CAIE A-Level Further Math AS 4.2 Inference Using Normal and T Distributions Questions

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: 6]

Maya is an athlete who competes in 1500 -metre races. Last summer her practice run times had mean 4.22 minutes. Over the winter she has done some intense training to try to improve her times. A random sample of 10 of her practice run times, x minutes, this summer are summarised as follows.

\sum x=42.05 \quad \sum x^{2}=176.83

Maya's new practice run times are normally distributed. She believes that on average her times have improved as a result of her training.

Test, at the 5% significance level, whether Maya's belief is supported by the data.

$\mathrm{H}_{0}: \mu=4.22 \quad \mathrm{H}_{1}: \mu<4.22$ $\bar{x}=\frac{42.05}{10}[=4.205], \quad s^{2}=\frac{1}{9}\left(176.83-\frac{42.05^{2}}{10}\right)\left[=\frac{13}{12000}=0.0010833\right]$ $t=\frac{4.205-4.22}{\sqrt{\frac{s^{2}}{10}}}$
Their 4.205, their $s^{2}$ .

t=-1.44
Condone sign.

Tabular value =1.833: ' 1.44 ' <1.833, accept $\mathrm{H}_{0}$
Compare their value with correct tabular value, signs
consistent, allow 'not significant'.

There is insufficient evidence to support Maya's belief.
Correct conclusion in context, following correct work,
level of uncertainty in language.
A0 if hypotheses wrong way round or missing.

6

[Maximum number: 6]

Shane is studying the lengths of the tails of male red kangaroos. He takes a random sample of 14 male red kangaroos and measures the length of the tail, $x \mathrm{~m}$ , for each kangaroo. He then calculates a 90% confidence interval for the population mean tail length, $\mu \mathrm{m}$ , of male red kangaroos. He assumes that the tail lengths are normally distributed and finds that $1.11 \leqslant \mu \leqslant 1.14$ .

Find the values of $\sum x$ and $\sum x^{2}$ for this sample.

$\bar{x}+1.771 \sqrt{\frac{s^{2}}{14}}=1.14$ or $\bar{x}-1.771 \sqrt{\frac{s^{2}}{14}}=1.11$
SOI
Allow incorrect t-value, not z-value.

[Add: $\left.\bar{x}=\frac{1}{2}(1.14+1.11)=1.125\right] \quad \sum x=15.75$
Does not depend on use of a t-value.

Subtract or substitute: $\sqrt{\frac{s^{2}}{14}}=\frac{1}{2}\left(\frac{1.14-1.11}{1.771}\right)$
Allow incorrect t-value, but not a z-value.

$s^{2}=0.00100[4]$ or s=0.0316[9] $\frac{450}{448063}$ , implied by correct final answer.

$\sum x^{2}=13 s^{2}+\frac{\left(\sum x\right)^{2}}{14}$

М1

OE

$\sum x^{2}=17.7$ (3)
CWO

6

[Maximum number: 6]

A factory produces small bottles of natural spring water. Two different machines, X and Y, are used to fill empty bottles with the water. A quality control engineer checks the volumes of water in the bottles filled by each of the machines. He chooses a random sample of 60 bottles filled by machine X and a random sample of 75 bottles filled by machine Y. The volumes of water, x and y respectively, in millilitres, are summarised as follows.

\sum x=6345 \quad \sum(x-\bar{x})^{2}=243.8 \quad \sum y=7614 \quad \sum(y-\bar{y})^{2}=384.9

$\bar{x}$ and $\bar{y}$ are the sample means of the volume of water in the bottles filled by machines X and Y respectively.

Find a 95% confidence interval for the difference between the mean volume of water in bottles filled by machine X and the mean volume of water in bottles filled by machine Y.

$s_{x}^{2}=\frac{243.8}{59}\left[=\frac{1219}{295}=4.132\right], \quad s_{y}^{2}=\frac{384.9}{74}\left[=\frac{3849}{740}=5.201\right]$
Implied by correct s or pooled estimate
$\frac{243.8+384.9}{60+75-2}=4.727$ $s^{2}=\frac{4.132}{60}+\frac{5.201}{75}[=0.1382]$ or s=0.3718}
Using their sample variances.
Pooled estimate M0.
CI: $\frac{6345}{60}-\frac{7614}{75} \pm 1.96 \times{ }^{\prime} 0.3718^{\prime}$
With a z-value
With 1.96 (with their s )
Pooled $\frac{6345}{60}-\frac{7614}{75} \pm 1.96 \times 2.174 \sqrt{\frac{1}{60}+\frac{1}{75}}$

[3.5[0],4.96] or(3.5, 4.96)
$4.23 \pm 0.729$ is A0, condone [4.96,3.5] etc.

6

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Covered syllabus points

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