Question

The federal government would like to estimate the proportion of households in the U.S. population that own their own homes. A pilot sample of 40 households found that 24 of them owned their homes. Determine the number of households that need to be sampled to construct a 99% interval with a margin of error equal to 3% to estimate the proportion.

Answer 1

Answer:

1770

Step-by-step explanation:

We need to estimate number of households to be sampled to construct a 99% confidence interval.

Number of households to be sampled=n=?

$n=\frac{(z_{\frac{\alpha}{2} })^2 pq }{E^{2} }$

$z_{\frac{\alpha}{2} } =z_{\frac{\0.01}{2} }=z_{0.005 }=2.576$

The proportion can be estimated as

p=x/n.

We know that 24 out of 40 households owns their home.

so, x=24 and n=40.

p=24/40

p=0.6

q=1-p=1-0.6=0.4

pq=0.6*0.4=0.24

E=0.03

$n=\frac{(z_{\frac{\alpha}{2} })^2 pq }{E^{2} }$

$n=\frac{2.576^2(0.24)}{0.03^2}$

$n=\frac{1.5926}{0.0009}$

n=1769.56.

n=1770.

Thus, the number of households that need to be sampled are 1770.