Question

You have been asked by your new boss at Belk's to do a survey of customers on their preference of the location of the customer service desk. She would like a 95% confidence interval with a margin of error of 0.09. How many people do you need to survey?

Answer 1

The required CI level is 95%, hence the respective z-value for the given CI level is 1.96

Here the ME for a CI of proportion is given.
As initial value of p is not given, we assume it as 0.5
p = 0.5
ME = 0.09

Now using the ME formula,
ME = z * sqrt(p*(1-p)/n)

n = (z/ME)^2 * p * (1-p)
n = (1.96/0.09)^2 * 0.5 * 0.5
n = 118.6
n = 119 (rounding to the whole number)