It is believed that 18% of U.S. homes have a direct satellite television receiver. How large a sample is necessary to estimate the true population of homes that have a direct satellite television receiver with 90% confidence and within 4 percentage points?

Answer:
Given:

The Population Proportion $(p)=0.18$
The Margin Of Error $(E)=0.04$
The Confidence Interval Level $=90\mathrm{\%}$

Solution:
The Significance Level $(\alpha ):$
$\alpha =1-0.90$
$=0.10$

The Critical Value $(z_c):$
$z_c=z_{\alpha /2}$
$=z_{0.10/2}$
$=1.645$

→ Here we use a margin of error formula to find out the sample size

$E=z_c\times \sqrt{\frac{p\times (1–p)}{n}}$

$0.04=1.645\times \sqrt{\frac{0.18\times (1–0.18)}{n}}$

$n=248.12$

$n\approx 248$

Final Answer:
The sample size $(n)=248$