CAIE A-Level Further Mathematics 4.2 Inference Using Normal and t-Distributions

Practise selecting one-sample, paired or two-sample inference, calculating the required variance and using the correct normal or t critical value in context.

Exam points

identify the sampling design before choosing a one-sample, paired or independent statistic
calculate unbiased, pooled or difference variance with the correct degrees of freedom
compare the test statistic or interval with its critical value and conclude in context

Covered syllabus points

4.2.1—Hypothesis tests 4.2.2—A pooled estimate of a population 4.2.3—Hypothesis tests 4.2.4—Mean confidence intervals 4.2.5—Difference confidence intervals

[Maximum number: 10]

A large number of children are competing in a throwing competition. The distances, in metres, thrown by a random sample of 8 children are as follows.
19.8
20.8
21.5
22.1
23.7
24.4
25.0
26.3

(a)

Assuming that distances are normally distributed, test, at the 5% significance level, whether the population mean distance thrown is more than 22.0 metres.

[ 7 ]

$\sum x=183.6, \sum x^{2}=4249.08, \bar{x}=22.95$

B1

$s^{2}=\frac{1}{7}\left(4249.08-\frac{183.6^{2}}{8}\right)=5.066$

M1

$\mathrm{H}_{0}: \mu=22.0, \quad \mathrm{H}_{1}: \mu>22.0$

B1

$t=\frac{22.95-22.0}{\sqrt{\frac{s^{2}}{8}}}=1.194$

M1A1

Compare t with correct tabular value 1.895

M1

Accept $\mathrm{H}_{0}$ : mean distance thrown is not more than 22.0 m

A1

5a 7

(b)

Find a 95\% confidence interval for the population mean distance thrown.

[ 3 ]

$22.95 \pm t \sqrt{\frac{s^{2}}{8}}$

M1

With t=2.365

B1

[21.1, 24.8]

A1

5b 3

[Maximum number: 12]

Nassa is researching the lengths of a particular type of snake in two countries, A and B.

(a)

He takes a random sample of 10 snakes of this type from country A and measures the length, $x \mathrm{~m}$ , of each snake. He then calculates a 90% confidence interval for the population mean length, $\mu \mathrm{m}$ , for snakes of this type, assuming that snake lengths have a normal distribution. This confidence interval is $3.36 \leqslant \mu \leqslant 4.22$ .

Find the sample mean and an unbiased estimate for the population variance.

[ 4 ]

$\bar{x}=\frac{1}{2}(4.22+3.36)=3.79$

B1

$4.22-3.36=\frac{2 t s}{\sqrt{10}}$

М1

Using z value implies M0

With t=1.833

A1

( s=0.7418 ) variance =0.55(0)

A1

4

(b)

Nassa also measures the lengths, $y \mathrm{~m}$ , of a random sample of 8 snakes of this type taken from country B. His results are summarised as follows.

\sum y=27.86 \quad \sum y^{2}=98.02

Nassa claims that the mean length of snakes of this type in country B is less than the mean length of snakes of this type in country A. Nassa assumes that his sample from country B also comes from a normal distribution, with the same variance as the distribution from country A.

Test at the 10\% significance level whether there is evidence to support Nassa's claim.

Additional Page

If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

[ 8 ]

$\mathrm{H}_{0}: \mu_{A}=\mu_{B} \quad \mathrm{H}_{1}: \mu_{A}>\mu_{B}$

B1

Not $\bar{x}$ $s^{2}=\frac{1}{7}\left(98.02-\frac{27.86^{2}}{8}\right)=0.142(507)$

M1

$=\frac{1}{7}(98.02-97.02)$

Pooled estimate $=\frac{9 \times 0.550316+7 \times 0.142507}{16}$

M1

= 0.372

A1

$t=\frac{\frac{27.86}{8}-3.79}{\sqrt{0.372} \sqrt{\frac{1}{10}+\frac{1}{8}}}=(-) 1.063$

M1A1

${ }^{\prime} 1.063^{\prime}<1.337$ oe

M1

Compare their value with 1.337

Accept $\mathrm{H}_{0}$
Nassa's claim is not supported

A1ft

Correct conclusion in context, level of uncertainty in language
used. No contradictions.

8

All question bank results loaded

Covered syllabus points

4.2 Inference using normal and t-distributions question 1

Question (a)

Question (b)

4.2 Inference using normal and t-distributions question 2

Question (a)

Question (b)