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.

[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.

$s_{x}^{2}=\frac{243.8}{59}\left[=\frac{1219}{295}=4.132\right], \quad s_{y}^{2}=\frac{384.9}{74}\left[=\frac{3849}{740}=5.201\right]$

B1

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}

M1

Using their sample variances.
Pooled estimate M0.

A1

CI: $\frac{6345}{60}-\frac{7614}{75} \pm 1.96 \times{ }^{\prime} 0.3718^{\prime}$

M1

With a z-value

A1

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)

A1

$4.23 \pm 0.729$ is A0, condone [4.96,3.5] etc.

6

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4.2.5—Difference confidence intervals question 1