Question

a business student is interested in estimating the 90% confidence interval for the proportion of students who bring laptops to campus. he wants a precise estimate and is willing to draw a large sample that will keep the sample proportion within nine percentage points of the population proportion. what is the minimum sample size required by this student, given that no prior estimate of the population proportion is available?

Answer 1

n=106

Explanation:

Given p = 0.5 and 1-p = q = 0.5Margin of error = 0.08 Confidence level = 90% Z score for 90% confidence level = 1.65As we know - Margin of error = z * Sqrt (pq/n)Substituting the given value, we get – 0.08 = 1.65 * Sqrt (0.5*0.5/n)Squaring both the sides and solving, we get n = 1.65^2*0.5^2/0.08^2n = 106.34 = 106