Question

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

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

$z_{\alpha /2}=1.96$

The endpoints of the confidence interval for $p_1-p_2$are then:

$({\hat{p}}_1-{\hat{p}}_2)\pm z_{\alpha /2}.\sqrt{\frac{{\hat{p}}_1(1-{\hat{p}}_1)}{n_1}+\frac{{\hat{p}}_2(1-{\hat{p}}_2)}{n_2}}=(0.28-0.25)\pm 1.96\sqrt{\frac{0.28(1-0.28)}{125}+\frac{0.25(1-0.25)}{68}}$

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

Thus the correct answer is (b).