Question

practice quiz 3
question 13 of 25 (1 point) | question attempt: 1 of unlimited
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?

part: 1 / 2
part 2 of 2
how large a sample is necessary if nothing is known about the population?

Explanation:

Step1: Determine the formula

When nothing is known about the population, we use \(p = 0.5\) (this gives the maximum value of \(p(1 - p)\)). The formula for sample size \(n\) in estimating a proportion is \(n=\dfrac{z^{2}_{\alpha/2}p(1 - p)}{E^{2}}\).
For a \(99\%\) confidence level, \(\alpha=1 - 0.99 = 0.01\), and \(\alpha/2=0.005\). From the standard normal distribution table, \(z_{\alpha/2}=z_{0.005} = 2.576\). The margin of error \(E = 0.03\) (since \(3\) percentage points \(=0.03\)).

Step2: Substitute values into the formula

Substitute \(z_{\alpha/2}=2.576\), \(p = 0.5\), \(1 - p=0.5\), and \(E = 0.03\) into the formula:

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Step3: Calculate the value of \(n\)

\(n=\dfrac{1.658944}{0.0009}\approx1843.27\)

Since the sample size \(n\) must be an integer, we round up to the next whole number.

Answer:

\(n = 1844\)