Question

a researcher wishes to estimate, with 99% confidence, the population proportion of families who eat fast food at least once per week. her estimate must be accurate within 4% of the population proportion. (a) no preliminary estimate is available. find the minimum sample size needed. (b) find the minimum sample size needed, using a prior study that found that 40% of the respondents said they eat fast food four to six times per week. (c) compare the results from parts (a) and (b). (a) what is the minimum sample size needed assuming that no prior information is available? n = (round up to the nearest whole number as needed.)

Explanation:

Step1: Determine the z - value

For a 99% confidence level, the z - value \(z_{\alpha/2}\) is 2.576. The margin of error \(E = 0.04\). When no preliminary estimate is available, we use \(p=0.5\) (this gives the maximum value of \(p(1 - p)\)).

Step2: Use the formula for sample size

The formula for sample size \(n\) when estimating a proportion is \(n=\frac{z_{\alpha/2}^{2}\times p(1 - p)}{E^{2}}\).
Substitute \(z_{\alpha/2}=2.576\), \(p = 0.5\), and \(E=0.04\) into the formula:

$$
LATEXBLOCK0
$$

Answer:

\(n = 1037\)