CAIE A-Level Further Mathematics 4.2.5 Difference Confidence Intervals

Practise forming confidence intervals for a difference of population means with pooled t or separate normal standard errors, then interpreting whether zero is included.

Exam points

choose pooled t or separate large-sample variance from the sample sizes and assumptions given
subtract the sample means in the stated order and attach the correct standard-error margin
use whether zero lies inside the interval to interpret evidence for a population-mean difference

[Maximum number: 7]

Kayla is investigating the lengths of the leaves of a certain type of tree found in two forests X and Y. She chooses a random sample of 40 leaves of this type from forest X and records their lengths, $x \mathrm{~cm}$ . She also records the lengths, $y \mathrm{~cm}$ , for a random sample of 60 leaves of this type from forest Y. Her results are summarised as follows.

\sum x=242.0 \quad \sum x^{2}=1587.0 \quad \sum y=373.2 \quad \sum y^{2}=2532.6

Find a 90% confidence interval for the difference between the population mean lengths of leaves in forests X and Y.

$\begin{aligned} s_{x}^{2}=\frac{1}{39}\left(1587.0-\frac{242.0^{2}}{40}\right)=3.15128\left(=\frac{1229}{390}\right) s_{y}^{2}=\frac{1}{59}\left(2532.6-\frac{373.2^{2}}{60}\right)=3.58129\left(=\frac{26412}{7375}\right) \end{aligned}$

M1 A1

Either unsimplified
Both correct

$s^{2}=\frac{3.15128}{40}+\frac{3.58129}{60}=0.13847$

M1 A1

Pooled variance is M0A0
or s=0.3721

$6.05-6.22 \pm z s$

M1

FT their s, must be a z value

$=0.17 \pm 1.645 \sqrt{0.13847}$

A1

With 1.645

=[-0.442,0.782] or [-0.782,0.442]

A1

7

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4.2.5—Difference confidence intervals question 1