It is believed that 45% 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 3 percentage points?

Answer:
Given:

The Population Proportion $ (p) = 0.45 $
The Margin Of Error $ (E) = 0.03 $
The Confidence Interval Level $ = 90\% $

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}} $

$ 0.03 = 1.645 \times \sqrt{\frac{0.45 \times (1 – 0.45)}{n}} $

$ n = 739.64 $

$ n \approx 740 $

Final Answer:
The sample size $ (n) = 740 $