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Chapter 10: Q.2 (page 664)

Thirty-five people from a random sample of 125workers from Company A admitted to using sick leave when they weren鈥檛 really ill. Seventeen employees from a random sample of 68workers from Company B admitted that they had used sick leave when they weren鈥檛 ill. 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 is

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

(b) localid="1650367573248" 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1.96(0.28)(0.72)125+(0.25)(0.75)68.

Step by step solution

01

Given Information

x1=35

n1=125

x2=17

n2=68

c=95%

02

Explanation

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

p^1=x1n1

=35125

=0.28

localid="1650367755652" p^2=x2n2

=1768

=0.25

For confidence level 1-伪=0.95, determine z伪/2=z0.025using table II (look up 0.025in 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:

localid="1650367861788" p^1-p^2卤z伪/2路p^11-p^1n1+p^21-p^2n2=(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

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