Question

Explain how to solve this please and thanks

Construct 90% and 95% confidence intervals for the population proportion.

In a survey of 674 US males ages 18 to 64; 396 say they have gone to the dentist in the past year

Answer 1

a)

p= 0.588

q= 0.412

n= 674

alpha=0.1 then Z(alpha/2)=1.645

Margin of error E= Z(alpha/2)*sqrt(p*q/n)

=1.645*sqrt(0.588*0.412/674)

=0.03119

90% Confidence interval for population proportion =(p-E,p+E)

rounded lower bound= 0.557

rounded upper bound= 0.619

b)

p= 0.588

q= 0.412

n= 674

alpha=0.05 then Z(alpha/2)= 1.96

Margin of error E=Z(alpha/2)*sqrt(p*q/n)

=1.96*sqrt(0.588*0.412/674)

=0.037158947

95% Confidence interval for population proportion =(p-E,p+E)

rounded lower bound= 0.5508

rounded upper bound= 0.6252

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