ were not accepted.

b.

## Question

In this question you may assume that these data are a random sample from a bivariate normal distribution, with population product moment correlation coefficient $$\rho$$.

Richard wishes to do some research on two types of exams which are taken by a large number of students. He takes a random sample of the results of 10 students, which are shown in the following table. Using these data, it is decided to test, at the 1% level, the null hypothesis $${H_0}:\rho = 0$$ against the alternative hypothesis $${H_1}:\rho > 0$$.

Richard decides to take the exams himself. He scored 11 on Exam 1 but his result on Exam 2 was lost.

Caroline believes that the population mean mark on Exam 2 is 6 marks higher than the population mean mark on Exam 1. Using the original data from the 10 students, it is decided to test, at the 5% level, this hypothesis against the alternative hypothesis that the mean of the differences, $${\text{d}} = {\text{exam 2 mark }} – {\text{ exam 1 mark}}$$, is less than 6 marks.

For these data find the product moment correlation coefficient, $$r$$.


a.

(i)     State the distribution of the test statistic (including any parameters).

(ii)     Find the $$p$$-value for the test.


b.

Using a suitable regression line, find an estimate for his score on Exam 2, giving your answer to the nearest integer.


c.

(i)     State the distribution of your test statistic (including any parameters).

(ii)     Find the $$p$$-value.

(iii)     State the conclusion, justifying the answer.


d.

## Markscheme

$$r = 0.804$$    A2

Note: Accept any number that rounds to 0.80.

[2 marks]

a.

(i)     $$t$$ distribution with 8 degrees of freedom     A1A1

(ii)     $$p{\text{ – value}} = 0.00254$$     A2

Notes: Accept any number that rounds to 0.0025.

Award A1 for 2-tail test giving an answer that rounds to 0.0051.

(iii)     $$p{\text{ – value}} < 0.01$$, so conclude that there is positive correlation     R1A1

Notes: Only award the A1 if the R1 is awarded.

Do not accept just “reject $${H_0}$$” or “accept $${H_1}$$”.

The words “positive correlation” must be seen.

[6 marks]

b.

regression line of $$y$$ (Exam 2 mark) on $$x$$ (Exam 1 mark) is     (M1)

$$y = 0.59407 \ldots x + 21.387 \ldots$$    (A1)

$$x = 11$$ gives $$y = 28$$ (to nearest integer)     A1

[3 marks]

c.

(i)     applying the $$t$$ test to the differences

$$t$$ distribution with 9 degrees of freedom     A1A1

(ii)     $$p{\text{ – value}} = 0.239$$     A2

Notes: Accept any number that rounds to 0.24.

Award A1 if subtraction done the wrong way round giving $$p{\text{ – value}} = 0.109$$.

(iii)     $$p{\text{ – value}} > 0.05$$, so accept $${H_0}$$ or $${u_d} = 6$$     R1A1

[6 marks]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

## Question

John rings a church bell 120 times. The time interval, $${T_i}$$, between two successive rings is a random variable with mean of 2 seconds and variance of $$\frac{1}{9}{\text{ second}}{{\text{s}}^2}$$.

Each time interval, $${T_i}$$, is independent of the other time intervals. Let $$X = \sum\limits_{i = 1}^{119} {{T_i}}$$ be the total time between the first ring and the last ring.

The church vicar subsequently becomes suspicious that John has stopped coming to ring the bell and that he is letting his friend Ray do it. When Ray rings the bell the time interval, $${T_i}$$ has a mean of 2 seconds and variance of $$\frac{1}{{25}}{\text{ second}}{{\text{s}}^2}$$.

The church vicar makes the following hypotheses:

$${H_0}$$: Ray is ringing the bell; $${H_1}$$: John is ringing the bell.

He records four values of $$X$$. He decides on the following decision rule:

If $$236 \leqslant X \leqslant 240$$ for all four values of $$X$$ he accepts $${H_0}$$, otherwise he accepts $${H_1}$$.

Find

(i)     $${\text{E}}(X)$$;

(ii)     $${\text{Var}}(X)$$.


a.

Explain why a normal distribution can be used to give an approximate model for $$X$$.


b.

