It is believed that 50% 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 95% confidence and within 2 percentage points?

Answer:
Given:

The Population Proportion $(p)=0.50$
The Margin Of Error $(E)=0.02$
The Confidence Interval Level $=95\mathrm{\%}$

Solution:
The Significance Level $(\alpha ):$
$\alpha =1-0.95$
$=0.05$

The Critical Value $(z_c):$
$z_c=z_{\alpha /2}$
$=z_{0.05/2}$
$=1.96$

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

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

$0.02=1.96\times \sqrt{\frac{0.50\times (1–0.50)}{n}}$

$n=2401$

Final Answer:
The sample size $(n)=2401$