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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鈥檛 really ill. Seventeen employees from a random

sample of 68 workers from Company B admitted that they had used sick leave when

they weren鈥檛 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鈥檛 ill?

(a) 0.03卤(0.28)(0.72)125+(0.25)(0.75)68

(b) 0.03卤1.96(0.28)(0.72)125+(0.25)(0.75)68

(c) 0.03卤1.645(0.28)(0.72)125+(0.25)(0.75)68

(d) 0.03卤1.96(0.269)(0.731)125+(0.269)(0.731)68

(e)0.03卤1.645(0.269)(0.731)125+(0.269)(0.731)68

Short Answer

Expert verified

The correct answer is :

(b) 0.03卤1.960.28(0.72)125+0.25(0.75)68

Step by step solution

01

Given information

We are given,

x1=35

n1=125

x2=17

n2=68

c=95%

02

Calculation

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

p^1=x1n1=35125=0.28

p^2=x2n2=1768=0.25

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

z伪/2=1.96

The endpoints of the confidence interval for p1-p2are then:

(p^1-p^2)卤z伪/2.p^1(1-p^1)n1+p^2(1-p^2)n2=(0.28-0.25)卤1.960.28(1-0.28)125+0.25(1-0.25)68

=0.03卤1.960.28(0.72)125+0.25(0.75)68

Thus the correct answer is (b).

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Most popular questions from this chapter

National Park rangers keep data on the bears that inhabit their park. Here is a histogram of the weights of bears measured in a recent year:

Which of the following statements is correct?

a. The median will lie in the interval (140,180), and the mean will lie in the interval(180,220).

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