Use this model to find the values of $$A$$ and $$B$$ such that $${\text{P}}(A < X < B) = 0.9$$, where $$A$$ and $$B$$ are symmetrical about the mean of $$X$$.


c.

Calculate the probability that he makes a Type II error.


d.

## Markscheme

(i)     $${\text{mean}} = 119 \times 2 = 238$$     A1

(ii)     $${\text{variance}} = 119 \times \frac{1}{9} = \frac{{119}}{9}{\text{ }}( = 13.2)$$     (M1)A1

Note: If 120 is used instead of 119 award A0(M1)A0 for part (a) and apply follow through for parts (b)-(d). (b) is unaffected and in (c) the interval becomes $$(234,{\text{ }}246)$$. In (d) the first 2 A1 marks are for $$0.3633 \ldots$$ and $$0.0174 \ldots$$ so the final answer will round to 0.017.

[3 marks]

a.

justified by the Central Limit Theorem     R1

since $$n$$ is large     A1

Note: Accept $$n > 30$$.

[2 marks]

b.

$$X \sim N\left( {238,{\text{ }}\frac{{119}}{9}} \right)$$

$$Z = \frac{{X – 238}}{{\frac{{\sqrt {119} }}{3}}} \sim N(0,{\text{ }}1)$$     (M1)(A1)

$${\text{P}}(Z < q) = 0.95 \Rightarrow q = 1.644 \ldots$$    (A1)

so $${\text{P}}( – 1.644 \ldots < Z < 1.644 \ldots ) = 0.9$$     (R1)

$${\text{P}}( – 1.644 \ldots < \frac{{X – 238}}{{\frac{{\sqrt {119} }}{3}}} < 1.644 \ldots ) = 0.9$$    (M1)

interval is $$232 < X < 244{\text{ }}({\text{3sf}}){\text{ }}(A = 232,{\text{ }}B = 244)$$     A1A1

Notes: Accept the use of inverse normal applied to the distribution of $$X$$.

Alternative is to use the GDC to find a pretend $$Z$$ confidence interval for a mean and then convert by multiplying by 119.

Either $$A$$ or $$B$$ correct implies the five implied marks.

Accept any numbers that round to these 3sf numbers.

[7 marks]

c.

under $${{\text{H}}_1},{\text{ }}X \sim N\left( {238,{\text{ }}\frac{{119}}{9}} \right)$$     (M1)

$${\text{P}}(236 \leqslant X \leqslant 240) = 0.41769 \ldots$$    (A1)

probability that all 4 values of $$X$$ lie in this interval is

$${(0.41769 \ldots )^4} = 0.030439 \ldots$$     (M1)(A1)

so probability of a Type II error is 0.0304 (3sf)     A1

Note: Accept any answer that rounds to 0.030.

[5 marks]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

## Question

Consider the recurrence relation

$${u_n} = 5{u_{n – 1}} – 6{u_{n – 2}},{\text{ }}{u_0} = 0$$ and $${u_1} = 1$$.

Find an expression for $${u_n}$$ in terms of $$n$$.


a.

For every prime number $$p > 3$$, show that $$p|{u_{p – 1}}$$.


b.

## Markscheme

the auxiliary equation is $${\lambda ^2} – 5\lambda + 6 = 0$$     M1

$$\Rightarrow \lambda = 2,{\text{ }}3$$     (A1)

the general solution is $${u_n} = A \times {2^n} + B \times {3^n}$$     A1

imposing initial conditions (substituting $$n = 0,{\text{ }}1$$)     M1

$$A + B = 0$$ and $$2A + 3B = 1$$     A1

the solution is $$A = – 1,{\text{ }}B = 1$$

so that $${u_n} = {3^n} – {2^n}$$     A1

[6 marks]

a.

$${u_{p – 1}} = {3^{p – 1}} – {2^{p – 1}}$$

$$p > 3$$, therefore 3 or 2 are not divisible by $$p$$     R1

hence by FLT, $${3^{p – 1}} \equiv 1 \equiv {2^{p – 1}}(\bmod p)$$ for $$p > 3$$     M1A1

$${u_{p – 1}} \equiv 0(\bmod p)$$     A1

$$p|{u_{p – 1}}$$ for every prime number $$p > 3$$     AG

[4 marks]

b.

[N/A]

a.

[N/A]

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.