Question

In a random sample of 500 residential units, it is found that 114 of them are insured by a homeownership insurance scheme. a Construct a 90% confidence interval for the true proportion of residential units that are insured by a homeownership insurance scheme. b Utilizing the information provided by the answer in part (a) about the possible range of the true population proportion, determine how large a sample is needed if you wish to be 99% confident that the sample proportion is correct to within ±0.08 of the true proportion of residential units that are insured by a homeownership insurance scheme?

Answer 1

a)

p= 114/500=0.228

q= 1-p=0.772

n= 500

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

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

=1.645*sqrt(0.228*0.772/500)

=0.03086

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

rounded lower bound= 0.197

rounded upper bound= 0.259

b)

p(hat)= 0.228

q(hat)= 0.772

MOE= 0.08

Zalpha/2= 2.576

we know the relation

E=Z(alpha/2)*sqrt(p(hat)*q(hat)/n)

n=p(hat)*q(hat)*(Z/E)^2

n=0.228*0.772*(2.576/0.08)^2

n= 182.5004294

by rounding it n=183

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