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Which of the following is the correct margin of error for a 99%confidence interval for the difference in the proportion of male and female college students who worked for pay last summer?

a. 2.5760.851(0.149)550+0.851(0.149)500

b. 2.5760.851(0.149)1050

c. 2.5760.880(0.120)550+0.820(0.180)500

d. 1.9600.851(0.149)550+0.851(0.149)500

e. 1.9600.880(0.120)550+0.820(0.180)500

Short Answer

Expert verified

Option (c) is correct.

Step by step solution

01

Given Information

It is given that x1=410

x2=484

n1=500

n2=550

c=0.99

02

Calculation

Sample proportion is p^1=x1n1=410500=0.82

p^2=x2n2=484550=0.88

When 1-α=0.99, using table za/2=z0.005

zscore is za/2=2.576

Margin of Error is E=zα/2×p^11-p^1n1+p^21-p^2n2

=2.576×0.82(1-0.82)500+0.88(1-0.88)550

=2.576×0.88(1-0.88)550+0.82(1-0.82)500

=2.576×0.88(0.120)550+0.82(0.180)500

Option (c) is correct.

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