Question

direct satellite television it is believed that 25% of u.s. homes have a direct satellite television receiver. round up your final answers to the next whole number. part: 0 / 2 part 1 of 2 how large a sample is necessary to estimate the true population of homes that have a direct satellite television receiver with 99% confidence and within 3 percentage points? n =

Explanation:

Step1: Determine the values of \(p\), \(q\), \(E\), and \(z\)

• Given \(p = 0.25\) (proportion of U.S. homes with a direct - satellite television receiver), then \(q=1 - p=1 - 0.25 = 0.75\)
• The margin of error \(E = 0.03\) (since 3 percentage points \(=0.03\))
• For a \(99\%\) confidence level, the z - value \(z_{\alpha/2}\) is \(z_{0.005}\). From the standard normal distribution table, \(z_{0.005}=2.576\)

Step2: Use the formula for sample size in proportion

The formula for sample size \(n\) when estimating a proportion is \(n=\frac{z_{\alpha/2}^{2}pq}{E^{2}}\)
Substitute the values into the formula:

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Answer:

Since we need to round up to the next whole number, \(n = 1383\)