Chapter 10: Q. T10.2 (page 697)

Thirty-five people from a random sample of $125$ workers from Company A admitted
to using sick leave when they weren’t really ill. Seventeen employees from a random
sample of $68$ workers from Company B admitted that they had used sick leave when
they weren’t ill. Which of the following is a $95 %$ confidence interval for the difference
in the proportions of workers at the two companies who would admit to using sick
leave when they weren’t ill?
(a) $0.03 \pm \sqrt{\frac{(0.28) (0.72)}{125} + \frac{(0.25) (0.75)}{68}}$
(b) $0.03 \pm 1.96 \sqrt{\frac{(0.28) (0.72)}{125} + \frac{(0.25) (0.75)}{68}}$
(c) $0.03 \pm 1.645 \sqrt{\frac{(0.28) (0.72)}{125} + \frac{(0.25) (0.75)}{68}}$
(d) $0.03 \pm 1.96 \sqrt{\frac{(0.269) (0.731)}{125} + \frac{(0.269) (0.731)}{68}}$
(e) $0.03 \pm 1.645 \sqrt{\frac{(0.269) (0.731)}{125} + \frac{(0.269) (0.731)}{68}}$

Short Answer

Expert verified

The correct answer is :

(b) $0.03 \pm 1.96 \sqrt{\frac{0.28 (0.72)}{125} + \frac{0.25 (0.75)}{68}}$

Step by step solution

Given information

We are given,

$x_{1} = 35$

$n_{1} = 125$

$x_{2} = 17$

$n_{2} = 68$

$c = 95 %$

Calculation

The sample proportion is the number of successes divided by the sample size:

${\hat{p}}_{1} = \frac{x_{1}}{n_{1}} = \frac{35}{125} = 0.28$

${\hat{p}}_{2} = \frac{x_{2}}{n_{2}} = \frac{17}{68} = 0.25$

For confidence level $1 - α$ $= 0.95$ , determine width="91" height="26" role="math">zα/2=z0.025using table $∥$ (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):

$z_{α / 2} = 1.96$

The endpoints of the confidence interval for $p_{1} - p_{2}$ are then:

$({\hat{p}}_{1} - {\hat{p}}_{2}) \pm z_{α / 2} . \sqrt{\frac{{\hat{p}}_{1} (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} (1 - {\hat{p}}_{2})}{n_{2}}} = (0.28 - 0.25) \pm 1.96 \sqrt{\frac{0.28 (1 - 0.28)}{125} + \frac{0.25 (1 - 0.25)}{68}}$