Aresearcher wants to construct a 99% confidence interval for the proportion of elementary school students in seward county who receive free or reduced-price school lunches. what sample size is needed so that the confidence interval will have a margin of error of at most 0.07?

The sample size can be calculated using the formula:

n = p (1-p) (Zc / m)^2

where:

p = proportion of students who received lunches= 0.5

Zc = taken from standard tables at 99% CI = 2.575

m = margin of error =  0.07

Substituting known values:

n = 0.5 (1-0.5) (2.575/0.07)^2

n = 338.3

n = 339 samples

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step-by-step explanation:

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